Appendix F: Exercises

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Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing
DigitalImaging.png
Series Geophysical References Series
Title Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing
Author Enders A. Robinson and Sven Treitel
Chapter 6
DOI http://dx.doi.org/10.1190/1.9781560801610
ISBN 9781560801481
Store SEG Online Store
Little strokes fell great oaks.

-Benjamin Franklin

1. Let a sandstone layer be 24 m thick with a compressional wave velocity of 3000 m/s. A downgoing wave strikes the top of the layer and is reflected with reflection coefficient 0.1. The part that is transmitted into the sandstone is reflected at the bottom of the sandstone layer with reflection coefficient –0.1. Please neglect amplitude losses in transmission through the top of the sandstone and neglect intrabed multiples.

(a) If the basic wavelet sampled at 4 ms has amplitudes given by 8, 7, –7, –5, 0, 4, 2, find the composite wavelet shape that is reflected by the sandstone layer. Explain why we can say in this case that the top and base of the sandstone are “resolved.”

(b) If the basic wavelet sampled at 4 ms now has amplitudes given by 4, 8, 10, 7, 0, –7, –6, –5, –1, 0, 2, 4, 1, 2, find the composite wavelet shape. Is the sand layer resolved in this case?

2. Let a shale bed be in the middle of a sandstone layer so that 12 m of sand lie above the shale and 12 m of sand lie below it. Let the thickness of the shale be such that the traveltime through it is the same as it would be through 12 m of sand. Let the reflection coefficient at the top of the shale be –0.1 and at the bottom of the shale be 0.1. For the wavelet given in exercise 1(a) above, find the composite reflected wavelet. This example shows that component reflections from several reflectors can interfere and give a tuned response. The ringing character results from the natural resonance.

3. Most of the energy in a seismic wavelet is associated with frequencies centered about the dominant frequency. The dominant period can be defined as the time between two major crests. The dominant frequency is the reciprocal of the dominant period. The basic equation for wavelength is

Calculate wavelengths for the following cases:

(a) Shallow rocks: 2000 m/s, 50 Hz [Answer: 40 m.]

(b) Deep rocks: 6000 m/s, 25 Hz [Answer: 240 m.]

(c) What is the dominant frequency in each of the wavelets in (a) and (b) of Exercise 1 above? How does resolution depend on this frequency? [Answer: 50 Hz and 27.8 Hz.]

4. In the following, assume that the dip is so small that it has negligible effect. The reflection coefficients shown below are scaled up and rounded off to make calculation easier. Assume a minimum-phase wavelet (dominant frequency 50 Hz) digitized at 0.004-s intervals: wavelet = (10, 9, –8, –9, 0, 5, 3, 0).

(a) Assume sandstone encased in shale. Calculate and plot the reflection wave shape when the sandstone is 20 m thick. The two-way normal-incidence traveltime through the sandstone is 0.012 s. Assume a shale-to-sandstone reflection coefficient of 0.1 (to make it easy to calculate). [Answer: (1, 0.9, –0.8, –1.9, –0.9, 1.3, 1.2, 0, –0.5, –0.3, 0).]

(b) Repeat for a sandstone that is 26.67 m thick. (The two-way time is now 0.016 s.) [Answer: (1, 0.9, –0.8, –0.9, –1.0, –0.4, 1.1, 0.9, 0, –0.5, –0.3, 0).]

(c) Assume two sandstones each 20 m thick separated by shale and encased in shale. Plot the reflection waveshape. The two-way traveltime through the internal shale layer is 0.008 s. [Answer: (1, 0.9, –0.8, –1.9, –0.9, 2.3, 2.1, –0.8, –2.4, –1.2, 1.3, 1.2, 0, –0.5, –0.3, 0).]

(d) Assume a dipping sandstone 41 m thick, with gas in the upper 14 m. Two-way time is 0.008 s through the gas sandstone and 0.016 s through the water sandstone. Assume a shale-to-gas-sandstone reflection coefficient of –0.2, a gas-sandstone-to-water-sandstone reflection coefficient of +0.3, and a water-sandstone-to-shale reflection coefficient of –0.1. [Answer: (–2, –1.8, 4.6, 4.5, –2.4, –3.7, –1.6, 0.6, 1.7, 0.9, 0, –0.5, –0.3, 0).]

(e) Repeat for gas in the upper 28 m. [Answer: (–2, –1.8, 1.6, 1.8, 3, 1.7, –4, – 3.6, 0.8, 2.4, 0.9, –0.5, –1.3, 0).]

(f), (g), and (h) Repeat situation (a), (c), and (d) for the same waveshape but half the frequency. Such a wavelet is (5, 10, 12, 9, 0, –8, –11, –9, –4, 0, 4, 5, 4, 3, 1, 0). [Answer: (f ): (0.5, 1, 1.2, 0.4, –1, –2, –2, –0.9, 0.4, 1.1, 1.3, 0.9, 0.4, –0.1, –0.4., –0.4, –0.3, –0.1, 0).] [Answer: (g): (0.5, 1, 1.2, 0.4, –1, –1.5, –1, 0.3, 0.8, 0.1, –0.7, –1.1, –0.5, 0.3, 0.7, 0.9, 0.6, 0.3, –0.1, –0.4, –0.4, –0.3, –1, 0).]

[Answer: (h): (–1, –2, –0.9, 1.2, 3.6, 4.3, 1.7, –1.6, –3.7, –3.6, –2, –0.2, 1.5, 1.8, 1.4, 0.9, –0.1, –0.5, –0.4, –0.3, –0.1, 0).]

5. You would like to clearly resolve the top and bottom reflections from certain geologic beds. Assume constant density. To do so, you decide to use a seismic signal with a dominant frequency corresponding to a wavelength no greater than the bed thickness b. Determine this frequency for the following beds. [Note: .]

(a)

(b)

(c)

At what frequencies would you expect constructive interference for the above beds? [Answer: 7.5 Hz, 12.5 Hz, 16 Hz.] At what frequencies would you expect destructive interference? [Answer: 6 Hz, 10 Hz, 20 Hz.]


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Appendix F: Exercises
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