Apéndice G: Ejercicios

Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing

Series	Geophysical References Series
Title	Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing
Author	Enders A. Robinson and Sven Treitel
Chapter	7
DOI	http://dx.doi.org/10.1190/1.9781560801610
ISBN	9781560801481
Store	SEG Online Store

In delay there lies no plenty.

—William Shakespeare

1. What two properties characterize a seismic wavelet? Draw a typical wavelet. Here we mean a physical wavelet as recorded by a geophone. What is an interpreter wavelet? On the same graph, draw a typical interpreter wavelet. What is wavelet processing?

2. The two-length wavelets (1, 2) and (2, 1) have the same autocorrelation. Find some other wavelets that have the same autocorrelation. These other wavelets will necessarily be of length greater than two.

3. Given the wavelets (1, –1, 1, –1), (1, 1, 1, 1), and (1, 2, 3, 4), find the autocorrelation of each wavelet and the crosscorrelation of each pair of wavelets.

4. Given the wavelet a = (4, –2, 1), according to our convention, the first coefficient (4) occurs at time n = 0. The time reverse of this wavelet would be the noncausal function (…, 0, 0, 1, –2, 4, 0, 0, …), where the last coefficient (4) occurs at n = 0. Compute the autocorrelation by convolving the wavelet with its time reverse. If we shift the time-reverse function forward by two time units, we obtain the (causal) wavelet $a^{R}$ = (1, –2, 4). Its first coefficient (1) occurs at n = 0. Compute a * a and show that it is the same as the shifted autocorrelation of a (where the time shift is two units to the right). Explain this in terms of transforms.

5. Given the following two-length (causal) wavelets

$a=\left({2,\ }-1\right){\ ,\ }a^{R}=\left(-{1,2}\right),$

$b=\left({3,\ }-2\right){\ ,\ }b^{R}=\left(-{2,3}\right),$

$c=\left({3,1}\right){\ ,\ }c^{R}=\left({1,\ 3}\right),$

$d=\left({4,\ }-1\right){\ ,\ }d^{R}=\left(-{1,4}\right),$

compute the following wavelets

$a*b{,\ }a^{R}*b^{R},$

$c*a{,\ }c*a,$

$c*a*a{,\ }a*c^{R}*b^{R},$

$d*d^{R}*a{,\ }a*d*d.$

6. Find the autocorrelation of each of the wavelets in exercise 5. Which autocorrelations are the same, and why?

7. Given the wavelets (3, 2, 1), (3, 2, –1), (3, –2, 1), (3, –2, –1), find the autocorrelation of each wavelet and the crosscorrelation of each pair of wavelets. Plot the results in a 4 × 4 table, with the autocorrelations on the diagonal and the crosscorrelations off the diagonal. Show that only the right-hand side of each correlation function is needed in such a table.

8. Find the autocorrelations and crosscorrelations of the following wavelets:

$\left({4,\ }-{3,2\ ,\ 1}\right){\ ,\ }\left({1,\ 0,\ }-{6,2}\right),$

$\left({8,\ }-{4,2,\ }-1\right),(7,5,3,1).$

Plot the results. Show that it is necessary to plot only the right halves of the correlation functions.

9. By inspection, say what you can about the autocorrelations and crosscorrelations of the two wavelets (1, 2, –8, 6), (6, –8, 2, 1). Confirm by computation.

10. Write out all the mathematical steps to establish that crosscorrelation of a and b can be accomplished by the convolution $a*b^{R}$.

11. (a) Convolve (2, 5, –2, 1) with (6, –1, –1). [Answer: (12, 28, –19, 3, 1, –1).]

(b) Crosscorrelate (2, 5, –2, 1) with (6, –1, –1). For what shift are these functions most nearly alike? [Answer: (–2, –7, 9, 31, –13, 6), maximum at shift 1.]

(c) Convolve (2, 5, –2, 1) with (–1, –1, 6); compare with the answer in (b) and explain the difference. [Answer: None.]

(d) Autocorrelate (6, –1, –1) and (3, –5, –2). The autocorrelation of a function is not unique to the given function. For example, additional wavelets having the same autocorrelation as the given ones are (–1, –1, 6) and (–2, –5, 3). Which member of the set is the minimum-delay wavelet? [Answer: (6, –5, 38, –5, –6). MD = (6, –1, –1).]

(e) What is the normalized autocorrelation of (6, –1, –1)? What is the normalized crosscorrelation for the result obtained in exercise 11b? What can you conclude from the magnitude of the largest value of this normalized crosscorrelation?

[auto = (–3/19, –5, 38, 1, –5/38, –3/19),

cross = (–1/s, –7/2s, 9/2s, 31/2s, –13/2s, 3/s) where $s={\sqrt {323}}$.]

12. Find the eight wavelets of length 4 that can be computed from the three given minimum-phase wavelets (10, 4), (5, –3), and (2, 1). [Answer: (100, 30, –34, –12) minimum-delay wavelet (–12, –34, 30, 100) maximum-delay wavelet (50, 90, –32, 24) (24, –32, 90, 50) (–60, 48, 78, 20) mixed-delay wavelets (20, 78, 48, –60) (–30, –22, 96, 40) (40, 96, –22, –30).]

13. Consider a convolution of two minimum-delay dipoles or their reverse.

Convolution Wavelet type

(2, 1) * (3, –1) (6, 1, –1) minimum delay

(1, 2) * (–1, 3) (–1, 1, 6) maximum delay

(2, 1) * (–1, 3) (–2, 5, 3) mixed delay

(1, 2) * (3, –1) (3, 5, –2) mixed delay

Plot the energy-buildup curves for the four wavelets.

14. Discuss the following properties of an autocorrelation. The autocorrelation

(a) has a positive maximum at zero lag that is equal to the mean power (b) contains no information on the phase relation between frequency components (c) approaches zero as the lag becomes large for random functions (d) is a real, finite, even function (e) has a positive frequency spectrum, called the power spectrum

15. White noise is a stationary stochastic process with a flat power spectrum. However, signals with finite energy exist that also have flat spectra. Such signals are the phase-shift signals. An important subclass is composed of the causal phase-shift signals, which also are called all-pass signals, or equivalently, all-pass operators. An all-pass operator is defined as a causal operator whose autocorrelation function is equal to the unit spike. An example of an all-pass operator would be the wavelet

$p{=\ }\left({1/2,\ 3/4,\ }-{3/8,3/16,\ }-{3/32,3/64,\ }-{3/128,\ }\cdots \right).$

Because the autocorrelation of an all-pass operator is equal to the unit spike, it follows that its energy spectrum is flat and equal to unity. Its magnitude spectrum therefore is also flat and equal to unity. An all-pass operator, in passing a signal from input to output, does not alter the amplitude spectrum of the signal but does add to the phase spectrum of the signal. As a matter of terminology, any operator with a flat amplitude spectrum is called a phase-shift operator, because such an operator cannot change the shape of the magnitude spectrum of an input signal but can change only the phase spectrum of the input signal. An arbitrary phase-shift operator can be two-sided - that is, it can have both a causal component and an anticausal component. An all-pass operator can be described as a causal phase-shift operator. One further fact about phase-shift operators is of interest: The inverse of a phase-shift operator is equal to the reverse of the phase-shift operator. Moreover, this fact can be true only for an operator with a flat magnitude spectrum - that is, for a phase-shift operator. A phase-shift operator is one whose autocorrelation function is equal to the unit spike. However, the autocorrelation of an operator is the same as the convolution of the operator with its time reverse. Hence, the convolution of a phase-shift operator with its time reverse is equal to the unit spike. Because by definition the convolution of an operator with its inverse is equal to the unit spike, it follows that the inverse of a time-shift operator is the same as its time reverse. Write down the inverse of the all-pass operator given above. The causal component of the phase-shift operator becomes the anticausal component of its inverse. Both the all-pass operator and its inverse are one-sided; the all-pass operator is one-sided in the sense of having only a causal component. As we have said, an all-pass operator is a phase-shift operator that is causal in real time; by the same token, the inverse of an all-pass operator is not causal in real time. An all-pass operator can be described as a causal filter that adds a certain phase spectrum to the phase spectrum of an input signal without changing its magnitude spectrum. The inverse to this all-pass operator can be described as an anticausal filter that subtracts the phase spectrum from the phase spectrum of an input signal without changing its magnitude spectrum. It can be shown that any wavelet always can be represented as the convolution of an all-pass operator with the minimum-delay wavelet with the same autocorrelation as the given wavelet.

16. Draw a diagram of a minimum-phase wavelet and a zero-phase wavelet. Would a symmetric causal wavelet of the same length and frequency content necessarily be zero phase?

17. Describe what wavelet below is a minimum-phase, linear-phase, maximum-phase, or zero-phase wavelet, where all the wavelets have the same amplitude spectrum.

(a) $1-{0.6}Z-{071}Z^{2}{+0.24}Z^{3}{+0.16}Z^{4}$. [Answer: Minimum phase.] (b) ${0.4+0.18}Z-{1.25}Z^{2}{+0.18}Z^{3}{+0.4}Z^{4}$. [Answer: Linear phase.] (c) ${0.16+0.24}Z-{0.71}Z^{2}-{0.6}Z^{3}+Z^{4}$. [Answer: Maximum phase.] (d) ${0.4}Z^{-2}+0.18Z^{-1}-{1.25+0.18}Z+0.4Z^{2}$. [Answer: Zero phase.]

18. Discuss: Why is a zero-phase wavelet anticipatory? [Answer: It begins before time zero.] Phase curves depend on the location of the time reference point. Other mixed-phase wavelets also can be constructed from the same component dipoles. An important feature of a minimum-phase wavelet is that its energy arrives as soon as possible, consistent with its amplitude spectrum. Consider a set of possible filters with identical amplitude spectra. The minimum-phase filter is the one that delays the energy the least. Therefore, it also is called a minimum-delay filter. If the input to a minimum-phase filter is itself minimum phase, then the output also will be minimum phase. Much of the filtering performed during digital processing is of the minimum-phase type. A minimum-phase wavelet is sometimes said to be front-loaded because its energy is concentrated in the front end of the wavelet. Maximum-phase or maximum delay is the other extreme, because energy is concentrated in the back end of the wavelet. Mixed-phase is an intermediate case, with energy concentrated at intermediate positions of the wavelet.

19. The zero-point reverse of a finite-length causal wavelet is purely noncausal. Show that the zero-point reverse can be made causal by shifting it in time so that all coefficients occur at time zero or later.

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También en este capítulo

• Ondículas
• Transformada de Fourier
• Transformada Z
• Retraso: Mínimo, mixto y máximo
• Ondículas de doble longitude
• Ilustración del espectro
• Retraso en general
• Energía
• Autocorrelación
• Representación canónica
• Ondículas de fase cero
• Ondículas simétricas
• Ondícula de Ricker

Vínculos externos

find literature about Appendix G: Exercises/es