User:Ageary/Chapter 9
Problems-in-Exploration-Seismology-and-their-Solutions.jpg | |
Series | Geophysical References Series |
---|---|
Title | Problems in Exploration Seismology and their Solutions |
Author | Lloyd P. Geldart and Robert E. Sheriff |
DOI | http://dx.doi.org/10.1190/1.9781560801733 |
ISBN | ISBN 9781560801153 |
Store | SEG Online Store |
Contents
- 1 Chapter 9 Data processing
- 2 9.1 Fourier series
- 3 9.2 Space-domain convolution; Vibroseis acquisition
- 4 9.3 Fourier transforms of the unit impulse and boxcar
- 5 9.4 Extension of the sampling theorem
- 6 9.5 Alias filters
- 7 9.6 The convolutional model
- 8 9.7 Water reverberation filter
- 9 9.8 Calculating crosscorrelation and autocorrelation
- 10 9.9 Digital calculations
- 11 9.10 Semblance
- 12 9.11 Convolution and correlation calculations
- 13 9.12 Properties of minimum-phase wavelets
- 14 9.13 Phase of composite wavelets
- 15 9.14 Tuning and waveshape
- 16 9.15 Making a wavelet minimum-phase
- 17 9.16 Zero-phase filtering of a minimum-phase wavelet
- 18 9.17 Deconvolution methods
- 19 9.18 Calculation of inverse filters
- 20 9.19 Inverse filter to remove ghosting; Recursive filtering
- 21 9.20 Ghosting as a notch filter
- 22 9.21 Autocorrelation
- 23 9.22 Wiener (least-squares) inverse filters
- 24 9.23 Interpreting stacking velocity
- 25 9.24 Effect of local high-velocity body
- 26 9.25 Apparent-velocity () filtering
- 27 9.26 Complex-trace analysis
- 28 9.27 Kirchhoff migration
- 29 9.28 Using an upward-traveling coordinate system
- 30 9.29 Finite-difference migration
- 31 9.30 Effect of migration on fault interpretation
- 32 9.31 Derivative and integral operators
- 33 9.32 Effects of normal-moveout (NMO) removal
- 34 9.33 Weighted least-squares
Chapter 9 Data processing
9.1 Fourier series
9.1a Show that the Fourier series coefficients, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle b_{n}} in equation (9.1a), are given by equations (9.1d,e).
Background
If is a periodic function, that is, one that repeats exactly at intervals (called the period) and has a finite number of maxima, minima, and discontinuities in each interval , then it can be expressed in a Fourier series in any one of the three following equivalent forms, being an integer:
( )
( )
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} =\sum\limits_{n=-\infty}^{+\infty} \alpha _{n} e^{\hbox {j}\omega _{n} t}, \end{align} } ( )
where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{0} =2\pi f_{0} =2\pi /T=} fundamental angular frequency, harmonic frequency, and the amplitudes and phases are given by the following equations where , except for equation (9.1h) where :
( )
( )
( )
( )
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \alpha _{n} =\frac{1}{T} \int_{-T /2}^{+T /2}g(t)e^{-{\hbox {j}}\omega _{n}t} \mathrm{d}t. \end{align} } ( )
Solution
We multiply both sides of equation (9.1a) by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle {\rm \; cos\;}\omega _{r} t} , being any nonzero positive integer, and integrate between the limits Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -T/2} and . (We omit the limits on the integrals in the following proofs since they are all the same.) Equation (9.1a) now becomes
By using the identities
the first integrand reduces to the sum of two cosines of the form while the second integrand becomes the difference between two sines of the form If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r\ne n} , the integrals are zero, because at the limits the arguments of the cosines and sines are equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 2p\omega _{0} T/2=2p\pi} , integral. When , the first integrand becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \cos^{2} \omega _{n} t=\frac{1}{2} (1+\cos\,2\omega _{n} t)} and the right-hand side equals Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a_{n} T/2} . When , the second integrand becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \cos \omega _{n} t\sin\omega _{n} t\,\mathrm{d}t=\frac{1}{2} \sin\,2\omega _{n}t\, \mathrm{d}t} , which again gives zero upon integration.
Thus we are left with
Solving for gives equation (9.1d).
To verify equation (9.1e), we multiply both sides of equation (9.1a) by and integrate, all integrals on the right side of the equation giving zero except for the one with the integrand . Proceeding as before, we arrive at equation (9.1e).
When we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n=0} in equations (9.1d,e) we get and
9.1b Verify equations (9.1f,g) for the Fourier series coefficients Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle c_{n}}
and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \gamma _{n}}
.
Solution
To verify equation (9.1f) for , we multiply both sides of equation (9.1b) by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \cos(\omega_{r} t-\gamma _{r})} , , and proceed as in part (a). All integrals vanish except when , and the result is equation (9.1f).
When , we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int g(t)\,\mathrm{d}t=\frac{c_{0}}{2} \int \mathrm{d}t=\frac{c_{0} T}{2} ,\\ \hbox{so}\quad \quad \quad \frac{c_{0}}{2} =\frac{1}{T} \int g(t)\,\mathrm{d}t = \mathrm{average\ value\ of}\ g(t). \end{align} }
To verify equation (9.1g) for , we compare terms in equations (9.1a) and (9.1b) that involve , . The result is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} a_{n}\cos \omega_{n} t+b_{n} \sin\omega _{n} t&=c_{n} \cos(\omega _{n} t-\gamma _{n})\\ &=c_{n}(\cos \omega_{n}t\,\cos\,\gamma_{n}+\sin \omega_{n}t\,\sin\, \gamma_{n}). \end{align} }
Equating coefficients of and , we get
When , , , .
9.1c Verify the coefficients for the exponential form, equation (9.1h).
Solution
To derive equation (9.1h), we use Euler’s equations (Sheriff and Geldart, 1995, problem 15.12a), namely,
Multiplying both sides of equation (9.1c) by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle e^{-\hbox {j}\omega _{r} t}} , , and integrating, we get a series of integrals with integrands such as . Using Euler’s formula, , so the integrals vanish for all values of except when ; for this value we get
which is equation (9.1h), . For ,
9.2 Space-domain convolution; Vibroseis acquisition
The techniques and concepts of convolution, aliasing, -transforms, and so on can be applied to other domains than the time-domain. Express the source and group patterns of Figure 9.2a as functions of (horizontal coordinate) and convolve the two to verify the effective pattern shown.
Background
The convolution of two functions and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle h(t)} , written , is defined as
( )
For digital functions (see problem 9.4), the integrals become sums:
( )
These equations state that convolution is equivalent to superposition in which each element of one function is replaced by the other function multiplied (weighted) by the element being replaced. The sum of all values at a given time is the value of at that instant.
Another explanation of convolution is the following: We note that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g(t)} reflected in the -axis; in other words, the curve of is the same as that of except that it is reversed in direction [keeping fixed]. The function is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g(t)} displaced units to the right. Thus, in equation (9.2a) the value of at the time is obtained by moving to the right units, reflecting it in the -axis, then summing (integrating) the products of corresponding abscissas. The result is the same whichever function is displaced and reflected.
Arrays are discussed in problem 8.6, aliasing in problem 9.4, and -transforms in Sheriff and Geldart, 1995, section 15.5.3.
Solution
Taking the spatial sampling interval as 16.5 m, the source pattern consists of [1,1,2,2,3,3,3, 3,2,2,1,1] for a total array length of 200 m. The geophone group is an array 100 m long; to convolve the source and receiver arrays, they should have the same spatial intervals, so we take six receivers spaced 16.5 m apart. To convolve the two arrays, we replace each element of the receiver array with the source array (taking the weights as unity). The result is that shown at the bottom of Figure 9.2a.
9.3 Fourier transforms of the unit impulse and boxcar
9.3a Because a unit impulse is zero except at , where it equals Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +1} , we can apply the Fourier transform equation (9.3c) and find that . Show that
Background
Equations (9.1a,b,c) apply to a harmonic function which repeats after every interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle T} . If we let increase, the repetitions occur at longer intervals and in the limit when the function becomes aperiodic, that is, it no longer repeats. To see the effect of this on equation (9.1c), we replace Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \alpha _{n}} in equation (9.1c) with the right-hand side of equation (9.1h). The result is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} g(t)=\sum\limits_{n=-\infty}^{+\infty} \left(\frac{1}{T}\int_{-T /2}^{+T /2} g(t)e^{-\hbox {j}\omega _{n}t} \mathrm{d}t\right)e^{\hbox {j}\omega_{n} t}. \end{align} } ( )
We now let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle T\to +\infty}
so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{0} \to 0}
, which causes the two adjacent frequencies, and , differing by , to approach each other. In the limit becomes a continuous variable , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 1/T}
becomes , the summation becomes an integral, and equation (9.3a) becomes
( )
It is convenient to represent the inner integral by a symbol and write equation (9.3b) as two equations:
( )
( )
The function [or ] is the Fourier transform of while is the inverse Fourier transform of [or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle G(f)}
]. The relation between and can be indicated by a double arrow:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} g(t)\leftrightarrow G(\omega)\quad \mathrm{or}\quad g(t)\leftrightarrow G(f). \end{align} } ( )
The unit impulse Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \delta(t)} , also called the Dirac delta, is by definition zero everywhere except at where it equals Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +1} ; similarly, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \delta(t-t_{0})} is zero except when , where it equals . In digital notation (see problem 9.4) we write , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \delta_{(t-t_{0})}} , but the meaning is the same.
The convolution is discussed in problem 9.2. The convolution theorem [see Sheriff and Geldart, 1995, equation (15.145)] states that
( )
where and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle H(\omega)}
are the Fourier transforms of and . For digital functions we use -transforms (see Sheriff and Geldart, 1995, section 15.5.3), the equivalent of equation (9.3f) being
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} g_{t} *h_{t} \leftrightarrow G(z)H(z). \end{align} } ( )
The boxcar, written , is a function whose value is everywhere zero except in the interval from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -t_{0}}
to , where its value is . A boxcar in the frequency domain is written or .
Solution
The transform of is given by equation (9.3c):
( )
9.3b Show that
( )
( )
Solution
Convolution involves replacing each element of one function with the other function and since involves only one nonzero element at zero time, replacing an element at with gives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g(t)} , thus proving equation (9.3i). Likewise, replacing an element at time with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g(t)} gives , thus proving equation (9.3j).
An alternative proof for both cases above can be obtained by using transforms. Because the transform of both and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \delta_{t}} is , equations (9.3f, g) show that the transforms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \delta(t)*g(t)} and give and , respectively.
To prove equations (9.3j), we first use equation (9.3h) and get
using Sheriff and Geldart, 1995, equation (15.136). Next we use Sheriff and Geldart, 1995, problem (15.27) to write
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \delta_{t-\tau} *g_{t} \leftrightarrow z^{\tau} G(z)\leftrightarrow g_{t-\tau}. \end{align} }
9.3c Show that a boxcar of height extending from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +f_{0}}
in the frequency domain has the transform
where area of the boxcar.
Solution
Using equation (9.3d) a boxcar in the frequency domain becomes in the time domain:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} g(t)&=\int_{-\infty}^{+\infty} h \mathrm{box}_{2f_0} (f)e^{\hbox {j}2\pi ft} \mathrm{d}f=h\int_{-f_0}^{+f_0} e^{\hbox {j}2\pi ft}\, \mathrm{d}f\\ &=\left (\frac{h}{2\pi jt}\right) e^{\hbox {j}2\pi ft} |_{+f_0}^{-f_0}=\frac{h}{2\pi jt} (e^{\hbox {j}2\pi f_0t} -e^{-{\hbox {j}}2\pi f_0t}).\\ &=\frac{h}{\pi t} \sin(2\pi f_{0} t)=\frac{2f_{0} h}{(2\pi f_{0} t)} \sin\,(2\pi f_{0} t)=A \sin c (2\pi f_{0} t), \end{align} } ( )
where and sinc Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle x=(\sin x)/x.}
9.3d Calculate the transform of the pair of displaced boxcars in Figure 9.3b. Discuss the relation between your result and equation (9.3k) for a single boxcar centered at the origin.
Solution
Equation (9.3d) gives for the transform of the pair of boxcars:
( )
where , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle A_{1} =2hf_{1}}
, and we have used Euler’s formulas (see Sheriff and Geldart, 1995, problem 15.12a) to get the sines. Thus, the transform is the difference between two sinc functions corresponding to the upper and lower limiting frequencies of the boxcars.
Note that equation (9.3Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \ell} ) can be regarded as giving the result of two boxcars, one extending from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +\omega _{2}} , the other extending from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +\omega _{1}} , the effect of the latter being subtracted from that of the former.
To compare equation (9.3l) with equation (9.3k), we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{1} =0} . Then sinc Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{1} t= \sin\,0/0=1} , so sinc Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{1} t=0} and becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(t)=2hf_{2} \sin \mathrm{c}\ \omega _{2} t. \end{align} }
Setting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{2} =f_{0}}
gives equation (9.3k).
9.4 Extension of the sampling theorem
Explain why sampling at 4 ms is sufficient to reproduce exactly a signal whose spectrum is within the range Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 0} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 125\ {\hbox {Hz}} =f_{N}} , but not if the bandwidth is shifted upward?
Background
To convert an analog (continuous) signal to the digital form, values of the signal are measured at a fixed interval , called the sampling interval. The result is a series of numbers representing the amplitudes and polarities of the signal at the times Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=n\Delta} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} is an integer (often we omit Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \Delta} and specify the time by giving only). We write for the digital function corresponding to .
Sampling is equivalent to multiplying the analog signal by a comb, a function consisting of an infinite series of unit impulses spaced at a fixed interval . The equation of comb(t) is
( )
The transform of comb(t) is another comb with impulses in the frequency domain at intervals :
( )
where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k=2\pi /\Delta}
[see Sheriff and Geldart, 1995, equation (15.155)].
The Nyquist frequency, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{N}} , is half the frequency of sampling Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 1/\Delta} , that is,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} f_{N} =1/2\Delta. \end{align} } ( )
While a signal that does not contain frequencies higher than Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{N}}
will be recorded accurately, a frequency higher than Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{N}}
by the amount (i.e., a frequency that is not sampled at least twice per cycle) will produce an alias frequency equal to (Sheriff and Geldart, 1995, section 9.2.2c).
The convolution theorem, equation (9.3f), has a converse [see Sheriff and Geldart, 1995, equation (15.146)]:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} g(t)h(t)\leftrightarrow \frac{1}{2\pi} G(\omega)*H(\omega). \end{align} } ( )
Solution
We examine first the case where the signal frequency spectrum extends from to 125 Hz, next consider modifications when the spectrum is shifted upward 125 Hz Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle (=f_{N}} for 4-ms sampling), and finally, the case where the signal is shifted upward by an amount .
Part (i) of Figure 9.4a shows the signal and its frequency spectrum, the latter being confined to the range Hz [the negative frequencies arise when we use Euler’s equations (Sheriff and Geldart, 1995, problem 15.12a) to transform equation (9.1a) into (9.1c). Part (ii) shows comb(t), with elements 4 ms apart, and its transform, a comb with elements spaced Hz apart. Digitizing the signal is equivalent to multiplying it by comb(t) [part (iii)], which is equivalent to convolving the spectrum with comb(f) [see equation (9.4b)]; this results in a repetition of the spectrum for each impulse in comb(f) [see equation (9.3j)]. To eliminate the repeated spectra, we multiply by a boxcar [part (iv)], thus getting back the original spectrum [part (v)]. Multiplication in the frequency domain is equivalent to convolution in the time domain, so we convolve with a sinc function in part (V) and recover the original time-domain signal.
If the signal is shifted upward by , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} is an integer, then the sampling interval is larger than 1/2 cycle for all frequencies and the resulting spectrum is that of the alias function rather than of the signal, as illustrated in Figure 9.4b. Because we no longer have the signal spectrum after sampling, we cannot recover the signal.
If the signal is shifted upward so that it overlaps the Nyquist frequency (Figure 9.4c), the sampling causes the frequencies above to overlap and distort the signal spectrum. Now we cannot separate the signal spectrum from the distorted spectrum and hence cannot recover the signal.
9.5 Alias filters
9.5a The standard alias filter such as that shown in Figure 9.5a has a 3-dB point at about half the Nyquist frequency and a very steep slope, so that noise above is highly attenuated relative to the passband of the system. Assuming an original flat spectrum, alias filtering with a 125-Hz, 72-dB/octave filter, and subsequent resampling from 2 to 4 ms (without additional alias filtering), graph the resulting alias noise versus frequency.
Background
The passband of a system is the range of frequencies that are unattenuated by passage through the system. The limits are usually taken as the frequencies for which the attenuation is 3 dB.
Solution
With the 3-dB point on the high-cut filter at 125 Hz and a 72 dB/octave slope, the attenuation at Hz is 75 dB. Frequencies higher than the Nyquist frequency, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{N} +\Delta f} , alias to appear as the frequencies , i.e., they fold back about , as shown in Figure 9.5a.
A 65-Hz 72-dB/octave filter is usually applied before resampling to prevent aliasing. If resampling to 4 ms is done without this additional filtering, for the resampled data is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 1/(2\times 0.004)=125} Hz; the aliased noise is shown by the foldback curves in Figure 9.5a.
9.5b Some believe that standard alias filters may be unnecessarily restrictive. The standard alias filter for 4-ms sampling is about 65 Hz, 72-dB/octave. Graph the alias noise versus frequency for a 90-Hz, 72-dB/octave filter for 4-ms sampling and draw conclusions.
Solution
The 90-Hz filter is shown in Figure 9.5a. Because the sampling interval is 4 ms, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{N}} is 125 Hz and the alias noise is given by the foldback curves shown. The 90-Hz filter gives a broader passband than the standard 65-Hz alias filter if signal frequencies above about 80 Hz and signals attenuated more than 60 dB are not important.
9.6 The convolutional model
9.6a The earth acts as a filter to seismic energy to change the waveshape so that what we record is a distorted version of the waveshape we send into the earth. Assume that the signal generated by a seismic source is so short that it can be represented adequately by an impulse; what aspects of passage through the earth will distort the impulse and what will be the effects?
Solution
The signal from a seismic source will be distorted: (a) by the region near the source where the stresses are so high as to produce nonlinear effects Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle s_{i}} , (b) by the near-surface region where absorption is especially large , (c) by absorption in passage through the main body of the earth Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle p_{i}} , and (d) perhaps by other factors. The first of these will change an impulsive signal to a waveform , which we can think of as a series of impulses with amplitudes that give the shape . Passage through the near surface will replace each of these impulses by multiplied by the scaled value , that is, . Absorption will change this series to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle s_{i} *n_{i} *p_{i}} (where the absorption is frequency dependent). Reflection from interfaces in the earth can also be thought of as additional convolutions changing the waveform to , etc. Thus the effect on the waveform can be thought of as a sequence of convolutions. In ordinary reflection seismic work the objective is to determine the reflectivity so we combine all the other waveshape-changing factors into an equivalent wavelet, generally called the embedded wavelet :
where . This concept is called the convolutional model.
9.6b What is the effect if the source is not impulsive?
Solution
A nonimpulsive source , such as the vibroseis, has the effect of replacing with , but otherwise the reasoning in part (a) is unchanged.
9.7 Water reverberation filter
9.7a Verify equation (9.7d) for the inverse filter for water reverberation by convolving it with equation (9.7a) for the water reverberation.
Background
Multiple reflections within a water layer can cause severe reverberation (ringing) when the reflection coefficient at the sea floor is sufficiently high. We take the reflection coefficients at the top and bottom of the water layer as and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle R} , and as the round-trip traveltime in the layer, being an integer and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \Delta} the sampling interval. Bounces can occur before or after a wave is reflected at deeper reflectors. We assign unit amplitude to a primary reflection with traveltime . A reflection from the same horizon that has suffered one bounce in the water layer has traveltime Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t+n\Delta} and amplitude Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -R} . Since the bounce may take place either before or after reflection at depth, there are two waves arriving at , the combined amplitude being . There are three waves that suffer two bounces: two bounces before the deep reflection, two after, and one before plus one after, so the multiple that arrives at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t+2n\Delta} has amplitude Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle +3R^{2}} . Reflections suffering three bounces arrive at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t+3n\Delta} with amplitude , and so on. The impulse response (see Sheriff and Geldart, 1955, 279) of the water layer is thus the infinite series
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} w_{t} =[1,\; -2R,\; 3R^{2} ,\; -4R^{3} ,\; 5R^{4} ,\ldots\ldots.]\quad \mathrm{when}\ n=1; \end{align} } ( )
( )
and so on for other values of .
A filter (see problem 7.11) that removes the effect of another filter is an inverse filter, and the inverse filter in this case is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle i_{t}} ; it satisfies the equation
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} w_{t} *i_{t} =\delta_{t}; \end{align} } ( )
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} i_{t} &=[1,\; 2R,\; R^{2} ,\; 0,\; 0,\; .\; .\; .]\qquad \mathrm{when}\ n=1 \end{align} } ( )
( )
[See Sheriff and Geldart, 1995, section 15.5.3 for a discussion of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle z} -transforms.]
Solution
For , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle i_{t} =\left[1,\; 2R,\; R^{2} \right]} , and . We wish to verify that . Using the procedure described in problem 9.2, we replace each element of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle i_{t}} by a scaled version of ; thus
1, | –Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 2R} , | +Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 3R^{2}} , | –, | +Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 5R^{4},\ldots} |
+, | –Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 4R^{2}} , | +, | –Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 8R^{4}, \ldots} | |
+, | –Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 2R^{3}} , | + | ||
= 1, | 0, | 0, | 0, | 0, … . |
We can proceed in the same way for other values of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} .
9.7b The spectrum of the water-layer filter is shown in Figure 9.7a for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n=1} ; the large peaks occur at the “singing frequency” Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle nf_{1}} , where is any odd number. Sketch the amplitude spectrum of the inverse filter.
Solution
In the time domain, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle w_{i} *i_{t} =\delta_{t}} ; in the frequency domain this becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle W(f)\times I(f)=1} . Thus the amplitude spectrum of is the inverse of that of , that is, for a given frequency,
as shown in Figure 9.7b.
9.7c Verify your sketch of the water-reverberation inverse filter by transforming
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \left[1, \; 2R,\; R^{2}\right]\leftrightarrow [1+2Re^{-{\hbox {j}}2\pi f{\Delta}} + R^{2} e^{-{\hbox {j}}4\pi f{\Delta}}, \end{align} } ( )
and calculating the value of the spectrum for , , , and .
Solution
We start with the -transform of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle i_{t}} , that is, . This is a complex quantity and we multiply by the conjugate complex (see Sheriff and Geldart, 1995, section 15.1.5) to get the magnitude squared. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle z=e^{\hbox {j}2\pi\Delta}} , the conjugate complex is and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle zz^{-1} =1} ; thus,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} |I(f)|^{2} &=(1+2Rz+R^{2} z^{2})(1+2Rz^{-1} +R^{2} z^{-2})\\ &=(1+Rz)^{2} (1+Rz^{-1})^{2} =[(1+Rz)(1+Rz^{-1})]^{2},\end{align}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mbox {so} \qquad \qquad |I(f)|=(1+Rz)(1+Rz^{-1})=[1+R^{2} +R(e^{+\hbox {j}2\pi f\Delta} +e^{-{\hbox {j}}2\pi f\Delta})]. \end{align} }
Substituting and using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} |I(f)|=[1.25+\cos\,(2\pi f\Delta)]. \end{align} }
Substituting , , and , we get the following values
The peaks and troughs in Figure 9.7a have the relative values 9, 1, 9; since these values are normalized, the agreement is exact.
9.8 Calculating crosscorrelation and autocorrelation
Four causal wavelets are given by , , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle c_{t} =[6,\; -7,\; 2]} , . Calculate the crosscorrelations and autocorrelations , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi _{ac}} , , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi _{aa}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi _{cc}} in both the time and frequency domains.
Background
A causal wavelet as defined in problem 5.21 has zero values when .
The crosscorrelation of and tells us how similar the two functions are when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle h_{t}} is shifted the amount relative to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t}} . The crosscorrelation is given by the equation
( )
This equation means that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle h_{t}} is displaced units to the left relative to , corresponding values multiplied, and the products summed to give Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi_{gh} (\tau)} . The result is the same if we move Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t}} to the right units, that is,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \phi _{gh} (\tau)=\phi_{hg} (-\tau). \end{align} } ( )
The functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi_{gh} (\tau)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t} *h_{t}} are closely related. Using two simple curves, it is easily shown that we can crosscorrelate by reversing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t}} and convolving the reversed function with , that is,
( )
(since convolution is commutative). Reversing a function in time changes the sign of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=n\Delta} , so the exponent of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle z} changes sign also, becoming and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle G(z)} becoming the conjugate complex . The convolution theorem (equation (9.3f)) now becomes the crosscorrelation theorem:
( )
If is the same as , we get the autocorrelation of and equations (9.8a,d) become
( )
the transform relation being the autocorrelation theorem. Since the two functions are the same, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi_{gg} (\tau)} does not depend upon the direction of displacement. The autocorrelation for equals the sum of the data elements squared, hence is called the energy of the trace;
( )
Both the autocorrelation and the crosscorrelation are often normalized; in the case of the autocorrelation, equation (9.8e) becomes
( )
The normalized crosscorrelation is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \phi_{gh} (\tau)_{\mathrm{norm}} =\phi_{gh} (\tau)/[\phi_{gg} (0)\phi_{hh} (0)]^{1/2}. \end{align} } ( )
Solution
Time-domain calculations:
Using equation (9.8a) to calculate , i.e., is first shifted to the left; we have:
So
where marks the value at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \tau=0} . Proceeding in the same way, we find that
To find we displace instead of and obtain , which equals .
Autocorrelations are found in the same way:
Frequency-domain calculations
The -transforms of , , and are , , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle C(z)=(6-7z+2z^{2})} . The conjugate complexes of these transforms are are , , . Using these transforms, we get
9.9 Digital calculations
Fill in the values in Table 9.9a.
Solution
Table 9.9b shows Table 9.9a completed.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=-2} | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=+4} | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=+5} | |||||||
---|---|---|---|---|---|---|---|---|---|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=-1} | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=+2} | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=+4} | |||||||
---|---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 2 | 1 | 1 | 0 | 0 | ||
0 | 0 | 0 | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -4} | 4 | 0 | 0 | |||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle c_t{=3a_{t-2}}} | 0 | 0 | 0 | 0 | 0 | 6 | 3 | 3 | |
1/2 | 1/2 | 1 | 0 | 0 | 0 | 0 | 0 | ||
0 | 0 | 0 | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 2\pi} | 0 | 0 | ||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle f_{t} =\left[-1,\; 1\right]} | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 3 | 1 | 0 | |||
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | |
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \phi_{ff}(t)} | 0 | 0 | 2 | 0 | 0 | 0 | 0 | ||
0 | 0 | 0 | Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -1} | 3 | 0 |
9.10 Semblance
Semblance is the ratio of the energy of a stack of traces to times the sum of the energies of the component traces, all summed over some time interval. Show that when the traces are identical.
Background
While crosscorrelation is a quantitative measure of the similarity of two traces, semblance is a measure of the similarity of a number of traces. Assuming traces, we sum (stack) the traces at time , square the sum to get the total energy, and sum the values over a time interval . The equation for is
( )
Solution
The sum of traces is and its energy is . If the traces are identical, the numerator of equation (9.10a) becomes . The denominator becomes . Since numerator and denominator are equal, .
9.11 Convolution and correlation calculations
9.11a Convolve with .
Background
Convolution is discussed in problem 9.2, crosscorrelation and autocorrelation in problem 9.8, minimum-phase wavelets in Sheriff and Geldart, 1995, section 9.4 and section 15.5.6a, -transforms in Sheriff and Geldart, 1995, section 15.5.3. One of the definitions of a minimum-phase function is that, when the function is expressed as the product of its factors in the form , the magnitudes of the roots must all be greater than unity.
Solution
To evaluate the convolution, we replace each element of with a scaled version of and then sum values for the same times.
6: | 12, | 30, | –12, | 6 | ||
–: | –2, | –5, | 2, | –1 | ||
–: | –2, | –5, | 2, | –1 | ||
Sum: | 12, | 28, | –19, | 3, | 1, | –1 |
The same result can be obtained by reversing one function and sliding it past the other, multiplying values at the same times and summing the results:
2 | 5 | –2 | 1 | Sum | ||||
t | –1 | –1 | 6 | |||||
0 | –1 | –1 | 6 | +12 | ||||
1 | –1 | –1 | 6 | +28 | ||||
2 | –1 | –1 | 6 | –19 | ||||
3 | –1 | –1 | 6 | +3 | ||||
4 | –1 | –1 | +1 | |||||
5 | –1 | –1 |
Still another method is to use -transforms; thus,
Transforming back to the time domain, we have
9.11b Crosscorrelate with . For what shift are these functions most nearly alike?
Solution
Using equation (9.8a) for , we write
2 | 5 | -2 | 1 | |||||
+2 | 6 | -1 | -1 | -2 | ||||
+1 | 6 | -1 | -1 | -7 | ||||
0 | 6 | -1 | -1 | +9 | ||||
-1 | 6 | -1 | -1 | +31 | ||||
-2 | 6 | -1 | -13 | |||||
-3 | 6 | +6 |
The functions are most alike for a shift of .
9.11c Convolve with . Compare with the answer in part (b) and explain.
Solution
Using the convolution method of replacing the first function by a scaled version of itself, the scale factors being the elements of the second function, we obtain
–2 | 5 | –2 | 1 | ||
–2 | –5 | 2 | –1 | ||
12 | 30 | –12 | 6 | ||
–2, | –7, | 9, | 31, | –13, | 6 |
Thus, the convolution is , the same result achieved by the correlation in part (b). Thus, .
9.11d Autocorrelate and . The autocorrelation of a function is not unique to that function (see Sheriff and Geldart, 1995, 552). Other wavelets having the same autocorrelation as the preceding are and . Which of the four is the minimum-phase wavelet?
Solution
We use -transforms to get the autocorrelations:
To determine which of the four wavelets are minimum-phase, we factor each of the transforms:
Roots | |
Since only the first wavelet has all roots of magnitude greater than 1, it is the only one that is minimum-phase.
9.11e What is the normalized autocorrelation of ? What is the normalized crosscorrelation in part (b)? What do you conclude from the magnitude of the largest value of this normalized crosscorrelation?
Solution
To normalize , we divide by [see equation (9.8g)] from part (d), giving
To normalize , we divide by [see equation (9.8h)]. We use equation (9.8f) to get . Using the values of from part (b), we get
The largest value comes when , so the two wavelets are most similar when is shifted one time unit to the right with respect to . The maximum value is , which means that and are quite similar, but not the same.
9.12 Properties of minimum-phase wavelets
9.12a Of the four wavelets given in problem 9.8, which are minimum-phase?
Background
Minimum-phase wavelets are discussed briefly in problem 9.11 and in more detail in Sheriff and Geldart, 1995, section 9.4 and section 15.5.6); -transforms are discussed in Sheriff and Geldart, 1995, section 15.5.3.
Solution
The four wavelets are: , , , and . The roots of the wavelets and are 2 and , so both are minimum-phase because the magnitudes of the roots are greater than unity. Since , the roots are 3/2 and 2; is therefore also minimum-phase. Finally, , the roots being and . Because , is mixed-phase.
9.12b Find and by calculating in the time domain.
Solution
9.12c Repeat part (b) except using transforms.
Solution
9.12d Find .
Solution
9.12e Does the largest value of a minimum-phase wavelet have to come at ?
Solution
The wavelet is minimum-phase if and . The ratio of the first two terms is and it has its minimum absolute value when and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle b} have the same signs. When and have the same sign and are both slightly larger than unity, the ratio is close to 1/2 and the second term is larger than the first. As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a} and/or increase, the ratio increases; the first and second terms are equal when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a=b=2} . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle b} differ significantly in magnitude, the second term can be larger than the first for large values of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a} or ; e.g., if , the ratio is when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle a=9} .
If and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle b} have opposite signs, the ratio cannot be smaller than 1 since the two terms in the denominator have opposite signs and the denominator cannot exceed the numerator.
When the wavelet has three factors, the ratio of the first to second term takes the form . When , , and are all close to unity and of the same polarity, the magnitude of the ratio is and the second term is larger than the first. Generalizing to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} terms the ratio can be .
9.12f Can a minimum-phase wavelet be zero at ?
Solution
If a wavelet is zero at , it is of the form (}, so
Since one of the roots is , the wavelet is not minimum-phase.
When we deal with an individual wavelet, we avoid the root by taking Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=0} as the time when the first nonzero value occurs.
9.13 Phase of composite wavelets
Using the wavelets
calculate the composite wavelets:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} W_{1}(z)+W_{2}(z) ;\quad W_{1}(z)+zW_{2}(z) ;\quad zW_{1}(z)+W_{2}(z) ;\quad z^{-1} W_{1}(z)+W_{2}(z). \end{align} }
Plot the composite wavelets in the time domain. All of these composite wavelets have the same frequency spectrum but different phase spectra because multiplication by shifts the phase. The results illustrate the effects of phase.
Background
It is shown in Sheriff and Geldart, 1995, section 15.1.5 that the phase of a complex quantity, , is , that is, (imaginary part)/(real part). Since , . If a wavelet has several elements, the imaginary part of the wavelet will be the sum of several sines and the real part will be the sum of several cosines. When the wavelet is multiplied by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle z^{n}} , both sums will change, hence the phase changes.
Solution
The convention is that (unless otherwise specified) the first nonzero element fixes the wavelet origin (see problem 9.12f). We start our plots in Figure 9.13a at [except for in order to show the complete waveforms.
The two wavelets Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle W_{1}} and are plotted first in Figure 9.13a and then the four sums of the wavelets. Note that two of the four factors that make up each of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle W_{1}} and are the same and the other two differ only in signs:
However, the waveshapes differ significantly, especially in apparent frequency. Clearly relatively small changes in the equation of a wavelet can produce significant changes in the waveshape.
Because delaying the second wavelet or advancing the first produces the same waveshape, composite wavelets and are the same except for a time shift. Wavelets and have distinctly different waveshapes from those of . We conclude that a shift of one component relative to the other has a significant effect on the location of peaks and troughs and hence on the waveshape.
9.14 Tuning and waveshape
9.14a The following wavelet (upper curve in Figure 9.14a) is approximately minimum-phase: , the sampling interval being 2 ms. Use km/s for the velocity in sand, km/s for the velocity in shale, and the shale-to-sand reflection coefficient (value scaled up and rounded off). Determine the reflected waveshape for sands 0, 2, 4, 6, 8, and 10 m thick encased in shale. (A thickness of 6 m is approximately a quarter wavelength.)
Background
Reflection and transmission coefficients are discussed in problem 3.6, minimum-phase and zero-phase wavelets in Sheriff and Geldart, 1995, section 15.5.6a,d, respectively.
Tuning, the build up because of constructive interference that occurs when the travelpath difference between the waves reflected at top and bottom of a bed produces a half-cycle shift, which occurs at a thickness of when reflection coefficients are of opposite polarity. The opposite effect of lowering the amplitude (detuning) occurs at a thickness of if the reflections from top and base of the bed have the same polarity.
Solution
Because for a shale-to-sand interface, the reflection at the sand-to-shale interface at the base of the sand is reversed in polarity compared to that from the top. The reflection energy is so 99% of the incident energy is transmitted into the sand; therefore we ignore transmission losses. The wave reflected at the base of the bed is delayed by the two-way traveltime, ms. Thus the reflection from the base is delayed by one time sample for each 2 m of sand thickness, i.e., time samples. Since the reflection coefficient is for both interfaces, we can take 0.1 as a scale factor and omit it in our calculations.
To get the composite wave, that is, the sum of the wavelets and , we displace by the amount and add it to . When , there is no reflection. The embedded wavelet is wavelet (i) shown in Figure 9.14a. Calculations for different sand thicknesses are shown in Table 9.14a. The embedded wavelet and the composite reflections are plotted in Figure 9.14b. The curves for , 2, 3 are roughly displaced copies of each other and have successively larger amplitudes. The curve at the tuning thickness has the maximum amplitude, the largest amplitudes of the and curves being slightly smaller. If measuring arrival time by timing the first peak, one would pick too early when the bed is thinner than the tuning thickness. The first trough of the curve for is broadened and a change of phase is clear in the first trough of the curve, indicating that the bed is thicker than the resolvable limit of .
For | ||||||||||||||||||
11, | 14, | 5, | –10, | –12, | –6, | 3, | 5, | 2, | 0, | –1, | –1, | 0 | ||||||
–11, | –14, | –5, | 10, | 12, | 6, | –3, | –5, | –2, | 0, | 1, | 1, | 0 | ||||||
Sum | 11, | 3, | –9, | –15, | –2, | 6, | 9, | 2, | –3, | –2 | –1, | 0, | 1, | 0 | ||||
For | ||||||||||||||||||
11, | 14, | 5, | –10, | –12, | –6, | 3, | 5, | 2, | 0, | –1, | –1, | 0 | ||||||
–11, | –14, | –5, | 10, | 12, | 6, | –3, | –5, | –2, | 0, | 1, | 1, | 0 | ||||||
Sum | 11, | 14, | –6, | –24, | –17, | 4, | 15, | 11, | –1, | –5, | –3, | –1, | 1, | 1, | 0 | |||
For | ||||||||||||||||||
11, | 14, | 5, | –10, | –12, | –6, | 3, | 5, | 2, | 0, | –1, | –1, | 0 | ||||||
–11, | –14, | –5, | 10, | 12, | 6, | –3, | –5, | –2, | 0, | 1, | 1, | 0 | ||||||
Sum | 11, | 14, | 5, | –21, | –26, | –11, | 13, | 17, | 8, | –3, | –6, | –3, | 0 | 1, | 1, | 0 | ||
For | ||||||||||||||||||
11, | 14, | 5, | –10, | –12, | –6, | 3, | 5, | 2, | 0, | –1, | –1, | 0 | ||||||
–11, | –14, | –5, | 10, | 12, | 6, | –3, | –5, | –2, | 0, | 1, | 1, | 0 | ||||||
Sum | 11, | 14, | 5, | –10, | –23, | –20, | –2, | 15, | 14, | 6, | –4, | –6, | –2 | 0, | 1, | 1 | 0 | |
For | ||||||||||||||||||
11, | 14, | 5, | –10, | –12, | –6, | 3, | 5, | 2, | 0, | –1, | –1, | 0 | ||||||
–11, | –14, | –5, | 10, | 12, | 6, | –3, | –5, | –2, | 0, | 1, | 1, | 0 | ||||||
Sum | 11, | 14, | 5, | –10, | –12, | –17, | –11, | 0, | 12, | 12, | 5, | –4, | –5, | –2, | 0, | 1, | 1, | 0 |
9.14b Repeat for the sand overlain by shale and underlain by limestone. Assume a sand-limestone reflection coefficient of .
Solution
The calculations (see Table 9.14b) are the same as in part (a) except that polarity at the sand/limestone interface is the same as at the shale/sand interface. The top curve in Figure 9.14b is for zero sand thickness so that the contact is shale/limestone with a reflection coefficient of . Timing the first peak would give erroneous depths for the , 2, and 3 curves and the curves show rather clear phasing, indicating that more than one reflector is involved. The amplitude of the first trough decreases as the sand thickens for the , 2, 3 curves and its character varies considerably for curves for , 4, 5.
9.14c Determine the waveshape for two sands, each 2 m thick and separated by 1.5 m of shale, the sequence being encased in shale; this illustrates a “tuned” situation. Compare the results with those for 4 m and 6 m of sand in part (a), that is, for the same net and gross thicknesses.
Solution
The composite reflected wave is the sum , calculated in Table 9.14c.
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||
−11, | −14, | −5, | 10, | 12, | 6, | −3, | −5, | −2, | 0, | 1, | 1, | 0 | ||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||
−11, | 14, | −5, | 10, | 12, | 6, | −3, | −5, | −2, | 0, | 1, | 1, | 0 | ||||
Sum | 11, | 3, | 2, | −12, | −10, | −9, | 7, | 8, | 6, | 0, | −4, | −2, | 0, | 0, | 1, | 0 |
For (2 m) | ||||||||||||||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
Sum | 11, | 25, | 19, | −5, | −22, | −18, | −3, | 8, | 7, | 2 | −1, | −2, | −1, | 0 | ||||
For (4 m) | ||||||||||||||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
Sum | 11, | 14, | 16, | 4, | −7, | −16, | −9, | −1, | 5, | 5, | 1, | −1, | −1, | −1, | 0 | |||
For (6 m ) | ||||||||||||||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
Sum | 11, | 14, | 5, | 1, | 2, | −1, | −7, | −7, | −4, | 3, | 4, | 1, | 0 | −1, | −1, | 0 | ||
For (8 m) | ||||||||||||||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
Sum | 11, | 14, | 5, | −10, | −1, | 8, | 8, | −5, | −10, | −6, | 2, | 4, | 2 | 0, | −1, | −1, | 0 | |
For (10 m) | ||||||||||||||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
11, | 14, | 5, | −10, | −12, | −6, | 3, | 5, | 2, | 0, | −1, | −1, | 0 | ||||||
Sum | 11, | 14, | 5, | −10, | −12, | 5, | 17, | 10, | −8, | −12, | −7, | 2, | 5, | 2, | 0, | −1, | −1, | 0 |
Single sands 4 and 6 m thick from part (a):
6 m | 11, | 14, | 5, | -21, | -26, | -11, | 13, | 17, | 8, | -3, | -6, | -3, | 0, | 1, | 1, | 0; |
4 m | 11, | 14, | -6, | -24, | -17, | 4, | 15, | 11, | -1, | -5, | -3, | -1, | 1, | 1, | 0. |
The composite wavelet for two 2-m sands is the upper curve in Figure 9.14d and those for single 4-m and 6-m sands are the lower curves, i.e., for the same net and same gross sand thicknesses. The composite curve for the two thin sands shows phasing where the reflections from the top and base of each sand interfere. The gross thickness of 6 m is above the tuning thickness (note the peak-to-trough time difference between the 4-m and 6-m sands, evidence that 6 m is larger than the tuning thickness). Where the gross thickness of an interval is smaller than a quarter wavelength, information as to the thickness of the different lithology (sand, in this case) is contained in amplitude rather than in time measurements (or waveshape changes).
9.14d Repeat part (a) with the low-frequency wavelet [6,11,14,14,10,5,–2,–10,–11,–12,–10,–6,0,3,4,5,4,3,1,0] [Figure 9.14a (ii)], which is the minimum-phase wavelet in Figure 9.14a(i), but stretched out so that it has about half the dominant frequency. Compare with the results of part (a) to illustrate the effect of frequency on resolution.
Solution
We proceed as in part (a) after replacing the former embedded wavelet with the new one. The results for the shale/sand/shale are shown in Table 9.14d and plotted in Figure 9.14e.
We compare Figures 9.14b and 9.14e which differ only in the frequencies, both having the same embedded wavelet waveshape. In Figure 9.14e the curves all have nearly the same shape but differ in amplitude except for the curve where the trough and second peak are broadened because is about equal to . The resolution is poorer than with the higher-frequency wavelet in Figure 9.14a(i).
For (2 m) | |||||||||||||||||||||||||
6, | 11, | 14, | 14, | 10, | 5, | –2, | –10, | –11, | –12, | –10, | –6, | 0, | 3, | 4, | 5, | 4, | 3, | 1, | 0 | ||||||
–6, | –11, | –14, | –14, | –10, | –5, | 2, | 10, | 11, | 12, | 10, | 6, | 0, | –3, | –4 | –5, | –4, | –3, | –1, | 0 | ||||||
Sum | 6, | 5, | 3, | 0, | 4, | 5, | 7, | 8, | 1, | 1, | 2, | 4, | 6, | 3, | 1, | 1, | 1, | 1, | 2, | 1, | 0 | ||||
For (4 m) | |||||||||||||||||||||||||
6, | 11, | 14, | 14, | 10, | 5, | –2, | –10, | –11, | –12, | –10, | –6, | 0, | 3, | 4, | 5, | 4, | 3, | 1, | 0 | ||||||
–6, | –11, | –14, | –14, | –10, | –5, | 2, | 10, | 11, | 12, | 10, | 6, | 0, | –3, | –4 | –5, | –4, | –3, | –1, | 0 | ||||||
Sum | 6, | 11, | 8, | 3, | 4, | 9, | 12, | 15, | 9, | 2, | 1, | 6, | 10, | 9, | 4, | 2, | 0, | 2, | 3, | 3, | 1, | 0 | |||
For (6 m) | |||||||||||||||||||||||||
6, | 11, | 14, | 14, | 10, | 5, | –2, | –10, | –11, | –12, | –10, | –6, | 0, | 3, | 4, | 5, | 4, | 3, | 1, | 0 | ||||||
–6, | –11, | –14, | –14, | –10, | –5, | 2, | 10, | 11, | 12, | 10, | 6, | 0, | –3, | –4, | –5, | –4, | –3, | –1, | 0 | ||||||
Sum | 6, | 11, | 14, | 8, | 1, | 9, | 16, | 20, | 16, | 10, | 0, | 5, | 12, | 13, | 10, | 5, | 1, | 1, | 4, | 4, | 3, | 1, | 0 | ||
For (8 m) | |||||||||||||||||||||||||
6, | 11, | 14, | 14, | 10, | 5, | –2, | –10, | –11, | –12, | –10, | –6, | 0, | 3, | 4, | 5, | 4, | 3, | 1, | 0 | ||||||
–6, | –11, | –14, | –14, | –10, | –5, | 2, | 10, | 11, | 12, | 10, | 6, | 0, | –3, | –4, | –5, | –4, | –3, | –1, | 0 | ||||||
Sum | 6, | 11, | 14, | 14, | 4, | –6, | –16, | –24, | –21, | –17, | –8, | 4, | 11, | 15, | 14, | 11, | 4, | 0, | –3, | –5, | –4, | –3, | –1, | 0 | |
For (10 m) | |||||||||||||||||||||||||
6, | 11, | 14, | 14, | 10, | 5, | –2, | –10, | –11, | –12, | –10, | –6, | 0, | 3, | 4, | 5, | 4, | 3, | 1, | 0 | ||||||
–6, | –11, | –14, | –14, | –10, | –5, | 2, | 10, | 11, | 12, | 10, | 6, | 0, | –3, | –4, | –5, | –4, | –3, | –1, | 0 | ||||||
Sum | 6, | 11, | 14, | 14, | 10, | –1, | –13, | –24, | –25, | –22, | –15, | –4, | 10, | 14, | 16, | 15, | 10, | 3, | –2, | –4, | –5, | –4, | –3, | –1, | 0 |
9.14e Repeat parts (a) and (b) using the zero-phase wavelet [1, 1, , , , , 10, 17, 10, , , , , 1, 1] [Figure 9.14a(iii)], which has nearly the same spectrum as wavelet (i).
Solution
The calculations are shown in Tables 9.14e and 9.14f and the reflection waveshapes in Figures 9.14f and 9.14g.
These curves have been plotted about their points of symmetry or asymmetry. The top curve in Figure 9.14f is the embedded wavelet, in Figure 9.14g it is for zero sand thickness. As with the minimum-phase wavelet, timing would be in error where the thickness is less than and the maximum amplitude in Figure 9.14f occurs where the thickness is . The peak-trough times increase for and in Figure 9.14f, and there is distinct phasing for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \tau=5} . In Figure 9.14g, there is a distinct change of shape at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \tau=3} and there is clear resolution and an amplitude minimum at .
For (2 m) | |||||||||||||||||||||
1, | 1, | –1, | –4, | –6. | –4, | 10, | 17, | 10, | –4, | –6, | –4, | –1, | 1, | 1, | 0 | ||||||
1, | 1, | 1, | 4, | 6. | 4, | –10, | –17, | –10, | 4, | 6, | 4, | 1, | –1, | –1, | 0 | ||||||
Sum | 1, | 0, | –2, | –3, | –2, | 2, | 14, | 7, | –7, | –14, | –2, | 2, | 3, | 2, | 0, | –1, | 0 | ||||
For (4 m) | |||||||||||||||||||||
1, | 1, | 1, | –4, | –6. | –4, | 10, | 17, | 10, | –4, | –6, | –4, | 1, | 1, | 1, | 0 | ||||||
–1, | –1, | 1, | 4, | 6. | 4, | –10, | –17, | –10, | 4, | 6, | 4, | 1, | –1, | –1, | 0 | ||||||
Sum | 1, | 1, | –2, | –5, | –5, | 0, | 16, | 21, | 0, | –21, | –16, | 0, | 5, | 5, | 2, | –1, | –1, | 0 | |||
For (6 m) | |||||||||||||||||||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t}} | 1, | 1, | –1, | –4, | –6. | –4, | 10, | 17, | 10, | –4, | –6, | –4, | –1, | 1, | 1, | 0 | |||||
–1, | –1, | 1 | 4, | 6. | 4, | –10, | –17, | –10, | 4, | 6, | 4, | 1, | –1, | –1, | 0 | ||||||
Sum | 1, | 1, | –1, | –5, | –7, | –3, | 14, | 23, | 14, | –14, | –23, | –14, | 3, | 7, | 5, | 1, | –1, | –1, | 0 | ||
For (8 m) | |||||||||||||||||||||
1, | 1, | –1, | –4, | –6. | –4, | 10, | 17, | 10, | –4, | –6, | –4, | –1, | 1, | 1, | 0 | ||||||
–1, | –1, | 1 | 4, | 6. | 4, | –10, | –17, | –10, | 4, | 6, | 4, | 1, | –1, | –1, | 0 | ||||||
Sum | 1, | 1, | –1, | –4, | –7, | –5, | 11, | 21, | 16, | 0, | –16, | –21, | –11, | 5, | 7, | 4, | 1, | –1, | –1, | 0 | |
For (10 m) | |||||||||||||||||||||
1, | 1, | –1, | –4, | –6. | –4, | 10, | 17, | 10, | –4, | –6, | –4, | –1, | 1, | 1, | 0 | ||||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -g_{t-4}} | –1, | –1, | 1 | 4, | 6. | 4, | –10, | –17, | –10, | 4, | 6, | 4, | 1, | –1, | –1, | 0 | |||||
Sum | 1, | 1, | –1, | –4, | –6, | –5, | 9, | 18, | 14, | 2, | –2, | –14, | –18, | –9, | 4, | 6, | 4, | 1, | –1, | –1, | 0 |
For (2 m) | |||||||||||||||||||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -g_{t-1}} | 1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | |||||
Sum | 1, | 2, | 0, | -5, | -10, | -10, | 6, | 27, | 27, | 6, | -10, | -10, | -5, | 0, | 2, | 1, | 0 | ||||
For (4 m) | |||||||||||||||||||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
1, | 1, | -1, | -4, | -6, | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
Sum | 1, | 1, | 0, | -3, | -7, | -8, | 4, | 13, | 20, | 13, | 4, | -8, | -7, | -3, | 0, | 1, | 1, | 0 | |||
For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \tau = 3} (6 m) | |||||||||||||||||||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -g_{t-3}} | 1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | |||||
Sum | 1, | 1, | -1, | -3, | -5, | -5, | 6, | 11, | 6, | 6, | 11, | 6, | -5, | -5, | -3, | -1, | 1, | 1, | 0, | ||
For (8 m) | |||||||||||||||||||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
Sum | 1, | 1, | 1, | 4, | 5, | 3, | 9, | 13, | 4, | 0, | 4, | 13, | 9, | 3, | 5, | 4, | 1, | 1, | 1, | 0 | |
For (10 m) | |||||||||||||||||||||
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t}} | 1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | |||||
1, | 1, | -1, | -4, | -6. | -4, | 10, | 17, | 10, | -4, | -6, | -4, | -1, | 1, | 1, | 0 | ||||||
Sum | 1, | 1, | -1, | -4, | -6, | -3, | 11, | 16, | 6, | -10, | -10, | 6, | 16, | 11, | -3, | -6, | -4, | -1, | 1, | 1, | 0 |
9.15 Making a wavelet minimum-phase
The wavelet is not minimum-phase. How would you change it to make it mimimum-phase without changing it any more than necessary? Give two methods.
Background
Minimum-phase wavelets are discussed in Sheriff and Geldart, 1995, section 9.4 and section 15.5.6a.
Solution
The transform of the wavelet is . The roots of are 0.9652 and , so to make the wavelet minimum-phase, we must change it so that the magnitudes of both roots are greater than unity. To increase both roots by the same factor, we multiply both roots by a number slightly greater than . If we use the multiplier 1.0500 the roots become 1.0135 and and the revised equation is
To obtain the same value at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t=0} as we had before, we multiply by and the transform becomes which gives the wavelet [, , 0.9371]. The modified wavelet is nearly the same as the original except that its “tail” has been decreased from 0.9915 to 0.9371, a 5.5% decrease.
A second method is to use a taper to reduce the tail of the wavelet. A commonly used taper is where is in milliseconds and is slightly less than unity, for example, 0.9950. Using this value for and assuming that the sampling interval is 2 ms, the values of the taper are 1.0000, , , that is, 1.0000, 0.9900, 0.9801 and we get . The roots of are now 0.9904, , both less than unity so we did not apply enough taper. We next try using , which reduces the wavelet to . The roots are now 1.0154 and and . Multiplying by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 0.9505/1.0184=0.9333} makes the wavelet [0.9505, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle -0.0116} , 0.9333].
Comparing the two methods, we see that the second method has changed the wavelet slightly more than the first. In practice, a wavelet will have many more than three elements and the second method will generally be more practical. To verify that the taper is large enough to achieve its objective, we have to find all of the roots.
9.16 Zero-phase filtering of a minimum-phase wavelet
Show that the result of passing a minimum-phase signal through a zero-phase filter is mixed phase.
Background
Zero-phase signals are discussed in Sheriff and Geldart, 1995, section 15.5.6d, where it is shown that the spectrum of a zero-phase signal comprises products of pairs of factors of the form , where can be complex. Because the imaginary part is zero, the phase is zero.
Solution
Let be the minimum-phase signal and the zero-phase filter. Time-domain filtering is accomplished by convolution, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{t} *f_{t}} ; in the frequency domain the result is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle W(z)F(z)} . The factors of occur in pairs of the form , and each pair has roots , . If , then . Thus, one member of each pair of roots is not minimum-phase and consequently the filtered signal is mixed-phase.
9.17 Deconvolution methods
Deconvolution, whose objective is undoing the results of a prior convolution, is a somewhat open-ended collection of methods, a number of which are briefly described in the background. List the assumptions of different methods, such as invariant wavelet, randomness of the reflectivity or of the noise, whether a source wavelet is the same as the recorded wavelet, and so on.
Background
Various deconvolution methods and their characteristics are described in the following text. Instead of giving a solution, we leave it up to the reader to list the assumptions.
A change in signal waveshape can be regarded as filtering, the effect being equivalent to convolution of the signal with a filter, . Deconvolution attempts to remove the effects of and obtain the original signal . Deconvolution in the time domain consists of convolution with an inverse filter , i.e., Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle h_{t} *i_{t}} (see problems 9.18 and 9.19); in the frequency doman its equivalent is multiplication of the transforms, (see Sheriff and Geldart, 1995, section 9.5 for more details). An inverse filter convolved with the filter yields an impulse, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle i_{t} *f_{t} \leftrightarrow I(\omega)F(\omega)=1\leftrightarrow \delta_{t}} . The objective of deconvolution often is to determine the shape of the embedded wavelet (see problem 9.6).
Most deconvolution methods are based on autocorrelations of individual traces. An autocorrelation measures the repetition in a time series. Presumably the embedded wavelet is repeated for every reflection and thus the early part of the autocorrelation is largely determined by the shape of the embedded wavelet (see Figure 9.17a). An embedded wavelet that is long will contribute significantly to several half-cycles of the autocorrelation. Because we generally want a short embedded wavelet, we want the autocorrelation’s amplitude to die off rapidly.
The wavelet changes shape as it travels through the earth. Nevertheless, most deconvolution methods assume constant waveshape, sometimes called stationarity. Obtaining a representative autocorrelation requires that an appreciable number of time samples be included in the calculation process, so usually 500 or more samples (1 s at 0.002 s sampling) are included. Because there is no phase information in an autocorrelation, the phase spectrum of the embedded wavelet cannot be recovered from it. It is often assumed that the embedded wavelet is minimum-phase.
Changes in the wavelet shape with time are generally accommodated by subdividing a trace into time segments, e.g., finding a deconvolution operator for the early portion of a trace and another for a later portion, assuming that the wavelet shape is constant during each portion. The portions analyzed often overlap, e.g., one autocorrelation is calculated for the data between 0.5 and 2.0 s, a second for the data between 1.5 and 3.0 s. Each deconvolution operator is then applied over its respective range, giving two outputs in the overlap region. These two outputs are then added together in different proportions in the overlap region, the output at 1.5 s being 100% of that given by the early operator and that at 2.0 s being 100% of that given by the latter operator, the proportions changing linearly during the overlap time. This is often called adaptive deconvolution.
While many deconvolution methods exist, they can be roughly divided into deterministic and statistical methods, but the division is often unclear because deterministic methods may employ statistics and statistical methods may utilize knowledege about the nature of the convolution to be undone. Deterministic deconvolution requires that we know or can reasonably assume the mechanism or properties of the convolution to be undone.
One type of deconvolution is dereverberation or deringing, whose objective is to remove the effects of energy bouncing repeatedly in a near-surface layer—usually the water layer with marine data. This requires a knowledge of the repetition time of the near-surface bounces and the relative amplitudes of successive bounces. In the marine case the repetition time is assumed to be given by the water depth and the amplitudes by the sea-floor reflectivity, which sometimes can be obtained by trials. One marine deghosting method used with ocean-bottom recording employs velocity geophones and hydrophones (see problem 7.10).
Deghosting is a type of deterministic deconvolution where we assume that the ghost is a replica of the original signal with amplitude reduced by the factor and delayed by , where is the reflection coefficient at the ghosting interface (note that is usually negative) for a wave approaching from below and is the two-way traveltime between the source and the ghosting interface. The transform of the signal plus ghost is then
( )
On land we usually can get fairly accurately from the depth of the source and the thickness and velocity of the LVL, but we must determine (or guess) the value of . In marine work, we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} from the depth of the water. Deghosting is often done in a recursive manner, called recursive deconvolution.
System deconvolution is sometimes deterministically carried out to remove the filtering action of recording instrumentation or data processing.
In spiking deconvolution we assume that the impulse response of the earth Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r_{t}} , whose elements are the reflection coefficients at the various interfaces, are randomly distributed, and, hence, the autocorrelation of has a nonzero value (a “spike”) only at zero time shift, that is,
( )
[“Random” here means unpredictable, i.e., one cannot predict the arrival time of a primary reflection based on the arrival times of earlier reflections.] If is the embedded wavelet, the geophone output is . We can use the method of least squares (least-squares filtering is discussed in problem 9.22) to find the optimum inverse filter that will give a result that has the properties we assume for (see Sheriff and Geldart, 1995, 296). Spiking deconvolution can be carried out in either the time or frequency domain. Because we consider as random, its spectrum contains all frequencies in equal abundance, that is, its spectrum should be flat; techniques for achieving this are called spectral flattening or flattening deconvolution. Flattening is usually done only over the passband where the signal is assumed to be dominant. Since the autocorrelation of is small except for zero shift, the inverse filter is
( )
where is the transform of the observed seismic trace. We can get the amplitude spectrum from the autocorrelation, but we also need to know the phase of in order to solve the problem. There is no phase information in the autocorrelation, and so we have to assume the phase to get a solution. We usually assume that is minimum-phase. Spiking deconvolution is often done to shorten the duration of the embedded wavelet, but it may make the data noisy if too much noise is included.
Predictive deconvolution uses information about the primary reflections to predict multiples produced by the same reflectors. Long-path multiples cause systematic repetition of a trace which produces significant values of the autocorrelation following the time delay associated with the multiple repetition time (see Figure 9.17a). Hence deconvolution to remove multiples is generally based on the portion of an autocorrelation trace after a lag time Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle L} . A predictive deconvolution filter does not begin to act until the time Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle L} . Autocorrelation elements that occur after are regarded as produced by multiples and a least-squares method can be used to find the filter that will predict the multiples. The predicted multiples can then be subtracted to get rid of the recorded multiples.
Entropy is a measure of the disorder or unpredictability of a system, the entropy increasing as the disorder increases. The autocorrelation of a time series is not unique, and a spectrum corresponds to a number of different time series. Maximum-entropy filtering attempts to select the time series that has the maximum entropy (maximum disorder).
We sometimes try to represent a seismic trace as a sequence minimizing the number of reflection events, i.e., involving only a few large reflections. This can be accomplished, for example, by minimizing the error in a least-fourth power sense rather than in a least-squares sense (as is done in Wiener filtering). This type of operation is sometimes called minimum-entropy (sparse-spike) deconvolution.
Homomorphic deconvolution (Sheriff and Geldart, 1995, 298) involves transformation into the cepstral domain, the cepstrum being defined by the relations
Where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \leftrightarrow} denotes the Fourier transformation. In the cepstral domain the geophone input Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle g_{t} =e_{t} *\omega _{t}} becomes . If varies more rapidly than Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \omega _{t}} , the two may be separable by frequency filtering.
9.18 Calculation of inverse filters
Assuming that the signature of an air-gun array is a unit impulse and that the recorded wavelet after transmission through the earth is , find the inverse filter that will remove the earth filtering. How many terms should the filter include?
Solution
The inverse filter (see problem 9.7) is a filter that will restore the source impulse, i.e.,
or in the frequency domain where ,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} G(z)I(z)=1. \end{align} }
Thus,
( )
where .
Since has the magnitude , the magnitude of for all values of , and we can expand equation (9.18a) using the binomial theorem [see equation (4.1b)] and Sheriff and Geldart, 1995, equation (15.43):
( )
We first find neglecting powers higher than :
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} -A&=-0.3333z+0.2500z^{2} +0.0833z^{3} ,\\ A^{2} &= +0.1111z^{2} -0.1667z^{3} +0.0069z^{4} ,\\ -A^{3} &= -0.0370z^{3} +0.0833z^{4} ,\\ A^{4} &= +0.0123z^{4} \\ \mathrm{Sum} &=-0.3333z+0.3611z^{2} -0.1204z^{3} +0.1025z^{4} \\ I(z)&=-\frac{1}{12} (1-A+A^{2} -A^{3} +A^{4})\\ &=-\frac{1}{12} (1-0.3333z+0.3611z^{2} -0,1204z^{3} +0.1025z^{4}). \end{align} }
We can verify the accuracy of by multiplying by . We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} I(z):1-0.3333z+0.3611z^{2} -0, 1204z^{3} +0.1025z^{4} ,\\ G(z):1+0.3333z-0.2500z^{2} -0.0833z^{3} \\ 1-0.3333z+0.3611z^{2} -0.1204z^{3} +0.1025z^{4} \\ +0.3333z-0.1111z^{2} +0.1204z^{3} -0.0401z^{4} +0.0342z^{5} \\ -0.2500z^{2} +0.0833z^{3} -0.0903z^{4} +0.0301z^{5} -0.0256z^{6}\\ {-0.0833z^{3} +0.0278z^{4} -0.0301z^{5} +0.0100z^{6} -0.0085z^{7}}\\ {1+0+0+0-0.0001z^{4} +0.0342z^{5} -0.0156z^{6} -0.0085z^{7}}. \end{align} }
We see that the inverse filter is exact as far as the term Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle z^{3}} and terms for higher powers are small. The overall effect is to create a small tail whose energy is 0.00149 or 0.1%.
To determine how the accuracy depends on the number of terms used in , we observe the effect on the pro