User:Ageary/Chapter 9

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Problems in Exploration Seismology and their Solutions
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


Chapter 9 Data processing

9.1 Fourier series

9.1a Show that the Fourier series coefficients, and 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:


(9.1a)
(9.1b)
(9.1c)

where fundamental angular frequency, harmonic frequency, and the amplitudes and phases are given by the following equations where , except for equation (9.1h) where :


(9.1d)
(9.1e)
(9.1f)
(9.1g)
(9.1h)

Solution

We multiply both sides of equation (9.1a) by , being any nonzero positive integer, and integrate between the limits 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 , the integrals are zero, because at the limits the arguments of the cosines and sines are equal to , integral. When , the first integrand becomes and the right-hand side equals . When , the second integrand becomes , 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 in equations (9.1d,e) we get and



9.1b Verify equations (9.1f,g) for the Fourier series coefficients and .


Solution

To verify equation (9.1f) for , we multiply both sides of equation (9.1b) by , , and proceed as in part (a). All integrals vanish except when , and the result is equation (9.1f).

When , we have



To verify equation (9.1g) for , we compare terms in equations (9.1a) and (9.1b) that involve , . The result is


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 , , 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 , written , is defined as


(9.2a)


For digital functions (see problem 9.4), the integrals become sums:


(9.2b)


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.

Figure 9.2a.  Vibroseis acquisition. (i) Field locations for six sweeps (shown displaced vertically); (ii), array pattern for four vibroseis trucks spaced 33 m apart moving forward 16.5 m for each sweep. Each receiver array consists of geophones spaced uniformly over 100 m. For the next location, an additional group of phones is added on one side and dropped on the other. The numbers in (ii) give the multiplicity.
Figure 9.2a.  Boxcar and its transform.
Figure 9.2b.  Two boxcars.

Another explanation of convolution is the following: We note that is 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 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 , 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 . 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 in equation (9.1c) with the right-hand side of equation (9.1h). The result is


(9.3a)


We now let so that , which causes the two adjacent frequencies, and , differing by , to approach each other. In the limit becomes a continuous variable , becomes , the summation becomes an integral, and equation (9.3a) becomes


(9.3b)


It is convenient to represent the inner integral by a symbol and write equation (9.3b) as two equations:


(9.3c)


(9.3d)


The function [or ] is the Fourier transform of while is the inverse Fourier transform of [or ]. The relation between and can be indicated by a double arrow:


(9.3e)

The unit impulse , also called the Dirac delta, is by definition zero everywhere except at where it equals ; similarly, is zero except when , where it equals . In digital notation (see problem 9.4) we write , , 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


(9.3f)


where and 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


(9.3g)


The boxcar, written , is a function whose value is everywhere zero except in the interval from to , where its value is . A boxcar in the frequency domain is written or .


Solution

The transform of is given by equation (9.3c):


(9.3h)


9.3b Show that


(9.3i)
(9.3j)


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 , thus proving equation (9.3i). Likewise, replacing an element at time with 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 is , equations (9.3f, g) show that the transforms of 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



9.3c Show that a boxcar of height extending from to 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:


(9.3k)

where and sinc


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:


(9.3l)


where , , 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 (9.3) can be regarded as giving the result of two boxcars, one extending from to , the other extending from to , the effect of the latter being subtracted from that of the former.

To compare equation (9.3l) with equation (9.3k), we set . Then sinc , so sinc and becomes



Setting 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 to , 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 , where is an integer (often we omit 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


(9.4a)

The transform of comb(t) is another comb with impulses in the frequency domain at intervals :


(9.4b)


where [see Sheriff and Geldart, 1995, equation (15.155)].

Figure 9.4a.  Sampling and recovering a waveform. The left column is time domain, the right is frequency domain.

The Nyquist frequency, , is half the frequency of sampling , that is,


(9.4c)


While a signal that does not contain frequencies higher than will be recorded accurately, a frequency higher than 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)]:


(9.4d)


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 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.

Figure 9.4b.  Repeat of Figure 9.4a for a 125–250 Hz signal.
Figure 9.4c.  Repeat of Figure 9.4a for spectrum overlapping the Nyquist frequency.

If the signal is shifted upward by , where 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.

Figure 9.5a.  Seismic filter responses.

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, , 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 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, 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 , (b) by the near-surface region where absorption is especially large , (c) by absorption in passage through the main body of the earth , 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 (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 , and as the round-trip traveltime in the layer, being an integer and 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 and amplitude . 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 has amplitude . Reflections suffering three bounces arrive at with amplitude , and so on. The impulse response (see Sheriff and Geldart, 1955, 279) of the water layer is thus the infinite series


(9.7a)


(9.7b)

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 ; it satisfies the equation


(9.7c)


(9.7d)


(9.7e)

[See Sheriff and Geldart, 1995, section 15.5.3 for a discussion of -transforms.]

Solution

For , , and . We wish to verify that . Using the procedure described in problem 9.2, we replace each element of by a scaled version of ; thus

1, , +, , +
+, , +,
+, , +
= 1, 0, 0, 0, 0, … .

We can proceed in the same way for other values of .

9.7b The spectrum of the water-layer filter is shown in Figure 9.7a for ; the large peaks occur at the “singing frequency” , where is any odd number. Sketch the amplitude spectrum of the inverse filter.

Solution

In the time domain, ; in the frequency domain this becomes . Thus the amplitude spectrum of is the inverse of that of , that is, for a given frequency,

as shown in Figure 9.7b.

Figure 9.7a.  Spectrum of water-layer filter; water depth, .
Figure 9.7b.  Inverse filter spectrum (dashed).

9.7c Verify your sketch of the water-reverberation inverse filter by transforming


(9.7f)

and calculating the value of the spectrum for , , , and .

Solution

We start with the -transform of , 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 , the conjugate complex is and ; thus,

Substituting and using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get

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 , , , . Calculate the crosscorrelations and autocorrelations , , , , and 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 is shifted the amount relative to . The crosscorrelation is given by the equation


(9.8a)

This equation means that is displaced units to the left relative to , corresponding values multiplied, and the products summed to give . The result is the same if we move to the right units, that is,


(9.8b)

The functions and are closely related. Using two simple curves, it is easily shown that we can crosscorrelate by reversing and convolving the reversed function with , that is,


(9.8c)

(since convolution is commutative). Reversing a function in time changes the sign of , so the exponent of changes sign also, becoming and becoming the conjugate complex . The convolution theorem (equation (9.3f)) now becomes the crosscorrelation theorem:


(9.8d)

If is the same as , we get the autocorrelation of and equations (9.8a,d) become


(9.8e)

the transform relation being the autocorrelation theorem. Since the two functions are the same, 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;


(9.8f)

Both the autocorrelation and the crosscorrelation are often normalized; in the case of the autocorrelation, equation (9.8e) becomes


(9.8g)

The normalized crosscorrelation is


(9.8h)

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 . 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 , , . 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.

Table 9.9a. Digital wavelets and operations.
Table 9.9b. Completed table.
0 0 0 2 1 1 0 0
0 0 0 4 0 0
0 0 0 0 0 6 3 3
1/2 1/2 1 0 0 0 0 0
0 0 0 0 0
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
0 0 2 0 0 0 0
0 0 0 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


(9.10a)

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 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 and/or increase, the ratio increases; the first and second terms are equal when . If and differ significantly in magnitude, the second term can be larger than the first for large values of or ; e.g., if , the ratio is when .

If and 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 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 as the time when the first nonzero value occurs.

9.13 Phase of composite wavelets

Using the wavelets

calculate the composite wavelets:

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 , 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 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 and are the same and the other two differ only in signs:

Figure 9.13a.  The composite wavelets.

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.

Figure 9.14a.  Wavelets used: (i) Minimum-phase wavelet [parts (a) to (c)]; (ii) low-frequency minimum-phase wavelet (part d); (iii) zerophase wavelet (part e).

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 .

Table 9.14a. Calculating composite wavelets for shale/sand/shale sequences.
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.

Figure 9.14b.  Shale/sand/shale reflections.

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.

Table 9.14c. Calculation of composite reflection for two thin sands in shale.
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
Table 9.14b. Calculating composite wavelets for shale/sand/limestone sequence.
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).

Figure 9.14c.  Shale/sand/limestone reflections.

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).

Figure 9.14d.  Effect of thin sands.
Table 9.14d. Calculation of reflection for shale/sand/shale sequence and low-frequency wavelet.
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