User:Ageary/Chapter 7

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

7.1 Radiolocation errors because of velocity variations

Ground conductivity affects radio-wave velocity because of currents induced in the earth. Range calculations assuming that travel is over normal seawater thus may be in error if this is not true. What errors are involved for travelpaths over the terrains in Table 7.1a?

Table 7.1a. Radio-wave velocities over various terrains.

Terrain	Velocity (km/s)
Seawater	299 670
Freshwater	299 250
Farmland	299 400
Dry sand	299 900
Mountains	298 800

Background

Radio-navigation methods are used to determine distances in marine surveys and the velocity is generally assumed to be that over saltwater.

Solution

Let $V_{S}$ be velocity over normal seawater and $V_{T}$ velocity over other terrains. If we use $V_{S}$ to calculate ranges, the error in range calculations will be $\Delta x=t\left(V_{S}-V_{T}\right)$ and the relative error is $\Delta x/x=1000\left(1-V_{T}/V_{S}\right)$ m/km. When $V_{T}<V_{S}$, the error will be positive, meaning that the calculated range is too great. The results are shown in Table 7.1b.

Table 7.1b. Location errors for various terrains.

Terrain	Velocity (km/s)	Error (m/km)
Seawater	299 670	0.0
Freshwater	299 250	1.4
Farmland	299 400	0.9
Dry sand	299 900	–0.8
Mountains	298 800	2.9

7.2 Effect of station angle on location errors

If the error in Shoran time measurements is $\pm 0.1\mu s$, what is the the size of the parallelogram of error in Figure 7.2a when (a) $\theta =30^{\circ }$ and (b) $\theta =150^{\circ }$? Take the velocity of radio waves as $3\times 10^{5}$ km/s.

Background

Shoran is a radio-navigation device which measures the 2-way traveltime between the point of observation and a fixed station. Using two fixed stations, the point of observation can be located by swinging arcs centered at the two stations; for large distances the arcs become nearly straight lines.

The traveltimes are subject to error $\pm \Delta t$, so the ranges are $V\left(t_{i}\pm \Delta t\right)$, $i=1$, 2. Swinging the four arcs corresponding to these time values, we get a parallelogram of error such as that in Figure 7.2a; the location lies somewhere inside this parallelogram.

Solution

In Figure 7.2a, the error in range $=AM=AN=AP=AQ$

${\begin{aligned}=\left(3\times 10^{8}\ \mathrm {m/s} \right)\left(1\times 10^{-7}\ \mathrm {s} \right)=30\ \mathrm {m} .\end{aligned}}$

Figure 7.2a  Parallelogram of error for traveltime uncertainty of $\Delta t=\pm 0.1\mu s$

.

The long diagonal $=2AR=2AQ/\sin 15^{\circ }=2\times 30/\sin 15^{\circ }=230$ m.

The short diagonal $=2AS=2\times 30/\cos 15^{\circ }=60$ m.

To get the figure for $\theta =150^{\circ }$ we merely reverse the arrow on ${\textit {AB}}$ or ${\textit {AC}}$; therefore the error values are the same as for $30^{\circ }$.

7.3 Transit satellite navigation

7.3a Determine the acceleration of gravity at the orbit of a Transit satellite 1070 km above the Earth, knowing that ${\textit {g}}$ at the surface of the Earth is 9.81 m/s$^{2}$, and that the gravitational force varies inversely as the square of the distance between the centers of gravity of the masses. The radius of the Earth is 6370 km.

Background

A satellite is in a stable orbit around the Earth when the gravitational force $({\textit {mg}})$ pulling it earthward equals the centrifugal force $mV^{2}/R$, where $g$ is the acceleration of gravity, $m$ and $V$ the satellite’s mass and velocity, and $R$ the radius of its orbit about the center of the Earth.

Solution

The radius of the satellite’s orbit is $\left(6370+1070\right)=7440$ km. Since $g$ is proportional to the force of gravity, at the satellite’s orbit,

${\begin{aligned}g=9.81(6370/7440)^{2}=7.19\ \mathrm {m/s} ^{2}.\end{aligned}}$

7.3b What is the satellite’s velocity if its orbit is stable?

Solution

For a stable orbit, the gravitational acceleration is balanced by the centripetal acceleration.

Thus,

${\begin{aligned}V^{2}/R=g=7.19\ \mathrm {m/s} ^{2},\\V=(7.19\times 10^{-3}\times 7440)^{1/2}=7.31\ \mathrm {km/s} .\end{aligned}}$

7.3c How long does it take for one orbit?

Solution

The length of the nearly circular orbit is $2\pi \times 7440$ km, so the time for one orbit is

${\begin{aligned}T=2\pi \times 7440/7.31=6395\ \mathrm {s} =106\ \mathrm {minutes} ,35\ \mathrm {seconds} .\end{aligned}}$

7.3d How far away is the satellite when it first emerges over the horizon?

Solution

In Figure 7.3a, ${\textit {O}}$ is the point of observation. The satellite first becomes visible when it reaches the tangent to the Earth at ${\textit {O}}$. The tangent is normal to the radius at ${\textit {O}}$, so

${\begin{aligned}x=(7440^{2}-6370^{2})^{1/2}=3840\,{\rm {km.}}\end{aligned}}$

Figure 7.3a  First visibility of satellite.

7.3e What is the maximum time of visibility on a single satellite pass?

Solution

In Figure 7.3a the angle subtended at the center of the Earth by ${\textit {x}}$ is

${\begin{aligned}\theta =\cos ^{-1}\left(6370/7440\right)=31.1^{\circ }.\end{aligned}}$

The satellite is visible while it traverses an arc subtending $2\theta =62.2^{\circ }$. Since the entire orbit is traversed in 6395 s, the time of visibility is $6395\left(62.2/360\right)=1105\ \mathrm {s} =18$ minutes, 25 seconds.

7.4 Effective penetration of profiler sources

7.4a Sieck and Self (1977) summarize “acoustic systems” as shown in Table 7.4a. For each of these systems, calculate the wavelengths and penetration depths given by Denham’s (1982) rule, that the maximum useable frequency is $f_{\max }=150/t$, where $t$ = traveltime. Reconcile your results with the stated purposes.

Table 7.4a. Acoustic systems.

System	Frequency (kHz)	Purposes
Fathometers	12–80	To map water bottom
Water-column bubble detectors	3–12	To locate bubble clusters, schools of fish
Side-scan sonar	38–250	To map bottom irregularities
Tuned transducers	3.5–7.0	To penetrate 30 m
Imploders	0.8–5.0	To penetrate 120 m, find gas-charged zones
Sparkers	0.04–0.15	To map 1000

Background

Acoustic systems include several devices that use sound waves to measure distances in water. Transducers emit sound waves, usually short pulses, the travel-times of which are used to measure distances. Fathometers are transducers that emit and record high-frequency pulses (usually about 100 kHz); they determine water depth from the two-way traveltime of the sea-floor reflection. Fathometers have little penetration, but similar devices using frequencies in the range $1-10$ kHz may penetrate as much as 30 m.

Side-scan sonar utilizes a towed transducer that emits high-frequency pulses and measures the traveltime of energy back-scattered by irregularities on the bottom.

The sparker is an energy source using the discharge of a large capacitor to create an electric arc between two electrodes in water, the sudden vaporization of the water being equivalent to an explosion.

Imploders create voids in the water and the effect of water rushing into the voids is to generate seismic waves; an example is the water gun which has two chambers, one being filled with air at high pressure, the other containing water; release of the air into the water chamber forces the water out at high velocity, thus creating voids into which the surrounding water collapses.

Denham (1982) devised an empirical rule relating the maximum useable frequency to the two-way traveltime $t$: $f_{\max }=150/t$, the useful frequency cutoff being determined by the increased absorption of higher frequencies and the background noise level. Note that high-frequency loss in water is very small, so the time is that below the seafloor.

Solution

We assume a value 1500 m/s for the velocity of sound in water, so $\lambda =1500/f$. Taking $z$ as the depth of penetration, Denham’s rule gives $z=Vt/2=112\times 10^{3}/f_{\max }$ where $f_{\max }$ is in kilohertz. The results are given in Table 7.4b.

Table 7.4b. Calculated wavelengths and depths of penetration.

System	Frequency (kHz)	Wavelength (m)	Penetration depth (m)
Fathometers	12–80	0.12–0.019	9–1
Bubble detectors	3–12	0.50–0.12	37–9
Side-scan sonar	38–250	0.039–0.006	3–0.4
Tuned transducers	3.5–7.0	0.43–0.021	32–16
Imploders	0.8–5.0	1.9–0.30	140–22
Sparkers	0.04–0.15	38–10	2800–750

Maximum penetration is given by the energy that the systems inject into the earth, higher frequency systems generally giving less energy. The bubble detector might locate a cluster of bubbles or a school of fish, but certainly not individual bubbles or individual fish.

7.4b Trade literature claims 30-cm resolution with imploders and 2–5 resolution with sparkers. How do these figures compare with the resolvable limits?

Solution

The resolvable limit (see problem 6.18) is $\lambda /4$ and using the shorter wavelengths in Table 7.4b gives resolvable limits of 0.5 to 0.08 m for imploders and 9.5 to 2.5 m for sparkers. Thus the claims in the trade literature are reasonable.

7.5 Directivity of linear sources

7.5a In Figure 7.5a, a linear vertical source ${\textit {MN}}$ (such as a column of explosives) of length $a\lambda$, $a$ being a constant and $\lambda$ the wavelength, is activated at all points simultaneously at time $t=0$. Taking the initial waveform as $Ae^{{\rm {j}}\left(\kappa r-\omega t\right)}$, show that the effect at point ${\textit {P}}$ is

${\begin{aligned}h\left(t\right)=Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\ {\rm {sinc}}\ \left(\pi a\sin \theta _{0}\right).\end{aligned}}$

(7.5a)

What is the array response?

Background

A distributed source can be thought of as an array. The array response is the ratio of the output of an array to the output when all of the elements are concentrated at the midpoint of the array.

Detonating cord is an explosive cord with a constant velocity of detonation; it is used to connect two charges in order to delay detonation of the second charge. Detonation of the first charge initiates detonation in the cord, which in turn detonates the second charge. By varying the length of cord, detonation of the second charge can be delayed a desired amount.

Solution

Although the explosive is exploded instantaneously, energy from different parts of the column arrive at ${\textit {P}}$ at different times because they must travel different distances. Denoting $a\lambda /2$ by ${\textit {c}}$, the total effect at ${\textit {P}}$ at time ${\textit {t}}$ will be

Figure 7.5a  Geometry for linear source.

${\begin{aligned}h(t)=\int _{z_{0-C}}^{z_{0+C}}Ae^{{\rm {j}}\left(\kappa r-\omega t\right)}\,{\rm {d}}z=Ae^{-{\rm {j}}\omega t}\int _{z_{0-C}}^{z_{0+C}}e^{{\rm {j}}\kappa r}\,{\rm {d}}z.\end{aligned}}$

To integrate we must get a relation between ${\textit {r}}$ and ${\textit {z}}$. We assume that $r_{0}\gg a\lambda$; then

${\begin{aligned}r\approx r_{0}+\left(z-z_{0}\right)\sin \theta _{0}=r_{0}(1-\sin ^{2}\theta _{0})+z\sin \theta _{0}\\\approx r_{0}\cos ^{2}\theta _{0}+z\sin \theta _{0}.\end{aligned}}$

(7.5b)

Thus,

${\begin{aligned}h\left(t\right)=Ae^{-{\rm {j}}\omega t}\int _{z_{0-C}}^{z_{0}+c}e^{{\rm {j}}\kappa \left(r_{0}\cos ^{2}\theta _{0}+z\sin \theta _{0}\right)}{\rm {d}}z\\=Ae^{{\rm {j}}\left(\kappa r_{0}\cos ^{2}\theta _{0}-\omega t\right)}\int _{z_{0-C}}^{z_{0}+c}e^{{\rm {j}}\kappa z\sin \theta _{0}}{\rm {d}}z\\=Ae^{{\rm {j}}\left(\kappa r_{0}\cos ^{2}\theta _{0}-\omega t\right)}\left({\frac {e^{{\rm {j}}{\kappa }\sin \theta _{0}}}{{\rm {j}}\kappa \sin \theta _{0}}}\left|_{z_{0}-c_{0}}^{z_{0}+c_{0}}\right.\right).\end{aligned}}$

Noting that $z_{0}\sin \theta _{0}=r_{0}\sin ^{2}\theta _{0}$ and $\kappa c=\pi a$, this becomes

${\begin{aligned}h\left(t\right)=2Ace^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\left({\frac {\sin \left(\pi a\sin \theta _{0}\right)}{\pi a\sin \theta _{0}}}\right)\\=Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\ \mathrm {sinc} \ (\pi a\sin \theta _{0}),\end{aligned}}$

(7.5c)

where sinc $x=\left(\sin x\right)/x$.

If the linear source is replaced by a concentrated source of equal strength at the center ${\textit {O}}$, the effect at ${\textit {P}}$ would be $Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}$. Dividing the right-hand side of equation (7.5c) by this quantity, we get for the array response ${\textit {F}}$

${\begin{aligned}F={\rm {sinc}}\ \left(\pi a\sin \theta _{0}\right).\end{aligned}}$

(7.5d)

7.5b An explosion initiated at the top of the explosive column ${\textit {MN}}$ in Figure 7.5a travels down the column with velocity $V_{e}$. Show that the array response is

${\begin{aligned}F=-{\rm {sinc}}[\pi a(\sin \theta _{0}-V_{r}/V_{e})],\end{aligned}}$

(7.5e)

where $V_{r}$ is the velocity in the rocks. Under what circumstances does equation (7.5e) reduce to equation (7.5d)?

Solution

In part (a), the entire column ${\textit {MN}}$ exploded at $t=0$. We now consider the case where the explosion starts at point ${\textit {M}}$ and travels down the column with velocity $V_{e}$, that is, the explosion starts at $z=z_{0}+c$ at $t=0$, where $c=a\lambda _{e}/2$. Writing $\kappa _{e}=\omega /V_{e}$, $\kappa _{r}=\omega /V_{r}$, $\gamma =V_{e}/V_{r}$ the wave generated by the element dz arrives at ${\textit {P}}$ with the phase $\omega [\left(z_{0}+c-z\right)/V_{e}+(r/V_{r})]$ Using equation (7.5b) the phase becomes

${\begin{aligned}\omega \left\{\left(z_{0}+c\right)/V_{e}+\left(r_{0}\cos ^{2}\theta _{0}\right)/V_{r}+z\left[\left(\sin \theta _{0}\right)/V_{r}-1/V_{e}\right]\right\}\\=\kappa _{e}[z_{0}+c+\gamma r_{0}\cos ^{2}\theta _{0}+z(\gamma \sin \theta _{0}-1)].\end{aligned}}$

Assuming a harmonic wave function $\psi _{P}$, we can write

${\begin{aligned}\psi _{P}=AK\int _{z_{0}+c}^{z_{0-C}}e^{{\rm {j}}mz}{\rm {d}}z,\end{aligned}}$

where $A={}$ amplitude, $K=e^{{\rm {j}}\kappa _{e}\left[\left(Z_{0}+c\right)+\gamma r_{0}\cos ^{2}\theta _{0}\right]\}}$, $m=\kappa _{e}\left(\gamma \sin \theta _{0}-1\right)$. Integrating, we obtain

${\begin{aligned}\psi _{P}=\left(AK/{\rm {j}}m\right)e^{{\rm {j}}mz}\left|_{z_{0}+c}^{z_{0-C}}\right.\\=\left(AK/{\rm {j}}m\right)e^{({\rm {j}}mz_{0})}\left[e^{-{\rm {j}}mc}-e^{{\rm {j}}mc}\right]=-\left(2AK/m\right)e^{{\rm {j}}mz_{0}}\sin \left(mc\right)\\=-\left(2AKc\right)e^{{\rm {j}}mz_{0}}\ {\rm {sinc}}\left(mc\right)\\=-2Ac\exp \left\{{\rm {j}}\kappa _{e}\left[\left(r_{0}+c\right)+\gamma r_{0}\cos ^{2}\theta _{0}+r_{0}\left(\gamma \sin \theta _{0}-1\right)\right]\right\}{\rm {sinc}}(mc)\\=-2Ac\exp \left\{{\rm {j}}\kappa _{e}\left(\gamma r_{0}+c\right)\right\}{\rm {sinc}}(mc).\end{aligned}}$

If we locate the same amount of explosive at $z_{0}$ and explode it at $t=0$, we get at ${\textit {P}}$

${\begin{aligned}\psi _{P}^{*}=2Ace^{-{\rm {j}}\kappa _{r}r_{0}}=2Ace^{-{\rm {j}}\kappa _{e}\gamma r_{0}},\end{aligned}}$

hence the array response is

${\begin{aligned}F=\psi _{P}/\psi _{P}^{*}=-e^{{\rm {j}}\kappa _{e}c}{\rm {sinc}}(mc).\end{aligned}}$

Omitting the first factor, which is independent of $\theta _{0}$ and $r_{0}$ and hence is merely a scale factor, we have

${\begin{aligned}F=-{\rm {sinc}}[c\kappa _{e}(\gamma \sin \theta _{0}-1)].\end{aligned}}$

But

${\begin{aligned}ca=\left(a\lambda _{e}/2\right)\left(2\pi /\lambda _{e}\right)=\pi a,\end{aligned}}$

so

${\begin{aligned}F=-{\rm {sinc}}[(\pi a(\sin \theta _{0}-V_{r}/V_{e})].\end{aligned}}$

(7.5f)

[The minus sign for sinc occurs here and not in equation (7.5d) because the direction of integration is opposite to that assumed in deriving equation (7.5d)]. For an instantaneous explosion the result is given by equation (7.5d), namely

${\begin{aligned}F={\rm {sinc}}\left(\pi a\sin \theta _{0}\right).\end{aligned}}$

Equation (7.5c) reduces to equation (7.5d) whenever $\left(V_{r}/V_{e}\right)\ll \sin \theta _{0}$. For $\theta _{0}\approx 90^{\circ }$, i.e., for rays traveling almost vertically downward, the required condition is that $V_{r}\ll V_{e}$. For most explosives, $V_{e}\approx 6{-}7$ km/s, so $V_{r}$ should be no more than about 1.5 km/s, the velocity of water, for the two equations to give nearly the same result.

7.5c Calculate the array response ${\textit {F}}$ for a column 10 m long, given that $\lambda _{r}=40$ m, $V_{e}=5.5$ km/s, $V_{r}=2.1$ km/s, and $\theta _{0}=0^{\circ }$, $30^{\circ }$, $60^{\circ }$, $90^{\circ }$.

Solution

${\begin{aligned}a=10/40=0.25,\;\;V_{r}/V_{e}=2.1/5.5=0.38;\end{aligned}}$

thus

${\begin{aligned}F={\rm {sinc}}\left[\left(\pi /4\right)\left(\sin \theta _{0}-0.38\right)\right]={\rm {sinc}}\,x.\end{aligned}}$

Table 7.5a. Array response as a function of direction.

$\theta _{0}$	$x$	$\sin x$	$F$
$0^{\circ }$	$-0.30$	$-0.29$	0.99
30	0.094	0.094	1.00
60	0.38	0.37	0.97
90	0.49	0.47	0.98

Differences in directivity are negligible when the charge length is much smaller than the wavelength.

7.5d If the column in part (c) is replaced by six charges, each 60 cm long and equally spaced to give a total length of 10 m, the charges being connected by spirals of detonating cord with detonation velocity 6.2 km/s, what length of detonating cord must be used between adjacent charges to achieve maximum directivity downward?

Solution

Let ${\textit {L}}$ be the length of detonating cord between successive charges; maximum directivity downward is achieved when the traveltime through the explosive column is the same as that in the adjacent rocks. In part (c) we were given $V_{e}=5.5$ km/s, $V_{r}=2.1$ km/s, so $10/21=6\times 0.60/5.5+5L/6.2$, hence $L=5.1$ m.

7.5e What are the relative amplitudes (approximately) of the waves generated by the explosives in part (d) at angles $\theta _{0}=0^{\circ }$, $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$ when $\lambda =40$ m?

Solution

An approximate solution can be obtained by assuming that the average velocity $V_{e}$ is equal to $V_{r}$; this means that the traveltime down through the 10 m column of explosives is the same as that for a wave in the adjacent 10 m of rock. In this case, $a=10/40=1/4$, $V_{r}/V_{e}=1.0$ and

${\begin{aligned}F={\rm {sinc}}\left[\left(\pi /4\right)\left(\sin \theta _{0}-1.0\right)\right]={\rm {sinc}}\ y.\end{aligned}}$

Table 7.5b. Relative amplitudes.

$\theta _{0}$	$y$	$\sin$	$|F|$
$0^{\circ }$	–0.79	–0.71	0.90
30	–0.39	–0.38	0.97
60	–0.11	–0.11	1.00
90	0.00	0.00	1.00

7.6 Sosie method

7.6a Imagine an impulsive source striking the ground at times $n\Delta$ apart, where ${\textit {n}}$ is a random number between 10 and 20, and $\Delta$ is the sampling interval. Assume reflections with arrival times and amplitudes as follows:

${\begin{aligned}{\begin{array}{cccccccc}t&\to &0&5\Delta &13\Delta &29\Delta &33\Delta &42\Delta \\{\rm {Amp}}&\to &+5&+2&-1&+3&+1&-2.\end{array}}\end{aligned}}$

Each new impact generates the reflection sequence.

One wants the reflections for each impact to add constructively, so Sosie processing involves adding the geophone responses for several impacts after each response has been shifted so that $t=0$ is the same for the responses from all the impacts. Add the reflection sequences as would be done in Sosie recording for 10, 20, and 30 impulses to see how the signal builds up as the multiplicity increases.

Table 7.6a. Random impact time intervals.

0,	16,	14,	12,	17,	19,	12,	16,	15,	11,
14,	11,	13,	20,	14,	18,	16,	14,	17,	10;
18,	18,	13,	11,	15;	10,	19,	12,	20,	17.

Figure 7.6a  Illustrating the buildup of reflection signals in the Sosie method. From top to bottom: the signal and sums for the first (i) 10, (ii) 20, and (iii) 30 impacts.

Background

The Sosie method often uses impactors as seismic sources. Each impactor strikes the ground at random intervals 5 to 10 times per second for approximately 3 minutes, a total of 900–1800 impacts, the times between impacts being much shorter than the reflection wavetrain. Each impact generates the reflection sequence. The reflections from each impulse are much smaller than the noise but the Sosie process builds up the reflections systematically while the noise partially cancels.

In digital recording, an analog signal is measured at fixed intervals $\Delta$, called the sampling interval; $\Delta$ is usually 1 or 2 ms. Table 6.22a is a table of random numbers.

Solution

We obtained 29 random numbers in the 10–20 range by reading systematically from Table 6.22a, using even or odd values to determine the tens place and every 4th digit to give the units place. We used these numbers as the time intervals between 30 successive impacts. The impact times we used are listed in Table 7.6a.

The time required for 30 impacts separated by about $15\Delta$ plus the record length means that we have about 500 samples. We have made the impact intervals about 1/3 of the length of the reflection wavetrain.

Table 7.6b lists for the first ten impulses the signal sequences starting at each impulse time, each sequence having length $44\Delta$, hence they overlap. $A\left(t\right)$ is the sum of the signal values occurring at time ${\textit {t}}$. Toward the right, subsets of $A\left(t\right)$ are listed, each being $44\Delta$ long, each subset constituting noise for the other subsets. Finally the subsets are summed. Figure 7.6a shows the reflection wavetrain and the result of summing the first 10, 20, and 30 impulses. (The data for the impulses 11–30 are not listed.)

Note the result for the two smallest reflections with amplitudes $\pm 1$, that is, the 3rd (negative) reflection at $t=13$ and the 5th at $t=33$; the 3rd shows up clearly, but the 5th is no larger than the “noise,” which in this exercise is that caused by the overlapping signals. The noise is attenuated relative to the reflections as more sequences are added.

Table 7.6b  Calculations for signal buildup in Sosie recording; no random noise added.

Figure 7.6b  Same as Figure 7.6a except for added random noise.

7.6b In part (a), we examined how the reflection sequence builds up. We now examine the effect of random noise by adding random noise values between $\pm 6$ to the signal. We obtain these from a random-number table by some scheme, such as systematically selecting pairs of numbers and changing the first member of each pair to a minus sign whenever it is odd and to a plus sign whenever it is even. The signal sequence has an rms value of 2.7, so adding noise values within the range $\pm 6$ (rms value 3.9) gives a signal/noise ratio of about 0.7.

Solution

The first page of our worksheet is shown as Table 7.6c. The first column is the sample number ${\textit {t}}$, the 2nd is the signal sequence from Tables 7.6a,b, the 3rd is the added noise, and the fourth the sum of the signal and noise. The signal plus noise subsets, each starting at the time of a source impact, are listed to the right followed by their sum.

The first 45 members of 30 subsets, each starting at one of the impulses, is shown in Table 7.6d. Summations for 10, 20, and 30 of the subsets are shown in Table 7.6e and in Figure 7.6b.

Examination of the numbers in the signal + noise columns in Table 7.6c indicates the difficulty of finding any of the signals. Table 7.6e lists the sums of the first subset, the first two subsets, then the first 4, 10, 20, and 30 subsets, with the times of the signals underlined. The first member of the signal sequence, which considerably exceeds the average noise, becomes distinguishable with the addition of only a few subsets, whereas the two smallest signal members (at times 13 and 33) still do not stand out even after summing 30 subsets. The broad, smeared-out peak between times 14 and 18, which corresponds roughly to the repetition rate of the impulses, would probably be erroneously interpreted as the interference of several reflections.

Table 7.6c  Calculations for signal buildup in Sosie recording; signal/noise $\approx 0.7$

Table 7.6c  Calculations for signal buildup in Sosie recording; signal/noise $\approx 0.7$ (cont.)

Table 7.6d  First members of 30 subsets; signal/noise $\approx 0.7$

Table 7.6d  First members of 30 subsets; signal/noise $\approx 0.7$ (cont.)

Table 7.6e  First members in sums for various numbers of impacts; $S/N\approx 0.7.$

7.7 Energy from an air-gun array

How much energy is released (approximately) by the air-gun array in Figure 7.7a when the input pressure is 2000 psi (14 MPa)? Assume that the change is adiabatic, that is, ${\mathcal {PV}}^{1.4}=$ constant, where ${\mathcal {P}}$, ${\mathcal {V}}$ are pressure and volume, that the final pressure is 2 atmospheres, and that the guns are far enough apart that they do not interact.

Background

An air gun consists of two chambers, both filled with air at high pressure. The two chambers are connected by a shuttle that is held in a closed position. When the restraining force is suddenly diminished the shuttle moves, allowing the air to vent into the water, creating the effect of an explosion. The energy release depends upon the change in air pressure and the volume of the chambers that discharge air into the water; the latter is usually given in cubic inches.

Solution

Energy released $=$ work done by expanding gas $=W=\int _{{\mathcal {V}}_{0}}^{{\mathcal {V}}_{1}}{\mathcal {P}}{\rm {d}}{\mathcal {V}}$. For an adiabatic change, ${\mathcal {PV}}^{1.4}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}$ or ${\mathcal {P}}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}/{\mathcal {V}}^{1.4}=\kappa /{\mathcal {V}}^{1.4}$. Thus,

${\begin{aligned}W=\kappa \int _{{\mathcal {V}}_{0}}^{{\mathcal {V}}_{1}}{\rm {d}}{\mathcal {V}}/{\mathcal {V}}^{1.4}=-\left(\kappa /0.4\right)\left(1/{\mathcal {V}}_{1}^{0.4}-1/{\mathcal {V}}_{0}^{0.4}\right)\\=2.5\left({\mathcal {P}}_{0}{\mathcal {V}}_{0}-{\mathcal {P}}_{1}{\mathcal {V}}_{1}\right)\quad \left({\rm {since}}\quad \kappa ={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}={\mathcal {P}}_{1}{\mathcal {V}}_{1}^{1.4}\right).\end{aligned}}$

We have

${\begin{aligned}{\mathcal {P}}_{0}=14\,{\rm {MPa}}=14\times 10^{6}\ \mathrm {N/m} ^{2},\\{\mathcal {P}}_{1}=2\,{\rm {atm.}}=2/\left(9.87\times 10^{-6}\right)\ {\rm {Pa}}=2.0\times 10^{5}\ {\rm {Pa}}=0.20\times 10^{6}\ {\rm {N/m}}^{2},\\{\mathcal {V}}_{0}=\left(305+200+125+95+75+60+50\right)\ {\rm {in}}^{3}=910\ {\rm {in}}^{3}\\=910\times (2.54\ {\rm {cm/in}})^{3}=1.49\times 10^{4}\ {\rm {cm}}^{3}=0.0149\ {\rm {m}}^{3}.\end{aligned}}$

Since ${\mathcal {P}}_{1}{\mathcal {V}}_{1}^{1.4}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}$, ${\mathcal {V}}_{1}={\mathcal {V}}_{0}({\mathcal {P}}_{0}/{\mathcal {P}}_{1})^{1/1.4},$

${\begin{aligned}V_{1}=0.0149\times (14\times 10^{6}/0.20\times 10^{6})^{0.71}=0.0149\times 70^{0.71}=0.30\ {\rm {m}}^{3}.\end{aligned}}$

Finally, $W=2.5\left(14\times 10^{6}\times 0.0149-0.20\times 10^{6}\times 0.30\right)=0.15\times 10^{6}$ J.

7.8 Dominant frequencies of marine sources

The dominant period of a marine seismic source is often determined by the source depth; this is the case when the second half-cycle of the downgoing wave is reinforced

Figure 7.7a  An air-gun array.

by the ghost reflection at the surface. Assuming that this is true for the source signatures in Figure 7.8a, determine their depths.

Background

For discussion of reinforcement depth in marine recording, see problem 3.5. The ghost is the wave reflected at the surface (or at the base of the LVL) with a reversal of phase (see problem 3.8).

When a gas at high pressure is suddenly generated by an underwater explosion or injected into the water by an air gun at sufficient depth that the gas does not quickly escape into the air, oscillation occurs as the gas expands until stopped by the water pressure, then contracts until the gas pressure becomes very high,

Figure 7.8a  Far-field waveshapes of marine sources. (a) Single 120 in3 air gun; (b) array of air guns of different sizes selected to attenuate bubble effects by destructive interference; (c) sleeve exploder; (d) Vaporchoc; (e) Maxipulse; (f) Flexichoc; (g) water gun; (h) 5 kJ sparker. Curves are intended to show features of waveshape, not amplitude relationships. ${\textit {B}}$ indicates bubble effects; the interval between successive bubbles decreases with time; I indicates implosion (after Sheriff and Geldart, 1995)

Table 7.8a. Source depth from frequency.

	$T$ (ms)	$d$ (m)
a) air gun	22	8
b) air-gun array	24	9
c) sleeve exploder	26	10
d) Vaporchoc	18	7
e) Maxipulse	24	9
f) Flexichoc	28	11
g) water gun	22	8
h) sparker	4	2

after which it expands again and the cycle is repeated. Each expansion is effectively a new activation of the source and so a new reflection sequence is generated. This phenomenon is called the bubble effect.

Air guns are described in problem 7.7, imploders, water guns, and sparkers in problem 7.4. Flexichoc$^{\rm {TM}}$, sleeve gun, Vaporchoc$^{\rm {TM}}$, and Maxipulse$^{\rm {TM}}$ are devices (now obsolete) designed to diminish the bubble oscillation. Flexichoc used the explosion of a small charge at the center of a steel cage that attenuated the bubble, the sleeve gun utilized the explosion of a propane-oxygen mixture inside a flexible chamber with the gases vented to the surface, Vaporchoc injected superheated steam into the water, and Maxipulse recorded the bubble oscillation for later processing to remove the bubble effects.

Solution

The phase of a ghost is reversed by reflection at the surface, hence to reinforce the downgoing wavelet the ghost must return to the source at time $T/2$, where ${\textit {T}}$ is the period measured in Figure 7.8a. Therefore the depth of the source $d=V_{w}T/4=380T$ m, taking $V_{w}={}$ velocity in water ${}=1500$ m/s. The results are shown in Table 7.8a.

7.9 Effect of coil inductance on geophone equation

If we wish to take into account the small inductance term $L\ {\rm {d}}^{3}i/{\rm {d}}t^{3}$ in the geophone equation (7.9a), show that, for a harmonic wave, it can be included approximately in the term involving the damping factor $h$ in equation (7.9b).

Background

The moving-coil electromagnetic geophone is widely used for land seismic work. The basic elements are a case with attached magnet and a coil suspended in the magnetic field by means of springs in such a way that, when an arriving seismic wave moves the case vertically, the inertia of the coil causes it to remain relatively stationary, that is, it moves in a different manner from that of the case; the relative motion between the coil and the magnetic field induces a voltage in the coil that is a replica of the seismic signal that caused the ground motion.

$^{\rm {TM}}$ Flexichoc and Vaporchoc are trademarks of Compagnie Géneralé de Géophysique; Maxipulse is a trademark of Western Geophysical Company.

Let $m$, $r$, $n$, and $i$ be the mass, radius, number of turns, and current in the coil, $R$ and $L$ the resistance and inductance of the coil; $H$ is the magnetic field strength, $\tau$ the mechanical damping factor (the damping force being $\tau \times$ vertical velocity of the coil relative to $H$), $K=2\pi nrH=$ electromagnetic force on the coil per unit current, $z$ the vertical displacement of the geophone case. Using these symbols, the geophone equation (see Sheriff and Geldart, 1995, 218–20) is

${\begin{aligned}L{\frac {{\rm {d}}^{3}i}{{\rm {d}}t^{3}}}+\left(R+{\frac {L\tau }{m}}\right){\frac {{\rm {d}}^{2}i}{{\rm {d}}t^{2}}}+\left({\frac {SL+\tau R+K^{2}}{m}}\right){\frac {{\rm {d}}i}{{\rm {d}}t}}+\left({\frac {SR}{m}}\right)i=K{\frac {{\rm {d}}^{3_{Z}}}{{\rm {d}}t^{3}}}\end{aligned}}$

(7.9a)

It is usually assumed that $L$ is small enough that it can be set equal to zero. In this case the geophone equation reduces to

${\begin{aligned}{\frac {{\rm {d}}^{2}i}{{\rm {d}}t^{2}}}+2h\omega _{0}{\frac {{\rm {d}}i}{{\rm {d}}t}}+\omega _{0}^{2}i=\left({\frac {K}{R}}\right){\frac {{\rm {d}}^{3}i}{{\rm {d}}t^{3}}},\end{aligned}}$

(7.9b)

where $\omega _{0}={\sqrt {S/m}}=2\pi f_{0}={}$ natural frequency of oscillation of the coil and

${\begin{aligned}2h\omega _{0}=\left({\frac {\tau }{m}}+{\frac {K^{2}}{mR}}\right).\end{aligned}}$

The quantity ${\textit {h}}$ is called the damping factor for the following reason. If $h=0$ in equation (7.9b), the equation becomes that for undamped simple harmonic motion; if the geophone is at rest and the coil is set in motion, it oscillates forever (theoretically) with undiminished amplitude. However, if $h\neq 0$, the amplitude decreases steadily by a fixed proportion each cycle; this decrease in amplitude is called damping [see also equation (2.18b)].

Solution

If in equation (7.9a) we set $L=0$ in all terms except the first, we get equation (7.9b) plus the additional term $L\left({\rm {d}}^{3}i/{\rm {d}}t^{3}\right)$. Assuming input of the form $i=i_{0}\cos \omega t$, we have ${\rm {d}}i/{\rm {d}}t=-\omega \sin \omega t$, ${\rm {d}}^{3}i/{\rm {d}}t^{3}=\omega ^{3}\sin \omega t$ hence ${\rm {d}}^{3}i/{\rm {d}}t^{3}=\omega ^{2}{\rm {d}}i/{\rm {d}}t$. Therefore we can take the term $L{\rm {d}}^{3}i/{\rm {d}}t^{3}$ in equation (7.9b) into account by changing the coefficient of ${\rm {d}}i/{\rm {d}}t$ to $(2h\omega _{0}+\omega ^{2}L)$.

7.10 Streamer feathering due to cross-currents

7.10a A 96-channel streamer with 25-m groups has the hydrophones spaced uniformly throughout its length. The lead-in and compliant sections together are 200 m in length and the tail section and buoy connection are 150 m long. Assume a source 100 m behind the ship, ship’s speed of 5.8 knots, and a 1.9 knot current perpendicular to the direction of travel.

What are the perpendicular and inline components of the distance to the farthest active group center with respect to the traverse direction?

Figure 7.10a  Marine streamer.

Background

Marine seismic work usually employs hydrophones to receive the seismic signals. These consist of piezoelectric crystals which generate a voltage difference between opposite faces when subjected to a pressure. The hydrophones are enclosed in a neutrally buoyant tube called a streamer (Figure 7.10a) which is towed by the ship. A number of hydrophones, spread over as much as 25 m, are connected together to form each group. Lead-in and compliant sections are inserted between the ship and the near end of the streamer; these increase the distance between the ship and the cable and reduce the effects of sudden changes which might jerk or break the streamer. A location sensor and radar target are usually placed on a tail buoy so that the end of the streamer can be located and the amount of feathering (sideways drift of the streamer) determined.

Common-midpoint (CMP) recording is discussed in problem 5.12.

Solution

We assume that the source is in the line between the ship and the streamer. We take the origin at the source and the $x$-axis along the line of traverse. With 200 m lead-in and compliant sections and active streamer length of $96\times 25=2400$ m, the farthest group center is $\left(200+2400-12\right)=2590$ m from the ship (assuming distances are to the group centers). The last group has a velocity of 5.8 knots along the $x$-axis and 1.9 knots along the $y$-axis, hence the streamer will be at the angle $\tan ^{-1}\left(1.9/5.8\right)=18^{\circ }$ (see Figure 7.10b). The coordinates of this group center with respect to the source are $x=2490\cos 18^{\circ }=2370$ m, $y=2490\sin 18^{\circ }=770$ m.

7.10b If the average velocity to a reflector 2.00 km below the source is 3.00 km/s and if the reflector dips $20^{\circ }$ perpendicular to the traverse direction, (i) by how much will the arrival time be changed for the far trace, and (ii) if this should be attributed to a change in velocity rather than cross-dip, what velocity would it imply?

Figure 7.10b  Drift of streamer in crosscurrent.

Solution

(i) We shall use a derivation similar to that in problem 4.2g except that the problem is now three dimensional, so we take the receiver coordinates as $\left(x,y,0\right)$. Because the dip is along the $y$-axis, the coordinates of the image point $I$ are $(0,2h\sin \xi ,2h\cos \xi )$, $h$ being the slant depth (2.00 km). We now have

${\begin{aligned}(Vt)^{2}=x^{2}+(y\pm 2h\sin \xi )^{2}+(2h\cos \xi )^{2}\\=x^{2}+y^{2}+4h^{2}\pm 4hy\sin \xi ,\end{aligned}}$

(7.10a)

the plus sign for the last term being used when the dip and feathering are in the same direction.

When the streamer is tracking behind the boat, $y=0$ and equation (7.10a) gives

${\begin{aligned}t=\left(1/3.00\right)(2.49^{2}+4^{2})^{1/2}=22.1^{1/2}/3.00=1.571\ \mathrm {s} .\end{aligned}}$

(7.10b)

When the feathering is in the downdip direction, $x=2.37$ km, $y=0.77$ km, and

${\begin{aligned}t=\left(1/3.00\right)(2.37^{2}+0.77^{2}+4^{2}+2\times 4\times 0.77\times \sin 20^{\circ })^{1/2}\\=24.32^{1/2}/3.00=1.644\ \mathrm {s} .\end{aligned}}$

(7.10c)

When the streamer is drifting updip, the sign of the $\sin \xi$-term is reversed, so

${\begin{aligned}t=\left(1/3.00\right)(2.37^{2}+0.77^{2}+4^{2}-2\times 4\times 0.77\times \sin 20^{\circ })^{1/2}\\=20.103^{1/2}/3.00=1.495\ \mathrm {s} .\end{aligned}}$

(7.10d)

(ii) At the source we observe the traveltime

${\begin{aligned}t_{0}=2h/V=4.00/3.00=1.333\ \mathrm {s} .\end{aligned}}$

(7.10e)

If we attribute the difference in traveltimes because of dip to a velocity change rather than to dip, we can determine the velocity as follows:

${\begin{aligned}(Vt)^{2}=x^{2}+y^{2}+4h^{2}=x^{2}+y^{2}+(Vt_{0})^{2},\end{aligned}}$

${\begin{aligned}{\text{so}}\qquad V^{2}\left(t^{2}-t_{0}^{2}\right)=x^{2}+y^{2}=\left(2.37^{2}+0.77^{2}\right)=6.21\ \mathrm {km} ^{2},\end{aligned}}$

${\begin{aligned}{\text{and}}\qquad V=2.49/(t^{2}-1.33^{2})^{1/2}\ \mathrm {km/s} .\end{aligned}}$

When the streamer is feathering in the down-dip direction,

${\begin{aligned}V=2.49/(1.64^{2}-1.33^{2})^{1/2}=2.49/0.96=2.57\ \mathrm {km/s} .\end{aligned}}$

For updip feathering,

${\begin{aligned}V=2.49/(1.45^{2}-1.33^{2})^{1/2}=2.49/0.694=3.56\ \mathrm {km/s} .\end{aligned}}$

Compared with the actual value of 3.00 km/s, these results are in error by 14% and 19% for the downdip and updip cases, respectively.

7.10c Assume that the amount of sideways drift of the streamer is ascertained by radar sighting on the tail buoy with an accuracy of only $\pm 3^{\circ }$; (i) how much change will this produce in locating the far group? (ii) How much change in arrival time will be associated with this uncertainty?

Solution

i) The radar sighting gives the angle of feathering $\theta$. The center of the last group is 2590 from the ship, so taking an origin at the ship, the $x'$ and $y'$ coordinates of the group are $x'=2590\cos \theta$, $y'=2590\sin \theta$; $\theta =18^{\circ }$. The errors in $x'$ and $y'$ due to $\Delta \theta =3^{\theta }=0.052$ radians are

${\begin{aligned}\Delta x'=\Delta x=-2590\sin \theta \Delta \theta =-\left(2590\times \sin 18^{\circ }\right)\times 0.052=-41\ \mathrm {m} ,\\\Delta y'=\Delta y=2590\cos \theta \Delta \theta =\pm 128\ \mathrm {m} .\end{aligned}}$

ii) When $\Delta \theta$ is positive, i.e., $\theta +\Delta \theta =21^{\circ }$, $\Delta x=-42$ m, $\Delta y=+128$ m. When $\Delta \theta$ is negative, the sign of $\Delta y$ is reversed. We use equation (7.10a) to find the traveltimes, that is,

${\begin{aligned}(Vt)^{2}=x^{2}+y^{2}+4h^{2}\pm 4hy\sin \xi ,\end{aligned}}$

where $x=2.49$ km, $y=0.77$ km, $h=2.00$ km, $V=3.00$ km/s, $\xi =21^{\circ }$, and we use the plus sign for the $\xi$-term when the cable is drifting downdip, the minus sign when the drift is updip. We have four cases to consider: the drift is down- or updip, and $\Delta \theta$ is $+3^{\circ }$ or $-3^{\circ }$.

Drift downdip

${\begin{aligned}\Delta \theta =+3^{\circ }:t=0.333[(2.49-0.04)^{2}+(0.77+0.13)^{2}\\\qquad \qquad +16+8\times \left(0.77+0.13\right)\sin 21^{\circ })]^{1/2}=1.676\ \mathrm {s} .\\\Delta \theta =-3^{\circ },t=0.333[2.45^{2}+(0.77-0.13)^{2}+16\\\qquad \qquad +8\times 0.64\sin 15^{\circ }]^{1/2}=1.639\ \mathrm {s} .\end{aligned}}$

Drift updip

${\begin{aligned}\Delta \theta =+3^{\circ },t=0.333(2.45^{2}+0.64^{2}+16-8\times 0.64\sin 20^{\circ })^{1/2}=1.515\ \mathrm {s} ,\\\Delta \theta =-3^{\circ },t=0.333(2.45^{2}+0.90^{2}+16-8\times 0.90\sin 20^{\circ })^{1/2}=1.504\ \mathrm {s} .\end{aligned}}$

7.10d The midpoint traces that are to be combined in a CMP stack will be distributed over what distance?

Solution

We determine the shifts from the line of traverse of groups #96 and #1, subtract these shifts, then divide by 2 to get the shifts of the midpoint. Thus, the shift relative to the ship of group center #96 is

${\begin{aligned}\Delta x=2590\left(\cos 18^{\circ }-1\right)=-129\ \mathrm {m} ,\qquad \Delta y=2590\sin 18^{\circ }=800\ \mathrm {m} .\end{aligned}}$

For group #1,

${\begin{aligned}\Delta x=212\left(\cos 18^{\circ }-1\right)=-10\ \mathrm {m} ,\qquad \Delta y=212\sin 18^{\circ }=66\ \mathrm {m} .\end{aligned}}$

The midpoints are shifted by half of these amounts, so the spread of the midpoints is: $\Delta x=\left(-128+10\right)/2=-59$ m, $\Delta y=\left(800-66\right)/2=367$ m.

7.11 Filtering effect of geophones and amplifiers

Use Figures 7.11a and 7.11b to determine the filter equivalent of a geophone with $f_{0}=10$ Hz and $h=0.7$, feeding into an amplifier with a 10–70 Hz bandpass filter and a 4-ms alias filter.

Background

A filter, whether analog or digital, is a device that attenuates certain ranges of frequencies present in a signal. A geophone is equivalent to a filter because the response determined by equation (7.9b) is frequency dependent; when the input is harmonic such that the vertical velocity of the geophone is ${\rm {d}}z/{\rm {d}}t=V_{0}\cos 2\pi ft$, the solution is [see Sheriff and Geldart, 1995, 220, equation (7.20)]

${\begin{aligned}i=\left(V_{0}/Z\right)\cos \left(2\pi ft-\gamma \right),\end{aligned}}$

(7.11a)

where the impedance $Z$ and the phase shift $\gamma$ are both functions of $\left(f/f_{0}\right)$, $f_{0}$ being the natural frequency of the geophone (see problem 7.9) and $f$ the frequency of the ground motion.

Figure 7.11a  Geophone frequency response versus damping factor $h$ (after Dennison, 1953).

Figure 7.11b  Seismic filter response.

The geophone sensitivity $\Gamma$ is a measure of the geophone output for a given ground velocity; it is defined by the relation

${\begin{aligned}\Gamma ={\hbox{(amplitude of output voltage)/(amplitude of ground velocity}}V_{0}).\end{aligned}}$

Because the numerator is proportional to $V_{0}$, $\Gamma$ is independent of $V_{0}$ and depends only upon the properties of the geophone and the ratio $\left(f/f_{0}\right)$ (see Sheriff and Geldart, 1995, 220, for more details).

Figure 7.11a shows $\Gamma$ (rationalized) as a function of $\left(f/f_{0}\right)$ for various values of the damping factor $h$ (see problem 7.9).

Seismic amplifiers include various types of filters. Band-pass filters pass a band of frequencies and discriminate sharply against frequencies outside the band, as shown in Figure 7.11b; the limits of the passband are usually taken as the frequencies at which the attenuation is 3 dB. Alias filters have a very steep high-frequency cutoff and are used to attenuate alias frequencies (see problem 9.4).

Solution

In Table 7.11a the column headed $\Gamma$ gives values of the geophone sensitivity for $h=0.7$ taken from Figure 7.11a; since the sensitivity is a ratio of amplitudes, we change the values to decibels $\left(=20\log _{10}\Gamma \right)$ in the 4th column. We obtain attenuation of the band-pass filter for the normalized frequencies 0.4, 0.6, etc., from Figure 7.11b using the 10-Hz low-cut and 70-Hz high-cut curves. The alias filter for 4-ms sampling rate is used to obtain the 6th column. The sum of the three attenuations is plotted in Figure 7.11c.

Table 7.11a. Combined filtering of geophone and amplifier.

Geophone				Amplifier filter
$f$(Hz)	$f/f_{0}$	$\Gamma$	$\Gamma$ (dB)	10–70 (dB)	alias (dB)	Sum (dB)
4	0.4	0.17	–15	–33	0	–48
6	0.6	0.35	–9	–19	0	–28
8	0.8	0.55	–5	–11	0	–16
10	1.0	0.70	–3	–6	0	–9
15	1.5	0.90	–1	–2	0	–3
20	2.0	0.98	0	0	0	0
40	4.0	1.00	0	0	0	0
60	6.0	1.00	0	–4	–1	–5
80	8.0	1.00	0	–8	–8	–16
100	10.0	1.00	0	–14	–30	–44

Figure 7.11c  Combined response of geophone and amplifier filters.

7.12 Filter effects on waveshape

Figure 7.12a illustrates filter effects. Evaluate the characteristics of (a) low-frequency cut, (b) high-frequency cut, (c) bandwidth, and (d) filter slope on (i) time delay to a point that could be timed reliably, (ii) apparent polarity, and (iii) ringing. The conclusions can be generalized for filters of other design types.

Background

Minimum phase is discussed in Sheriff and Geldart, 1995, Sections 9.4 and 15.5.6.

Solution

The first three columns in Figure 7.12a illustrate the effect of increasing the low-frequency cut and the last three columns do the same for the high-frequency cut. Broadly speaking, (i) as low frequencies are increasingly removed, the wavelet becomes more ringy (oscillatory) and energy shifts later in the wavelet; (ii) as high frequencies are removed, the pulse is lengthened without adding more cycles; (iii) as the filter slope increases, the energy is pushed later in the wavelet, resulting in pulse shape changes as evidenced by decreased ratios of first trough to first peak amplitudes.

Figure 7.12a  Impulse response of minimum-phase filters. In the last column, $\infty$ means that high-cut filters were not used (“out”). The filter passbands are given above.

1. Importance of low-frequency cut on
   1. time-delay: the backward drift of energy means that the first pickable point may be shifted to later time;
   2. apparent polarity: the backward energy drift also means a weaker first half-cycle, increasing the chance of missing it and picking the next half-cycle with reversed polarity;
   3. ringing: increases ringing.
2. Importance of high-frequency cut on
   1. time delay: lengthening the pulse delays the first pickable point;
   2. apparent polarity: no appreciable effect;
   3. ringing: no appreciable effect.
3. Importance of bandwidth: the 6–60 Hz passband is roughly 3.5 octaves, the 6–30 Hz is 2.5 octaves, the 18–60 Hz is 1.5 octaves;
   1. time delay: with narrower bandwidth, energy shifts backward within the wavelet, making it more difficult to pick early cycles, thus increasing time delays;
   2. apparent polarity: the same effect increases the likelihood of mistaking wavelet polarity;
   3. ringing: ringing increases with narrower bandwidth.
4. Importance of filter slope on
   1. time delay: increasing slope increases the delay;
   2. polarity: since the ratio of amplitudes of the first peak to the first trough changes, apparent polarity can be affected;
   3. ringing: increases ringing.

7.13 Effect of filtering on event picking

Figure 7.13a shows changes in wave-shape produced by the analog filtering in modern digital instruments. What can you conclude about the effects on picking?

Solution

The filtering on traces (b) and (c) is normal alias filtering for 2 ms and 4 ms sampling. Comparison of traces (a), (b), and (c) shows that lowering the high-frequency cut increases the delay. Comparing (c) and (d), we see that increasing the filter slope also increases the delay. Comparing traces (e) and (f) shows that changing the high-frequency slope changes the pulse shape but has little effect on the delay. Changing the low-frequency cut has a similar effect as seen by comparing (c) and (g). Narrowing the bandwidth results in shifting energy backward within the wavelet, decreasing the ability to pick early within the wavelet and increasing the likelihood of misinterpreting polarity.

Figure 7.13a  Far-field air-gun signals through various instrument filters: (a) no extra filtering; (b) out-24 Hz, 72 dB/octave; (c) out-62 Hz, 72 dB/octave; (d) out-62 Hz, 18 dB/octave; (e) 8–124 Hz with slopes of 18 and 72 dB/octave, respectively; (f) 18–124 Hz with 18 and 36 dB/octave slopes; (g) 8–62 Hz with 36 and 72 dB/octave slopes. Timing marks are 10 ms apart.

7.14 Binary numbers

1. Express the decimal numbers 19 and 10 as binary numbers.
2. Add the binary numbers together and convert the sum to a decimal number.
3. Multiply the two binary numbers and convert the product to decimal form.

Background

Mathematical operations are carried out in binary arithmetic in the same way as in decimal arithmetic. As an example of the similarity between decimal and binary arithmetic, when we add 75 to 129, we add 9 to 5 in the right-hand column and get 4 plus 1 to be carried to the next column. In binary addition, we have only the digits 1 and 0; when we add 1 to 1 we get 0 plus 1 to carry to the next column. In subtraction, when we subtract 1 from 0 we must borrow 1 from the next nonzero column to the left. Multiplication is identical in the two systems. To express a decimal number as a binary number, we subtract the largest power of 2 that is less than the decimal number, then follow the same procedure with the remainders until we end up with 1 or 0.

Solution

1. $19=2^{4}+2^{1}+2^{0}=10011;10=2^{3}+2^{1}=1010.$
2. $10011+1010=11101=2^{4}+2^{3}+2^{2}+2^{0}=16+8+4+1=29.$
3. $10011\times 1010=$ ${\begin{array}{c}10\,011\\1\,010\\\hline 100\,110\\10\,011\,0\\\hline 10\,111\,110=2^{7}+2^{5}+2^{4}+2^{3}+2^{2}+2^{1}+0\\\qquad =128+32+16+8+4+2\\=190.\end{array}}$