# Tuning and waveshape

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.14a

The following wavelet (upper curve in Figure 9.14a) is approximately minimum-phase: ${\displaystyle \left[11,\;14,\;5,\;-10,\;-12,\;-6,\;3,\;5,\;2,\;0,\;-1,\;-1,\;0\right]}$, the sampling interval being 2 ms. Use ${\displaystyle V_{sd}=2.0}$ km/s for the velocity in sand, ${\displaystyle V_{sh}=1.5}$ km/s for the velocity in shale, and the shale-to-sand reflection coefficient ${\displaystyle =0.1}$ (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 ${\displaystyle \lambda /4}$ when reflection coefficients are of opposite polarity. The opposite effect of lowering the amplitude (detuning) occurs at a thickness of ${\displaystyle \lambda /4}$ if the reflections from top and base of the bed have the same polarity.

### Solution

Because ${\displaystyle R=+0.1}$ 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 ${\displaystyle R^{2}=1\%}$ 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, ${\displaystyle \tau =2z/V_{sd}=2z/2000=z}$ ms. Thus the reflection from the base is delayed by one time sample for each 2 m of sand thickness, i.e., ${\displaystyle \tau =z/2}$ time samples. Since the reflection coefficient is ${\displaystyle \pm 0.1}$ 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 ${\displaystyle g_{t}}$ and ${\displaystyle -g_{t-\tau }}$, we displace ${\displaystyle -g_{t}}$ by the amount ${\displaystyle \tau }$ and add it to ${\displaystyle g_{t}}$. When ${\displaystyle z=0}$, 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 ${\displaystyle g_{t}}$ and the composite reflections are plotted in Figure 9.14b. The curves for ${\displaystyle \tau =1}$, 2, 3 are roughly displaced copies of each other and have successively larger amplitudes. The ${\displaystyle \tau =3}$ curve at the tuning thickness has the maximum amplitude, the largest amplitudes of the ${\displaystyle \tau =2}$ and ${\displaystyle \tau =4}$ 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 ${\displaystyle \tau =4}$ is broadened and a change of phase is clear in the first trough of the ${\displaystyle \tau =5}$ curve, indicating that the bed is thicker than the resolvable limit of ${\displaystyle \lambda /4}$.

 For ${\displaystyle \tau =1(2m)}$ ${\displaystyle g_{t}}$ 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 ${\displaystyle -g_{t-1}}$ –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 ${\displaystyle \tau =2(4m)}$ ${\displaystyle g_{t}}$ 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 ${\displaystyle -g_{t-2}}$ –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 ${\displaystyle \tau =3(6m\approx \lambda /4)}$ ${\displaystyle g_{t}}$ 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 ${\displaystyle -g_{t-3}}$ –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 ${\displaystyle \tau =4(8m)}$ ${\displaystyle g_{t}}$ 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 ${\displaystyle -g_{t-4}}$ –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 ${\displaystyle \tau =5(10m)}$ ${\displaystyle g_{t}}$ 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 ${\displaystyle -g_{t-4}}$ –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

## Problem 9.14b

Repeat for the sand overlain by shale and underlain by limestone. Assume a sand-limestone reflection coefficient of ${\displaystyle +0.1}$.

### 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 ${\displaystyle +0.2}$. Timing the first peak would give erroneous depths for the ${\displaystyle \tau =1}$, 2, and 3 curves and the ${\displaystyle \tau =2,3}$ 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 ${\displaystyle \tau =1}$, 2, 3 curves and its character varies considerably for curves for ${\displaystyle \tau =3}$, 4, 5.

Figure 9.14b.  Shale/sand/shale reflections.

## Problem 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 ${\displaystyle g_{t}-g_{t-1}+g_{t-2}-g_{t-3}}$, calculated in Table 9.14c.

 ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle -g_{t-1}}$ −11, −14, −5, 10, 12, 6, −3, −5, −2, 0, 1, 1, 0 ${\displaystyle g_{t-2}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle -g_{t-3}}$ −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 ${\displaystyle \tau =1}$ (2 m) ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle g_{t-1}}$ 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 ${\displaystyle \tau =2}$ (4 m) ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle g_{t-2}}$ 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 ${\displaystyle \tau =3}$ (6 m ${\displaystyle \approx \lambda /4}$) ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle g_{t-3}}$ 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 ${\displaystyle \tau =4}$ (8 m) ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle g_{t-4}}$ 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 ${\displaystyle \tau =5}$ (10 m) ${\displaystyle g_{t}}$ 11, 14, 5, −10, −12, −6, 3, 5, 2, 0, −1, −1, 0 ${\displaystyle g_{t-5}}$ 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.

## Problem 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 ${\displaystyle \tau =5}$ curve where the trough and second peak are broadened because ${\displaystyle \tau =5}$ is about equal to ${\displaystyle \lambda /4}$. The resolution is poorer than with the higher-frequency wavelet in Figure 9.14a(i).

Figure 9.14d.  Effect of thin sands.
 For ${\displaystyle \tau =1}$ (2 m) ${\displaystyle g_{t}}$ 6, 11, 14, 14, 10, 5, –2, –10, –11, –12, –10, –6, 0, 3, 4, 5, 4, 3, 1, 0 ${\displaystyle -g_{t-1}}$ –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 ${\displaystyle \tau =2}$ (4 m) ${\displaystyle g_{t}}$ 6, 11, 14, 14, 10, 5, –2, –10, –11, –12, –10, –6, 0, 3, 4, 5, 4, 3, 1, 0 ${\displaystyle -g_{t-3}}$ –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 ${\displaystyle \tau =3}$ (6 m) ${\displaystyle g_{t}}$ 6, 11, 14, 14, 10, 5, –2, –10, –11, –12, –10, –6, 0, 3, 4, 5, 4, 3, 1, 0 ${\displaystyle -g_{t-3}}$ –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 ${\displaystyle \tau =4}$ (8 m) ${\displaystyle g_{t}}$ 6, 11, 14, 14, 10, 5, –2, –10, –11, –12, –10, –6, 0, 3, 4, 5, 4, 3, 1, 0 ${\displaystyle -g{t-4}}$ –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 ${\displaystyle \tau =5}$ (10 m) ${\displaystyle g_{t}}$ 6, 11, 14, 14, 10, 5, –2, –10, –11, –12, –10, –6, 0, 3, 4, 5, 4, 3, 1, 0 ${\displaystyle -g_{t-5}}$ –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

## Problem 9.14e

Repeat parts (a) and (b) using the zero-phase wavelet [1, 1, ${\displaystyle -1}$, ${\displaystyle -4}$, ${\displaystyle -6}$, ${\displaystyle -4}$, 10, 17, 10, ${\displaystyle -4}$, ${\displaystyle -6}$, ${\displaystyle -4}$, ${\displaystyle -1}$, 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 ${\displaystyle \lambda /4}$ and the maximum amplitude in Figure 9.14f occurs where the thickness is ${\displaystyle \lambda /4}$. The peak-trough times increase for ${\displaystyle \tau =4}$ and ${\displaystyle \tau =5}$ in Figure 9.14f, and there is distinct phasing for ${\displaystyle \tau =5}$. In Figure 9.14g, there is a distinct change of shape at ${\displaystyle \tau =3}$ and there is clear resolution and an amplitude minimum at ${\displaystyle \tau =3}$.

Figure 9.14e.  Shale/sand/shale reflection with low-frequency wavelet in Figure 9.14a(ii).
Figure 9.14f.  Shale/sand/shale reflection, zero-phase wavelet.
Figure 9.14g.  Shale/sand/limestone reflection, zero-phase wavelet.
 For ${\displaystyle \tau =1}$ (2 m) ${\displaystyle g_{t}}$ 1, 1, –1, –4, –6. –4, 10, 17, 10, –4, –6, –4, –1, 1, 1, 0 ${\displaystyle -g_{t-1}}$ 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 ${\displaystyle \tau =2}$ (4 m) ${\displaystyle g_{t}}$ 1, 1, 1, –4, –6. –4, 10, 17, 10, –4, –6, –4, 1, 1, 1, 0 ${\displaystyle -g_{t-2}}$ –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 ${\displaystyle \tau =3}$ (6 m) ${\displaystyle g_{t}}$ 1, 1, –1, –4, –6. –4, 10, 17, 10, –4, –6, –4, –1, 1, 1, 0 ${\displaystyle -g_{t-3}}$ –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 ${\displaystyle \tau =4}$ (8 m) ${\displaystyle g_{t}}$ 1, 1, –1, –4, –6. –4, 10, 17, 10, –4, –6, –4, –1, 1, 1, 0 ${\displaystyle -g_{t-4}}$ –1, –1, 1 4, 6. 4, –10, –17, –10, 4, 6, 4, 1, –1, –1, 0 Sum 1, 1, –1, –4, –7, –5, 11, 21, 16, 0, –16, –21, –11, 5, 7, 4, 1, –1, –1, 0 For ${\displaystyle \tau =5}$ (10 m) ${\displaystyle g_{t}}$ 1, 1, –1, –4, –6. –4, 10, 17, 10, –4, –6, –4, –1, 1, 1, 0 ${\displaystyle -g_{t-4}}$ –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 ${\displaystyle \tau =1}$ (2 m) ${\displaystyle g_{t}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 ${\displaystyle -g_{t-1}}$ 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 ${\displaystyle \tau =2}$ (4 m) ${\displaystyle g_{t}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 ${\displaystyle -g_{t-2}}$ 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 ${\displaystyle \tau =3}$ (6 m) ${\displaystyle g_{t}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 ${\displaystyle -g_{t-3}}$ 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 ${\displaystyle \tau =4}$ (8 m) ${\displaystyle g_{t}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 ${\displaystyle -g_{t-4}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 Sum 1, 1, 1, 4, 5, 3, 9, 13, 4, 0, 4, 13, 9, 3, 5, 4, 1, 1, 1, 0 For ${\displaystyle \tau =5}$ (10 m) ${\displaystyle g_{t}}$ 1, 1, -1, -4, -6. -4, 10, 17, 10, -4, -6, -4, -1, 1, 1, 0 ${\displaystyle -g_{t-5}}$ 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