1. Environmental and Exploration Geophysics II
Amplitude, Frequency and Bandwidth and their relationship to Seismic Resolution
tom.h.wilson
Department of Geology and Geography
West Virginia University
Morgantown, WV

2. Transmissivity Factors
Divergence
Partitioning of Energy
Absorption
Refletion interference
Sca
tter
ing

3. M. Hewitt, 1981

4. M. Hewitt, 1981

5. M. Hewitt, 1981

6. M. Hewitt, 1981

7. M. Hewitt, 1981

8. <more absorptive … less absorptive>
M. Hewitt, 1981

9. When surface irregularities have wavelengths similar to the wavelengths comprising the wavelet, significant scattering will occur. The surface appears rough to the wavelet.
When surface irregularities are small compared to the wavelengths comprising the wavelet,scattering losses are small. The surface “appears smoother to the incident wavelet.

10. Ideally we would like to have
We developed a simple relationship between source amplitude and wavelet amplitude at a distance r from the source.
Ideally we would like to have
To do this we must design an operator to remove or correct for the various loss mechanisms
M. Hewitt, 1981

11. Application of this operator
Design the operator
Application of this operator
yields
Hence the ratio of amplitudes A1/A2 = the ratio of reflection coefficients R1/R2. The corrected amplitude Arc is proportional to reflection coefficient.
M. Hewitt, 1981

12. The relative amplitude differences shown in the high fidelity section (left-top) provide more accurate information about actual differences in reflection coefficient and thus lithology than does the section below.

13. Sampling
From the start you have been aware that seismic data is recorded as a series of measurements made at constant time intervals - the sample interval .
How do we decide what the sample rate should be? How often should ground motion be sampled to accurately represent it?
The example of a rotating wheel with spokes serves as the best example of the possible effect of sample rate on the conclusions of our observations concerning the motion of the wheel.

14. Start with a wheel that rotates at constant frequency and then sample it at varying rates.
Assume the period of rotation () is 1 second and that we sample the motion of the wheel every 1/4 second.
In this case observed = actual

15. Let t = 1/2 second
We estimate  = 1 but we are unable to determine the direction of wheel rotation.

16. Our estimate of  = 3 seconds is in error - by 300%
Let t = 3/4 second
Our estimate of  = 3 seconds is in error - by 300%

17. When t = 1 second, the wheel appears to remain at the starting point but actually makes a single rotation between samples.

18. In the second case, we deal with situations where the sample interval t is constant and the period of rotation  varies. Take the case where t = 1/2 second and  =4 seconds.
We have no problem in this case accurately delineating the period of rotation.
Note that f = 1/  = 1/4 second, and that f t =1/8th. f t is the fraction of a cycle the wheel has moved in time t.

19. When t equals 1/2 we also correctly identify rotation periods of 1
When t equals 1/2 we also correctly identify rotation periods of 1.5 seconds.
f t = 1/3rd

20. As before, when  = 1, and t = 1/2, we are still able to identify the period as 1 second.

21. However, when  = 3/4 sec, the frequency f =1/  = 1
However, when  = 3/4 sec, the frequency f =1/  = 1.33 cycles per second. Between consecutive samples, the wheel will turn through ft cycles (check units). ft =2/3rds cycle between samples. The output period is 1.5 seconds and the output frequency is 2/3rds cycle per second.
Our estimate of  in this case is twice its actual value.

22. When  = 2/3 seconds, the frequency of rotation, f=1. 5 cycles/sec
When  = 2/3 seconds, the frequency of rotation, f=1.5 cycles/sec. Thus the wheel turns through ft or 3/4 cycles between samples. The wheel appears to move counterclockwise with frequency equal to 1/2 cycle per second.
When  = 1/2 second, the frequency of rotation, f=2 cycles/sec. f t =1. Thus the wheel turns through one cycle between samples. The wheel appears fixed. The output frequency is 0.

23. Summary Table tau frequency Samples/cycle (1/ft) Cycles/sample (ft)
 out
f out 2 4 1
1.33
1.5
2/3 

24. A plot of input versus output (sampled) frequency reveals saw tooth like variations of output frequency as input frequency increases. The output frequency never increases beyond 1 cycle per second in the case where motion is sampled every 1/2 second.

25. The Nyquist frequency is the highest observable frequency at a given sample rate.

26. Higher input frequencies appear to have lower frequency in the output recording. This phenomena is referred to as frequency or signal aliasing.

27. Frequency content
of seismic signals

28. The range of frequencies present in the wavelet controls its ability to resolve the top and bottom of a layer of given thickness.
Recall our general introduction to the concept of the wavelet earlier in the semester.
The wavelet or transient mechanical disturbance generated by the source can be thought of as a superposition or summation of sinusoids with varying frequency and amplitude.
Hilterman, 1985

29. The examples below illustrate the effect of increasing the frequency range or bandwidth of the wavelet.
O. Ilmaz, 1987

30. The following simple example helps illustrate the concept of an amplitude spectrum. Below is a signal consisting of two sinusoids.

31. Each sinusoid is associated with a specific frequency
Each sinusoid is associated with a specific frequency. There are two frequency components. The 32 sample per cycle component has a frequency of 4 and the 8 samples per cycle component has a frequency of 16. The amplitude of the 32 sample/cycle component is twice that of the 8 sample/cycle component.
The frequency spectrum (above) of the “signal” at the top of the previous slide is an equivalent representation of the signal.

32. Frequency domain
Time domain
O. Ilmaz, 1987

33. Individual frequency components
Amplitude spectrum
Phase spectrum
Individual frequency components
Time-domain wavelets
Hilterman, 1985
Zero Phase
Minimum Phase

34. Extracting information about wavelet frequency content from an isolated reflection event.
The dominant period (c) of the response corresponds to the time from one peak to the next or from one trough to the next. The reciprocal of this dominant period is a measure of the dominant frequency (fc) of the signal or wavelet spectrum.
The reciprocal of the half-width of the response-envelop (b) provides an estimate of the bandwidth (fb) of the signal spectrum.
Hilterman, 1985

35. The dominant frequency and bandwidth measured from the time-domain representation of the signal wavelet can be used to provide a sketch of the wavelet spectrum.
Just as importantly these measures can be related directly to the resolution properties of the seismic wavelet.
Hilterman, 1985

36. Exxon in-house course notes
Let’s come back to this issue in a minute, but first let’s pull some ideas together to develop a basic understanding of how the seismic signal arises in terms of reflection coefficients and wavelets.
Exxon in-house course notes

37. Exxon in-house course notes

38. Reflection interference
Exxon in-house course notes

39. Exxon in-house course notes
As the two layers move closer and closer together we get to a point where the second cycle in the wavelet reflected from the top of the layer overlaps with the arrival of the lead cycle in the wavelet reflected from the base of the layer. This occurs at two-way time equal to 1/2 the dominant period of the wavelet.
Exxon in-house course notes

40. Exxon in-house course notes
At this point there is maximum constructive interference between the reflections from the top and bottom of the layer. The composite reflection event (at right above) attains peak amplitude.
Exxon in-house course notes

41. The peak period of the wavelet can be determined using peak-to-trough times which can be thought of as corresponding to one half the dominant period of the wavelet. Multiply those times by two to get the dominant period.

42. trough
Side lobe
peak
Maximum constructive interference illustrated for the zero phase wavelet. The peak-to-trough time equals c/2.

43. Exxon in-house course notes
Once the separation in time drops to less than half the dominant period of the wavelet destructive interference in the reflections from the top and bottom of the layer will occur.
However, as the layer continues to thin, the dominant period of the composite reflection event does not drop below 1/c. However, the amplitude of the composite continues to drop. But not the period.
Exxon in-house course notes

44. Maximum Constructive Interference
Seismic Wavelet
trough
Side lobe
peak
Two-way interval time separating reflection coefficients is c/2
The peak-to-trough time equals c/2.

45. These amplitude relationships are summarized below in the model seismic response of a thinning layer similar to that which you will generate in lab today.

46. The amplitude difference - trough-to-peak remains constant for two-way travel times much greater than half the dominant period.
As the top and bottom of the layers merge closer and closer together, the lead cycle in the reflection from the base of the layer overlaps with the follow-cycle in the reflection from the top and the amplitude of the composite reflection event begins to increase.
Thickness =Vp/4

47. Layer thickness is simply Vt/2, where t is the two-way interval transit time. Tuning occurs at two-way times equal to one-half the dominant period (c/2). If the interval velocity of the layer in question is known, the dominant period can be converted into the tuning thickness.

48. Difference of arrival time between the reflections from the top and bottom of the layer decreases abruptly at about 8 milliseconds.
8 milliseconds represents the two-way travel time through the layer; it is also the time at which tuning occurs and is half the dominant period of the seismic wavelet.
8 milliseconds is c/2 and the two way time through the layer. Thus, c/4 is the one-way time through the layer.

49. c/4, the one-way time through the layer, equals 4 milliseconds
c/4, the one-way time through the layer, equals 4 milliseconds. The interval velocity in the layer is 11,300 f/s. Hence, the thickness of the layer at this point is ~45 feet. This is the tuning thickness or minimum resolvable thickness of the layer obtainable with the given seismic wavelet.

50. What is the amplitude spectrum of wavelet #5?
Ilmaz, 1987

51. Spectral bandwidth, wavelet duration in the time domain and resolution
Spectral bandwidth, wavelet duration in the time domain and resolution. C is only one parameter that affects resolution. b is also an important parameter.
Hilterman, 1985

52. The minimum phase wavelet has its energy concentrated toward the front end of the wavelet. The amplitude of the disturbance decays exponentially. This wavelet is a causal wavelet and the location of the reflection coefficient is placed at the wavelet onset, which can be difficult for the interpreter to pick.

53. The zero phase wavelet is symmetrical
The zero phase wavelet is symmetrical. This wavelet is centered over the reflection coefficient. The zero phase wavelet is produced through data processing and is not generated naturally. It is non causal - half of the wavelet arrives before the reflector appears in time. It is easy for an interpreter to pick reflection times using the zero phase wavelet since highest amplitude occurs at the reflection boundary.

54. The exploration data is in a zero phase format.
The zero-phase wavelet is also considered to have higher resolving power. It is generally more compact than the equivalent minimum phase wavelet and is, overall, easier to interpret.
The exploration data is in a zero phase format.
Hilterman, 1985

55. The default wavelet in Struct is the Ricker wavelet
The default wavelet in Struct is the Ricker wavelet. The Ricker wavelet is zero phase.
Hilterman, 1985

56. Physical nature of the seismic response
Hilterman, 1985
Physical nature of the seismic response

57. Exxon in-house course notes
The output is a superposition of reflections from all acoustic interfaces

58. Exxon in-house course notes

60. Subsurface structure - North Sea
One additional topic to consider in general is wavelet deconvolution and how wavelet shape can affect geologic interpretations …. Consider the following structural model
Subsurface structure - North Sea
Neidel, 1991

61. Potential hydrocarbon trap?
Below is the synthetic seismic response computed for the North Sea model.
Potential hydrocarbon trap?
Neidel, 1991

62. Consider the effect of wavelet shape on the geologic interpretation of seismic response. In the case shown below, the primary reflection from the base of the Jurassic shale crosses a side-lobe in the wavelet reflected from the overlying basal Cretaceous interval.
Neidel, 1991

63. Deconvolution is a filter operation which compresses and simplifies the shape of the seismic wavelet. Deconvolution improves seismic resolution and simplifies interpretation.

64. Neidel, 1991
North Sea Seismic display after deconvolution. The geometrical interrelationships between reflectors are clearly portrayed.

65. Consider the following problem -
You are given the seismic wavelet shown below.
Using the estimation procedure discussed in class today measure the appropriate feature on the above seismic wavelet and answer the following questions:
What is the minimum resolvable thickness of a layer having an interval velocity of 10,000fps? Show work on your handout
What is the phase of the wavelet? Why do you say that?

66. Questions?
If you haven’t already, finish reading chapter 4.