Environmental and Exploration Geophysics II Amplitude, Frequency and Bandwidth and their relationship to Seismic Resolution tom.h.wilson wilson@geo.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV 1
M. Hewitt, 1981 M. Hewitt, 1981 2
M. Hewitt, 1981 M. Hewitt, 1981 3
M. Hewitt, 1981 <more absorptive less absorptive> M. Hewitt, 1981 4
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. We developed a simple relationship between source amplitude and wavelet amplitude at a distance r from the source. As Ar = T e r αr R Ideally we would like to have A r R M. Hewitt, 1981 To do this we must design an operator to remove or correct for the various loss mechanisms 5
As Ar = T e r αr Design the operator R re αr T Application of this operator A rc re = T yields αr A A T r rc = s e A R s αr R M. Hewitt, 1981 Hence the ratio of amplitudes A 1 /A 2 = the ratio of reflection coefficients R 1 /R 2. The corrected amplitude A rc is proportional to reflection coefficient. The relative amplitude differences shown in the high fidelity section (lefttop) provide more accurate information about actual differences in reflection coefficient and thus lithology than does the section below. 6
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. Start with a wheel that rotates at constant frequency and then sample it at varying rates. Assume the period of rotation (τ) is 1 second and that we sample the motion of the wheel every 1/4 second. In this case τ observed = τ actual 7
Let Δt = 1/2 second We estimate τ = 1 but we are unable to determine the direction of wheel rotation. Let Δt = 3/4 second Our estimate of τ = 3 seconds is in error - by 300% 8
When Δt = 1 second, the wheel appears to remain at the starting point but actually makes a single rotation between samples. In the second case, we deal with situations where the sample interval Δt is constant and the period of rotation τ varies. Take the case where Δt = 1/2 second and τ =4 seconds. We have no problem in this case accurately delineating the period of rotation. Note that f = 1/ τ = 1/4 second, and that f Δt =1/8th. f Δt is the fraction of a cycle the wheel has moved in time Δt. 9
When Δt equals 1/2 we also correctly identify rotation periods of 1.5 seconds. f Δt = 1/3rd As before, when τ = 1, and Δt = 1/2, we are still able to identify the period as 1 second. 10
However, when τ = 3/4 sec, the frequency f =1/ τ = 1.33 cycles per second. Between consecutive samples, the wheel will turn through fδt cycles (check units). fδt =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 τ in this case is twice its actual value. When τ = 2/3 seconds, the frequency of rotation, f=1.5 cycles/sec. Thus the wheel turns through fδt or 3/4 cycles between samples. The wheel appears to move counterclockwise with frequency equal to 1/2 cycle per second. When τ = 1/2 second, the frequency of rotation, f=2 cycles/sec. f Δt =1. Thus the wheel turns through one cycle between samples. The wheel appears fixed. The output frequency is 0. 11
tau frequency Samples/cycle (1/fΔt) Cycles/sample (fδt) τ out f out 2 ½ 4 ¼ 2 ½ 1 1 2 ½ 1 1 ¾ 1.33 1.5 2/3 1.5 2/3 2/3 1.5 1.33 ¾ 2 ½ ½ 2 1 1 0 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. 12
The Nyquist frequency is the highest observable frequency at a given sample rate. Higher input frequencies appear to have lower frequency in the output recording. This phenomena is referred to as frequency or signal aliasing. 13
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 14
The examples below illustrate the effect of increasing the frequency range or bandwidth of the wavelet. O. Ilmaz, 1987 The following simple example helps illustrate the concept of an amplitude spectrum. Below is a signal consisting of two sinusoids. 15
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. Frequency domain Time domain O. Ilmaz, 1987 16
Amplitude spectrum Phase spectrum Individual frequency components Time-domain wavelets Zero Phase Minimum Phase Hilterman, 1985 Extracting information about wavelet frequency content from an isolated reflection event. The dominant period (τ 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 (f c ) of the signal or wavelet spectrum. The reciprocal of the half-width of the response-envelop (τ b ) provides an estimate of the bandwidth (f b ) of the signal spectrum. Hilterman, 1985 17
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 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 18
Exxon in-house course notes Exxon in-house course notes 19
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 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 20
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. trough Side lobe peak Maximum constructive interference illustrated for the zero phase wavelet. The peak-to-trough time equals τ c /2. 21
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/τ c. However, the amplitude of the composite continues to drop. But not the period. Seismic Wavelet Maximum Constructive Interference trough Side lobe peak Two-way interval time separating reflection coefficients is τ c /2 The peak-totrough time equals τ c /2. 22
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. 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 =Vt/2 23
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 (τ c /2). If the interval velocity of the layer in question is known, the dominant period can be converted into the tuning thickness. 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 τ c /2 and the two way time through the layer. Thus, τ c /4 is the one-way time through the layer. 24
τ 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. What is the amplitude spectrum of wavelet #5? Ilmaz, 1987 25
Spectral bandwidth, wavelet duration in the time domain and resolution. τ C is only one parameter that affects resolution. τ b is also an important parameter. Hilterman, 1985 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. 26
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. 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 27
The default wavelet in Struct is the Ricker wavelet. The Ricker wavelet is zero phase. Hilterman, 1985 Hilterman, 1985 Physical nature of the seismic response 28
The output is a superposition of reflections from all acoustic interfaces Exxon in-house course notes Exxon in-house course notes 29
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 30
Below is the synthetic seismic response computed for the North Sea model. Potential hydrocarbon trap? Neidel, 1991 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 31
Deconvolution is a filter operation which compresses and simplifies the shape of the seismic wavelet. Deconvolution improves seismic resolution and simplifies interpretation. Neidel, 1991 North Sea Seismic display after deconvolution. The geometrical interrelationships between reflectors are clearly portrayed. 32
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? If you haven t already, finish reading chapter 4. 33