Exam 2 Review Flashcards
What is needed in order to image a layer?
A contrast in physical properties between the layers
Define Seismic Impedance
A measure of how sharp the contrast between two layers is and thus how well we can detect a layer boundary using the reflection technique.
What happens to the seismic energy as it encounters a boundary?
Some of the energy gets reflected but some of the energy is also transmitted through to the bottom layer.
Seismic Impedence Equation I=
pv, where p is the denstiy of the layer and v is the velocity of the layer
What are the two cases of seismic impedance?
1) The density and/or velocity of Layer 2 is large
2) p2v2 with respect to layer 1 is low
Describe the first case of impedance
The density and/or velocity of Layer 2 is large - specifically the product of density x velocity (p2v2) is large with respect to that of layer 1. Hence the impedance is large and very little seismic energy will transmit through to the bottom layer.
What is good about the first case of impedance?
It is good for detecting this layer boundary with reflection techniques since most of the energy will get reflected back.
What is bad about the first case of impedance?
It is very bad for having any chance of detecting lower layers
What is the second case of impedance?
If p2v2 is low with respect to Layer 1, then we have a weak reflection from the layer boundary which may be difficult to detect.
What is the reflection coefficient good for?
We can directly quantify how much energy is reflected vs. transmitted.
Reflection Coefficient Equation:
R= (I2-I1)/(I2+I1) = (p2v2 - p1v1)/(p2v2 + p1v1)
What does the reflection coefficient equation only work for?
It only works for vertically incident waves! (DRAW DIAGRAM L09-2)
Incoming wave energy =
reflected energy + transmitted energy
Reflection Coefficient = R = (in words)
(Amplitude of reflected ray)/(Amplitude of incident ray)
Transmission Coefficient = T = (in words)
(Amplitude of transmitted ray)/(Amplitude of incident ray)
Transmission Coefficient (TC) = T =
T = (2*I1)/(I2 + I1) = (2p1v1)/(p2v2 + p1v1)
In general, we cannot resolve layering if lambda ______ layer thickness
if wavelength is greater than layer thickness, can’t resolve layering
What else do you need, aside from a high frequency source, to construct good seismic pulses?
a large bandwidth
define bandwidth
many frequencies
Draw diagram for Reflected wave travel times
L07-2
For reflected arrivals, the angle of reflection is ______ to the angle of the incident ray
equal, sigma1=sigma2
Total length of ray, 2L= (reflected)
sqrt(4h^2+x^2)
Travel time, reflected = T =
2L/v1 = sqrt(4h^2+x^2)/v1
Equation of line T^2, reflected linearized =
T^2 = (2h/v1)^2 + (1/v1)^2*x^2
Slope of T^2 reflected line =
(1/v1)^2
Intercept of T^2 reflected line =
(2h/v1)^2
What is x for a vertically incident reflection?
x=0
What is t0 the constant for a vertically incident reflection?
t0=2h/v1
What shape is this line?
hyperbola, if equation is rewritten this can be seen: T^2/t0^2 - x^2/(v1^2*t0^2) = 1
What does the travel time curve asymptotically approach?
A line with the slope 1/v1
Why does the travel time curve asymptotically approach this line?
Asymptotically approaches a line with the slope 1/v1, and it does this because increasing the distance between the source and receiver results in not much difference between the direct & reflected wave (draw diagrams that we drew in class review)
Draw and label the diagram of travel time curve for rays in a layer over a halfspace
See L07-6
Where is the source and receiver for the zero offset experiment?
Source and receiver at the same location
What do the seismograms reflect if velocity is constant with each layer?
If velocity is constant with each layer, then time scales linearly with depth, so that the seismograms reflect structure at depth
What are some potential challenges with zero-offset experiments?
Noise
time to propagate over layer thickness and back is less than the time length of source
Heterogeneity
Uncertainties in velocity structure
Describe the first possible problem with zero offset experiments: (with illustration and solution)
Single records can be noisy: reverberations of energy from near source and near receiver layers can contaminate observations, or, the reflector you wish to image may simply be too weak to generate an observed reflection
(see picture L07-8)
Solution: Use data from non-zero offset geometries and stack (sum up) the data.
Describe the second possible problem with zero offset experiments: (with illustration and solution)
The time to propagate over layer thickness and back (t1) is less than the time length of source (ts)
(see picture L07-8)
Solution: Deconvolution: strip the source signal off the seismogram
Describe the third possible problem with zero offset experiments: (with illustration and solution)
Heterogeneity may scatter energy to different (non-vertical) directions, biasing results that assume only vertical propagation paths.
(see picture L07-8)
Solution: Forward model: do a grid of scatterers, summing the data for each possibility (“migration”)
Describe the fourth possible problem with zero offset experiments: (with illustration and solution)
Uncertainties in velocity structure make it difficult to establish depth to reflector (the classic trade off between v and z)
(see picture L07-8)
Solution: Get help from other means, e.g., drill holes and conduct refraction studies.
Signal to Noise ratio =
amplitude of signal/amplitude of noise
Basic idea of common mid-point stacking
The basic idea is that, if we can choose where we put our recorders and sources of energy, then we can build in some sampling redundancy in our approach, and ultimately stack multiple recordings to improve the signal-to-noise ratio (SNR) of our measurements.
Collect data that shares the same central reflection point off an interface
The common midpoint method is based on finding…
all source-receiver combinations that have the same bounce point - or common midpoint
Draw a diagram of what the graph first looks like when plotting for the common midpoint method
see last figure L08-2
What does the graph trace out for the common midpoint method?
hyperbola
What do you do once you have the initial graph for the common midpoint method?
Make a stack for each common midpoint, but first, we need to apply a timing correction for the hyperbolic travel-time moveout with distance.
timing correction for common midpoint method:
t(x)~~t0*(1 + 0.5(x/vt0)^2)
Define moveout:
the time difference between arrivals at two different distances