Gamma X-rays Ultra- violet Infra- red Micro- waves Radio Unit G9 Waves.
1 Outline Full space, half space and quarter space Traveltime curves of direct ground- and air-...
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Transcript of 1 Outline Full space, half space and quarter space Traveltime curves of direct ground- and air-...
1
Outline
•Full space, half space and quarter space
•Traveltime curves of direct ground- and air- waves and rays
•Error analysis of direct waves and rays
•Constant-velocity-layered half-space
•Constant-velocity versus Gradient layers
•Reflections
•Scattering Coefficients
2
-X
z
A layered half-space
X
3
A layered half-space
with constant-velocity layers
1V
2V
3V
Eventually, …..
1 2 3 4V V V V
4V
4
A layered half-space
with constant-velocity layers
1V
2V
3V
Eventually, …..
1 2 3 4V V V V
4V
5
A layered half-space
with constant-velocity layers
1V
2V
3V
Eventually, …..
1 2 3 4V V V V
4V
6
A layered half-space
with constant-velocity layers
1V
2V
3V
………...after successive refractions,
1 2 3 4V V V V
4V
7
A layered half-space
with constant-velocity layers
1V
2V
3V
…………………………………………. the rays are turned back top the surface
1 2 3 4V V V V
4V
8
Outline
•Full space, half space and quarter space
•Traveltime curves of direct ground- and air- waves and rays
•Error analysis of direct waves and rays
•Constant-velocity-layered half-space
•Constant-velocity versus gradient layers
•Reflections
•Scattering Coefficients
9
Constant-velocity layers vs. gradient-velocity layers
1V 1V
1 0 mZV V constant1V
“Each layer bends the ray along part of a circular path”
10
Outline
•Full space, half space and quarter space
•Traveltime curves of direct ground- and air- waves and rays
•Error analysis of direct waves and rays
•Constant-velocity-layered half-space
•Constant-velocity versus gradient layers
•Reflections
•Scattering Coefficients
11
12
Direct water arrival
13
Hyperbola
x
y
2 2
2 21y x
a b
22
21 xy ab
As x -> infinity,
Y-> X. a/b, where a/b is the
slope of the asymptote
x
asym
ptot
e
14
Reflection between a single layer and a half-space below
P
O X/2 X/2
hV1
Travel distance = ?
Travel time = ?
15
Reflection between a single layer and a half-space below
P
O X/2 X/2
hV1
Travel distance = ?
Travel time = ?
Consider the reflecting ray……. as follows ….
16
Reflection between a single layer and a half-space below
P
O X/2 X/2
hV1
Travel distance =
Travel time =
2OPCDDDDDDDDDDDDD D
2velocityOPCDDDDDDDDDDDDD D
17
Reflection between a single layer and a half-space below
22
22
x h
velocity
Traveltime =
22 2
21
44xT h
V
2 22
2 21 1
4x hTV V
22 2
021
xT TV
(6)
18
Reflection between a single layer and a half-space below and
D-wave traveltime curves
asymptote
1V
constant1V
Matlab code
19
#1 At X=0, T=2h/V1
Two important places on the traveltime hyperbola
constant1V *
T0=2h/V1
h
Matlab code
20
#1As X--> very large values, and X>>h ,
then (6) simplifies into the equation of straight line with slope dx/dT = V1
22 2
021
xT TV
(6)
0 0T
If we start with
as the thickness becomes insignificant with respect to the source-receiver distance
21
22
21
xTV
1
xTV
1
1T xV
By analogy with the parametric equation for a hyperbola, the slope of this line is 1/V1 i.e.
a/b = 1/V1
22
What can we tell from the relative shape of the hyperbola?
Increasing velocity (m/s)
Increasing thickness (m)
1000
3000
50
250
23
“Greater velocities, and greater thicknesses flatten the shape of the hyperbola, all else remaining constant”
24
Reflections from a dipping interface
#In 2-D
Matlab code
Direct waves
1030
25
Reflections from a 2D dipping interface
#In 2-D:
“The apex of the hyperbola moves in the geological, updip direction to lesser times as the dip increases”
26
Reflections from a 3D dipping interface
#In 3-D
Azimuth (phi)
Dip
(the
ta)
strike
27
Reflections from a 3D dipping interface
#In 3-D
Matlab code
Direct waves
090
28
Reflections from a 2D dipping interface
#In 3-D:
“The apparent dip of a dipping interface grows from 0 toward the maximum dip as we increase the azimuth with respect to the strike of the dipping interface”
29
Outline
•Full space, half space and quarter space
•Traveltime curves of direct ground- and air- waves and rays
•Error analysis of direct waves and rays
•Constant-velocity-layered half-space
•Constant-velocity versus Gradient layers
•Reflections
•Scattering Coefficients
30
Amplitude of a traveling wave is affected by….
•Scattering Coefficient
Amp = Amp(change in Acoustic Impedance (I))
•Geometric spreading
Amp = Amp(r)
•Attenuation (inelastic, frictional loss of energy)
Amp = Amp(r,f)
31
Partitioning of energy at a reflecting interface at Normal
Incidence
Incident Amplitude = Reflected Amplitude + Transmitted Amplitude
Reflected Amplitude = Incident Amplitude x Reflection Coefficient
TransmittedAmplitude = Incident Amplitude x Transmission Coefficient
Incident Reflected
Transmitted
32
Partitioning of energy at a reflecting interface at Normal
Incidence
Scattering Coefficients depend on the Acoustic Impedance changes across a boundary)
Acoustic Impedance of a layer (I) = density * Vp
Incident Reflected
Transmitted
33
Nomenclature for labeling reflecting and transmitted
rays
N.B. No refraction,
normal incidence
P1`
P1` P1’
P1`P2` P1`P2`P2’
P1`P2`P2’P1
’
P1`P2`P2’ P2`
34
Amplitude calculations depend on transmission and reflection coefficients which depend on whether ray is traveling down or up
N.B. No refraction,
normal incidence
1
R12
T12T12 R23
T12 R23 T21Layer 1
Layer 2
Layer 3
T12 R23 R21
35
R12 = (I2-I1) / (I1+I2)
T12 = 2I1 / (I1+I2)
R21 = (I1-I2) / (I2+I1)
T21 = 2I2 / (I2+I1)
Reflection Coefficients
Transmission Coefficients
36
Example of Air-water reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
Air
Water
Layer 1
Layer 2
37
Example of Air-water reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
R12 = (I2-I1) / (I1+I2)
38
Example of Air-water reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
RAirWater = (IWater-IAir) / (IAir+IWater)
R12 = (I2-I1) / (I1+I2)
39
Example of Air-water reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
RAirWater = (IWater-IAir) / (IAir+IWater)
R12 = (I2-I1) / (I2+I1)
RAirWater = (IWater-0) / (0+IWater)
RAirWater = 1
40
Example of Water-air reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
Air
Water
Layer 1
Layer 2
41
Example of Water-air reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
R21 = (I1-I2) / (I1+I2)
42
Example of Water-air reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
RWaterAir = (IAir-IWater) / (IAir+IWater)
R22 = (I1-I2) / (I1+I2)
43
Example of Water-air reflection
Air: density =0; Vp=330 m/s
water: density =1; Vp=1500m/s
RWaterAir = (IAir-IWater) / (IAir+IWater)
R21 = (I1-I2) / (I1+I2)
RWaterAir = (0-IWater) / (0+IWater)
RWaterAir = -1 ( A negative reflection coefficient)
44
Effect of Negative Reflection Coefficient on a reflected pulse
45
Positive Reflection Coefficient (~0.5)
46
“Water-air interface is a near-perfect reflector”
47
In-class Quiz
Air
Water0.1m steel plate
What signal is received back from the steel plate by the hydrophone (triangle) in the water after
the explosion?
1 km
48
In-class Quiz
WaterLayer 1
Layer 2
Layer 3
R12 at time
t1
T12 R23 T21
at time t2
0.1m steel plate
49
Steel: density = 8; Vp=6000 m/s
water: density =1; Vp=1500m/s
RWaterSteel = (Isteel-Iwater) / (Isteel+Iwater)
R12 = (I2-I1) / (I1+I2)
50
Steel: density = 8; Vp=6000 m/s; I=48,000
water: density =1; Vp=1500m/s; 1500
RWaterSteel = (Isteel-Iwater) / (Isteel+Iwater)
R12 = (I2-I1) / (I1+I2)
RWaterSteel = (46,500) / (49,500)
RWaterSteel = 0.94
51
RSteel water= (Iwater-Isteel) / (Iwater+Isteel)
R21 = (I1-I2) / (I1+I2)
Steel: density = 8; Vp=6000 m/s; I=48,000
water: density =1; Vp=1500m/s; 1500
RSteel water= (-46,500) / (49,500)
Rsteel water = -0.94
52
Steel: density = 8; Vp=6000 m/s ; I=48,000
water: density =1; Vp=1500m/s; I=1500
T WaterSteel= 2IWater/ (Iwater+Isteel)
T12 = 2I1/ (I1+I2)
T WaterSteel= 3000/ (49,500)
T WaterSteel= 0.06
53
T SteelWater= 2ISteel/ (Iwater+Isteel)
T21 = 2I2/ (I1+I2)
Steel: density = 8; Vp=6000 m/s ; I=48,000
water: density =1; Vp=1500m/s; I=1500
T SteelWater= 96,000/ (49,500)
T SteelWater= 1.94
54
For a reference incident amplitude of 1
At t1: Amplitude = R12 = 0.94
At t2: Amplitude = T12R23T21
= 0.06 x -0.94 x 1.94
= -0.11 at t2
t2-t1 = 2*0.1m/6000m/s in steel
=0.00005s
=5/100 ms
55
Summation of two “realistic” wavelets
56
Either way, the answer is yes!!!
57
Outline-2
•AVA-- Angular reflection coefficients
•Vertical Resolution
•Fresnel- horizontal resolution
•Headwaves
•Diffraction
•Ghosts•Land•Marine
•Velocity layering•“approximately hyperbolic equations”•multiples
58
“As the angle of incidence is increased the amplitude of the
reflecting wave changes”
Variation of Amplitude with angle (“AVA”) for the fluid-over-fluid case
(NO SHEAR WAVES)
2
22 1
1
2
22 1
1
sincos 1
( )sin
cos 1
VI I
VR
VI I
V
(Liner, 2004; Eq.
3.29, p.68)
(7)
59
theta
V1,rho1
V2,rho2
P`
P`P’reflected
Transmitted and refracted
P`P`
For pre-critical reflection angles of incidence (theta < critical angle), energy at an interface is partitioned between returning reflection and transmitted refracted wave
60Matlab Code
61
What happens to the equation 7 as we reach the critical angle?
1 11 2
2sin ;critical
VV V
V
62
critical
angleV1,rho1
V2,rho2
P`
P`P’
At critical angle of incidence,angle of refraction = 90 degrees=angle of reflection
63
2
22 1
1
2
22 1
1
sincos 1
( )sin
cos 1
VI I
VR
VI I
V
At criticality, 1 1
2sinc
VV
1R
The above equation becomes:
64
critical
angleV1,rho1
V2,rho2
P`
P`P’
For angle of incidence > critical angle;
angle of reflection = angle of incidence and there are no refracted waves i.e. TOTAL INTERNAL REFLECTION
65
2 1
2
2
2
1
1
2
2
1
sin1cos
( )
ci
1oss n
VV
V
I I
R
I IV
The values inside the square root signs can be negative, so that the numerator, denominator and reflection coefficient
become complex numbers
66
A review of the geometric representation of complex numbers
Real (+)Real (-)
Imaginary (-)
Imaginary (+)
a B (IMAGINARY)
Complex number = a + ib
i = square root of -1
(REAL)
67
Think of a complex number as a vector
Real (+)Real (-)
Imaginary (-)
Imaginary (+)
a
Cb
68
Real (+)
Imaginary (+)
a
Cb
1tan ba
1. Amplitude (length) of vector
2 2a b2. Angle or phase of vector
69
1. Why does phase affect seismic data? (or.. Does it really matter that I
understand phase…?)
2. How do phase shifts affect seismic data? ( or ...What does it do to my
signal shape?
IMPORTANT QUESTIONS
70
1. Why does phase affect seismic data? (or.. Does it really matter that I
understand phase…?)
Fourier Analysis
frequency
Power or Energy or Amplitud
e
frequency
Phase
71
1. Why does phase affect seismic data?
Signal processing through Fourier Decomposition breaks down seismic data into not only its frequency components (Real portion of the seismic data) but into the phase component (imaginary part). So, decomposed seismic data is complex.
If you don’t know the phase you cannot get the data back into the time domain. When we bandpass filter we can choose to change the phase or keep it the same (default)
Data is usually shot so that phase is as close to 0 for all frequencies.
72
2. How do phase shifts affect seismic data?
IMPORTANT QUESTIONS
cos(2 ) sin 22
ft ft
2
is known as the phase
A negative phase shift ADVANCES the signal and vice versa
The cosine signal is delayed by 90 degrees with respect to a sine signal
Let’s look at just one harmonic component of a complex signal
73
cos(2 ) sin 22
ft ft
If we add say, many terms from 0.1 Hz to 20 Hz with steps of 0.1 Hz for both cosines and the
phase shifted cosines we can see:
Matlab code
74
Reflection Coefficients at all angles- pre and post-critical
Matlab Code
75
NOTES: #1
At the critical angle, the real portion of the RC goes to 1. But, beyond it drops. This does not mean that the energy is dropping. Remember that the RC is complex and has two terms. For an estimation of energy you would need to look at the square of the amplitude. To calculate the amplitude we include both the imaginary and real portions of the RC.
76
NOTES: #2
For the critical ray, amplitude is maximum (=1) at critical angle.
Post-critical angles also have a maximum amplitude because all the energy is coming back as a reflected wave and no energy is getting into the lower layer
77
NOTES: #3
Post-critical angle rays will experience a phase shift, that is the shape of the signal will change.
78
Approximating reflection events with hyperbolic shapes
We have seen that for a single-layer case:
2
2 20 2
1
xT x TV
(rearranging equation 6)
V1 h1
79
Approximating reflection events with hyperbolic shapes
From Liner (2004; p. 92), for an n-layer case we have:
2 2 41 2 3 ...T x c c x c x
1V
2V
3V
For example, where n=3, after 6 refractions and 1 reflection per ray we have the above scenario
1 2 3 4V V V V 4V
h1
h2
h3
h4
80
Approximating reflection events with hyperbolic shapes
Coefficients c1,c2,c3 are given in terms of a second function set of coefficients, the a series, where am is defined as follows:
2 3
12
mn
m i ii
a V h
For example, in the case of a single layer we have:
One-layer case (n=1)
1
2 3
1 1
111
22
i ii
na V h
hV
2 3
2 1 11
212 2
i ii
na V h V h
2 31
1
3
3 131
2 2i ii
na V h V h
81
Two-layer case(n=2)
2
1
1
2 2 3
1
2 3 2 3
1
1 1
1 22 2n
i iih ha V h V V
2 3 2 3 2 32
1 1 2
2 2 2
2 212 2n
i iia V h V h V h
2
3
2 3 3
1
3 31 1 2 2
2 2n
i iia V h V h V h
1 2
1 2
2h h
V V
1 1 222 h hV V
82
22 3
21 1 1
2mn
i iic a V z
The “c” coefficients are defined in terms of combinations of the “a” function, so that:
1
2
2
ac a
22
31 342
4a
ca aa
83
One-layer case (n=1)
1
2
11
20
2h
Vc T
2
2
1
1Vc
30c
2
2 20 2
1
xT x TV
84
C2=1/Vrms (See slide 14 of “Wave in Fluids”)
Two-layer case (n=2)
1 1 222 h hV V
2 1 1 2
1 21
220
2V h V hVV
c T
2
2 20 2
RMS
xT x TV
02c T
02c T2 2
1 1 2 2
1 2
2V h V h
V V
2 2
1 20 1 22T V t V t
1/2
2
1( )
1
j
irms j j
i
V tV
t
85
Two-layer case (n=2)
30c
What about the c3 coefficient for this case?
Matlab Code
86
Four-layer case (n=4) (Yilmaz, 1987 ;Fig. 3-
10;p.160;
For a horizontally-layered earth and a small-spread hyperbola3
0c
Matlab code
87
1/2
2
1( )
1
j
i irms j j
i
V tV
t
( )
j
i ii
average j
ii
V tV j
t
2
1/22
1
1 1
backus
i i
i i
j
i i
j j V ti i i V
V tV j
V t
( , ) ( ) ( )backus average rmsi i i ibackusV V V V V V V
88
1/2
2
1( )
1
j
irms j j
i
iV tV
t
( )
j
i ii
average j
ii
V tV j
t
Very important for basic seismic processing. Can be
obtained directly from seismic field data or GPR field data. Errors ~10%
Mean velocity; traditional
89
V=330 m/s, rho =0
z=100000m
s = 200m; V=1000 m/s, rho
=1.6V=1500 m/s, z= 500m rho =1.8
i=1
i=2
i=3=j
90
layer V rho z(m) time thickness(s)V*t V*V*t rho*V*t z*z z/(VVvrho)1 330 0.013 100000 303.0303 100000 33000000 1300 1E+10 70.636432 1000 1.6 200 0.2 200 200000 320 40000 0.0001253 1500 1.8 500 0.333333 500 750000 900 250000 0.000123
sums TIMES Vt VVt rhoVt ZZ z/(Vvrho)100700 303.5636 100700 33950000 2520 1E+10 70.63668
VrmsVrms 111838.2Vrms 334.4221
Vavg 331.7262
VbacVbac 56180Vbac 237.0232
Excel macro
91
1/2
2 2
1( )
1
j
i irms j j
i
V tV
t
( )
j
i ii
average j
ii
V tV j
t
2 2
1/22
1
1 1
backus
i i
i i i
j
i i
j j V ti i i V t
V tV j
V t
( , ) ( ) ( )backus average rmsi i i ibackusV V V V V V V