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

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


with constant-velocity layers

1V

2V

3V

Eventually, …..

1 2 3 4V V V V

4V

4



1V

2V

3V

Eventually, …..

1 2 3 4V V V V

4V

5



1V

2V

3V

Eventually, …..

1 2 3 4V V V V

4V

6



1V

2V

3V

………...after successive refractions,

1 2 3 4V V V V

4V

7



1V

2V

3V

…………………………………………. the rays are turned back top the surface

1 2 3 4V V V V

4V

8

Outline





•Constant-velocity versus gradient layers

•Reflections


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





•Constant-velocity versus gradient layers

•Reflections


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


P

O X/2 X/2

hV1

Travel distance = ?

Travel time = ?

Consider the reflecting ray……. as follows ….

16


P

O X/2 X/2

hV1

Travel distance =

Travel time =

2OPCDDDDDDDDDDDDD D

2velocityOPCDDDDDDDDDDDDD D

17


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


#In 3-D

Azimuth (phi)

Dip

(the

ta)

strike

27


#In 3-D

Matlab code

Direct waves

090

28


#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





•Constant-velocity versus Gradient layers

•Reflections


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




R12 = (I2-I1) / (I1+I2)

38




RAirWater = (IWater-IAir) / (IAir+IWater)

R12 = (I2-I1) / (I1+I2)

39




RAirWater = (IWater-IAir) / (IAir+IWater)

R12 = (I2-I1) / (I2+I1)

RAirWater = (IWater-0) / (0+IWater)

RAirWater = 1

40

Example of Water-air reflection



Air

Water

Layer 1

Layer 2

41




R21 = (I1-I2) / (I1+I2)

42




RWaterAir = (IAir-IWater) / (IAir+IWater)

R22 = (I1-I2) / (I1+I2)

43




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


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


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


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

Documents

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