Generalized Differential Semblance Optimization

Sanzong Zhang and Gerard SchusterKing Abdullah University of Science and Technology

Motivation

0

60 18

Z (k

m)

X (km)

Differential Semblance Inversion

0

60 18

Z (k

m)

X (km)

Problem: DSO sometimes has trouble achieving sufficient resolution

Solution: Generalized DSO = Subsurface Offset Inversion+DSOGeneralized Differential Semblance Inversion

Marmousi

Marmousi

Outline

Traveltime+waveform Inversion Generalized DSO Inversion

ε = ½∑[Dm Dh]2ε = ½∑[Dd Dt] 2

Motivation

Numerical Tests

Summary

Wave Eq. Traveltime+Waveform Inversion(Zhou et al., 1995; Luo+GTS, 1991)

Tim

e

ε = ½∑[wxDtx]2

x

Traveltime

+ ½∑[DtDd] 2

x, t

Waveform

WTW Misfit

Dtx

High wavenumber

1830 m

d(x,t)

(1-a) a

2.3

1.2

km/s

1.5

1.8

Low wavenumber

305 m

0 m

1830 mCourtesy Ge Zhan

a= 0 traveltime tomo. a= 1 FWI

e=½∑[DtxDd]2

e=½∑[DhDm]2

+ ½∑[DmDh]2

z, Dh

MVA, DSO General Differential Semblance Optimization

Tim

eε = ½∑[ Dh]

2

z

Subsurface Offset DSO

DhZ

d(x,t)

+Dh-Dh

Low wavenumber Intermediate wavenumber

Dtx

Dm

ObjectiveFunctions

Weight with offset DhWeight with amplitude Dm

z,Dhε = ½∑[Dm Dh] 2General

DSO ObjectiveFunction

General DSO Gradient

g(x) =

Low wavenumber

x,Dh ∂c(x)∂Dh ∑[ Dm Dh

2+ DmDh2 ∂Dm

∂c(x)]

Intermediate wavenumber

+Dh-Dh

Sub. offset CIG

Z

Migration

Dm

z,Dh ∂c(x)

∂(DmDh)= ∑ 2

g(x) ½

(1-a) a

(Stork, 1992; Symes & Kern, 1994;Sava & Biondi, 2004 ;Almomim, 2011;Zhang et al, 2012)

Tim

e

DhZ

d(x,t)

+Dh-Dh

Dtx

Dm+Dh-Dh

Sub. offset CIG

Z

Migration

0

60 18

Z (k

m)

X (km)

DSO Inversion0

60 18

Z (k

m)

X (km)

General DSO Inversion

MVA, DSO General Differential Semblance Optimization(Stork, 1992; Symes & Kern, 1994;Sava & Biondi, 2004 ;Almomim, 2011;Zhang et al, 2012)

+ ½∑[DmDh]2

z, Dhε = ½∑[ Dh]

2

z

Subsurface Offset DSO

Low wavenumber Intermediate wavenumber

ObjectiveFunctions

Dm (1-a) a

Outline

Traveltime+waveform Inversion Generalized DSO Inversion

ε = ½∑[Dm Dh] 2ε = ½∑[Dd Dt] 2

Motivation

Numerical Tests

Summary

Numerical Examples0

60 18

Z (k

m)

X (km)

0

10

t (s)

X (km)0 14

15 Hz Ricker wavelet 242 shots , 70 m spacing 700 receivers, 20 m spacing

(a) True velocity model

(b) CSG

Numerical Examples0

60 18

Z (k

m)

X (km)

Initial velocity model0

60 18

Z (k

m)

X (km)

True velocity model

0

60 18

Z (k

m)

X (km)

Inverted model (DSO)0

60 18

Z (k

m)

X (km)

Inverted model (Gen. DSO)

4.5

1

0

60 18

Z (k

m)

X (km)

Initial velocity model0

60 18

Z (k

m)

X (km)

Inverted model

0

30 9

Z (k

m)

X (km)

Initial velocity model0

30 9

Z (k

m)

X (km)

Inverted model (DSO)

Result Comparison

2

4

4.5

1

(Shen et al., 2001)

Numerical Examples RTM image (DSO)Z

(km

)0

60 18X (km)

RTM image (General DSO)

Z (k

m)

0

60 18X (km)

ε = a½∑[Dm Dh] +2 2

LSMGeneral DSOb½∑[ Dd]

LSM

LSRTM image (General DSO)

Z (k

m)

0

60 18X (km)

RTM image (General DSO)

Z (k

m)

0

60 18X (km)

ε = a½∑[Dm Dh] +2 2

LSMGeneral DSOb½∑[ Dd]

LSM

Numerical ExamplesAngle gathers (DSO)

Z (k

m)

0

60 18X (km)

Angle gathers (Gen. DSO) Gatthers)

Z (k

m)

0

60 18X (km)

Outline

Traveltime+waveform Inversion Generalized DSO Inversion

ε = ½∑[Dm Dh] 2ε = ½∑[Dd Dt] 2

Motivation

Numerical Tests

Summary

Summary Low+Intermediate Inversion = General DSO Inversion

Marmousi tests: DSO vs General DSO

Extension: Low+Int.+High wavenumber General DSO

ε = ½∑[Dm Dh] 2

ε = a½∑[Dm Dh] +2

b½∑[ Dd] 2

LSMGeneral DSO

Summary Limitations 1. No coherent events in CIGs, then unsuccessful 2. Expensive 3. Infancy, still learning how to walk 4. Low+intermediate wavenumber unless LSM or FWI

ThanksSponsors of the CSIM (csim.kaust.edu.sa)

consortium at KAUST & KAUST HPC