lesson 01 pendahuluan inversi geofisika
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8/15/2019 Lesson 01 Pendahuluan Inversi Geofisika
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Pendahuluan Inversi
Geofsika
Pelajaran 01
Defnisi & Jenis PemodelanInversi
Dr. Sugeng Pribadi2016
6!16 1
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"e#erensi
• $nalisis Da%a Geofsika Su'ri(an%o )I 200*
• Inversi Geofsika +endra Grandis I,- 200
• ,omograf $ndri Dian /ugraha I,-
• Inverse ,heor( -ill enke olombia universi%(
ldeo.olumbia.edu• Inverse Problems3 45ersies ,aran%ola 200
• Seismi Soure ehanism 7ugi 7agi )niv. ,sukuba Ja'an
• 4ar%h8uake $nal(sis +uruka9a 200* IIS44 Ja'an
• Geo:%omograf +$GI
• a%ema%ika )niversi%( o# Shaum
• Probabili%as & S%a%is%ika Danang Junaedi D4; )niv Sumu%.
• Pengan%ar a%riks
• Pemodelan a%lab
6!16 Pendahuluan Gelombang 2
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<u%line
• Defnisi & Jenis Pemodelan Inversi
• ,eknik Pendukung – a%rik Inversi
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e%ode Geofsika
• )n%uk mem'erkirakan model ba9ah:'ermukaan berdasarkan da%a hasilobservasi
• Parame%er observasi= 'arame%er model – medan gravi%asi = ra'a% massa
– medan magne% = suse'%ibili%as magne%ik
–medan lis%rik = resis%ivi%as – 9ak%u %em'uh gel seismik = kee'a%angelombang
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6!16 !
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Lecture 1
Describing Inverse Problems
h%%'3999.ldeo.olumbia.edu
olombia )niversi%(
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SyllabusLecture 01 Describing Inverse Problems
Lecture 02 Probability and Measurement Error, Part 1
Lecture 03 Probability and Measurement Error, Part 2
Lecture 04 !e L2 "orm and Sim#le Least S$uaresLecture 0% & Priori In'ormation and (eig!ted Least S$uared
Lecture 0) *esolution and +eneralied Inverses
Lecture 0- .ac/us+ilbert Inverse and t!e rade '' o' *esolution and ariance
Lecture 0 !e Princi#le o' Maimum Li/eli!ood
Lecture 05 Ineact !eories
Lecture 10 "onuni$ueness and Localied &verages
Lecture 11 ector S#aces and Singular alue Decom#ositionLecture 12 E$uality and Ine$uality 6onstraints
Lecture 13 L1 , L? "orm Problems and Linear Programming
Lecture 14 "onlinear Problems7 +rid and Monte 6arlo Searc!es
Lecture 1% "onlinear Problems7 "e8ton9s Met!od
Lecture 1) "onlinear Problems7 Simulated &nnealing and .ootstra# 6on'idence Intervals
Lecture 1- :actor &nalysisLecture 1 arima :actors, Em#ircal rt!ogonal :unctions
Lecture 15 .ac/us+ilbert !eory 'or 6ontinuous Problems; *adon9s Problem
Lecture 20 Linear #erators and !eir &d<oints
Lecture 21 :r=c!et Derivatives
Lecture 22 Eem#lary Inverse Problems, incl> :ilter Design
Lecture 23 Eem#lary Inverse Problems, incl> Eart!$ua/e Location
Lecture 24 Eem#lary Inverse Problems, incl> ibrational Problems
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Pur#ose o' t!e Lecture
distinguis! 'or8ard and inverse #roblems
categorie inverse #roblems
eamine a 'e8 eam#les
enumerate di''erent /inds o' solutions to inverse #roblems
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gravitational accelerationstravel time o' seismic 8aves
density
seismic velocity
"e8ton9s la8 o' gravity
seismic 8ave e$uation
0 a n a k a h m o d e l % e
o r i &
d a % a @ @
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t!ings t!at are measured in ane#eriment or observed in nature?
data, d A Bd1, d2, … dNC ,
t!ings you 8ant to /no8 about t!e 8orld ?
model #arameters, m A Bm1, m2, …mMC ,
relations!i# bet8een data and model #arameters
$uantitative model @or theoryA
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gravitational accelerationstravel time o' seismic 8aves
data, d A Bd1, d2, … dNC ,
model #arameters, m A Bm1, m2, …mMC ,
$uantitative model @or theoryA
density
seismic velocity
"e8ton9s la8 o' gravity
seismic 8ave e$uation
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a b m A variablemodel , A variabel da%a%erika%E
F G A variabel bebas
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Buantitative Model
Buantitative Model
mes
%
d're
mes% dobs
:or8ard !eory
Inverse !eory
estimates #redictions
observationsestimates
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Buantitative Model
Buantitative Model
m%ru
e
d're
dobs
mes%
due to
observational
error
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Buantitative Model
Buantitative Model
m%ru
e
d're
dobs
mes%
due to
observational
error due to error #ro#agation
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Cnderstanding t!e e''ects o'
observational error is central to Inverse !eory
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Pemodelan Geofsika
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Pemodelan
• Inverse modeling dimulai dari da%aobservasi seismik (ang %ersebar ke'enjuru bumi. Per%imbangan koreksi
s%asiun medium (ang dile9a%i dan'erilaku gelombang dihasilkan'endeka%an model di lokasi sumber.
• on%ohn(a3 %omograf mekanismegem'abumi s%ruk%ur la'isan bumieks'lorasi dll.
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or9ard odeling
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ujii 200*E
i%%ing model redE
9i%h observed da%ablueE
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'hase s(n%he%inearl( f% %oobserva%ion
seismogramPribadi 201E
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MATRIK INVERSI
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Matriks 4 x 2
Baris x kolom
−
=
1)
1213
41
A
.aris
olom
a%riks skalar
=
1E4%E4
E43E44 A
= 420
32124 A
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Hukum perkalian matriks :
A(BC) = (AB)C
A(B+C) = AB+AC
(B+C)A = BA+CA
A(B-C)=AB-AC(B-C)A = BA-CA
A(BC) = (aB)C= B(aC)
AI = IA = A
A2 = A A
A = A2 A
A4 = A A
A! = A4 A
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Perkalian ma%riks
=
0
1
3
B[ ]123= A
"entukan #asil A$B %an B$A
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Perkalian ma%riks
=
0
1
3
B
[ ] [ ] [ ]11A0E1@A1E2@A3E3@
0
1
3
E123E =++=
= B A
[ ]123= A
[ ]
=
=
=
000
123
3)5
1E02E03E0
1E12E13E1
1E32E33E3
123E
0
1
3
E A B
"entukan #asil A$B %an B$A
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PERPANGKATAN MATRIKS
3636
"entukan #asil 2A& + A'
−=
02
11 A
−
−=
−
−⋅=
44
2)
22
1322
2 A
−
−=
−
−⋅=
))
51%
22
3%33
3 A
−
−=
−
−+
−
−=+ 1010
-5))51%
442)32 32 A A
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Matriks bujursangkar (persegi) a%ala# matriks an
*erukuran n x n
Matriks nol a%ala# matriks an setiap entri atau elemennaa%ala# semuanya bilangan nol
i,at-si,at %ari matriks nol :-A+=A. /ika ukuran matriks A = ukuran matriks
-A$=. *eitu /ua $A=0
=
13
41 A
=
00
00
00
23 xO
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Matriks Diagonal a%ala# matriks persei an semua elemendiatas dan dibawah diagonalnya adalah nol. 1inotasikan
se*aai 10
Contoh :
Matriks Skalar a%ala# matriks %iaonal an semua elemen
pa%a diagonalnya sama
=%00020
001
33 x D
=
%00
0%0
00%
33 x D
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Matriks Identitas a%ala# matriks skalar an elemen-elemenpa%a diagonal utamanya bernilai 0
i,at-si,at matriks i%entitas :
A$I=A
I$A=A
Matriks Segitiga !tas a%ala# matriks persei an elemen %i
*aa# %iaonal utamana *ernilai nol Matriks Segitiga "awah a%ala# matriks persei an elemen
%i atas %iaonal utamana *ernilai nol
=
100
010
001
D
=
)00
210
%42
A
=
1%2
043
001
B
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NOTASI DETERMINAN
Pada matriks 2x2 cara menghitung nilaideterminannya adalah :
Contoh :
=
2221
1211
aa
aa A
21122211Adet@ aaaa A −=
= 31
%2
A 1%)Adet@
=−= A
2221
1211Adet@
aa
aa A =
31
%2
Adet@ = A
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METODE SARRUS
Pada matriks 3x3 cara menghitung nilaideterminannya adalah menggunakan Metode Sarrus
Metode Sarrus hanya untuk matrix berdimensi 3x3
122133112332132231322113312312332211Adet@ aaaaaaaaaaaaaaaaaa A −−−++=
=333231
232221
131211
aaa
aaa
aaa
A
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Contoh :
Nilai Determinan dicari menggunakan metodeSarrus
det(A) = (-2·1 ·-1) + (2 ·3 ·2) + (-3 ·-1 ·0) – (-3 ·1 ·2) –(-2 ·3·0)-(2 ·-1 ·-1)
= 2 +12+0+6-0-2= 18
−−
−−
−−=
02
11
22
102
311
322
A
METODE SARRUS
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METODE MINOR
=
333231
232221
131211
aaa
aaa
aaa
A
Unsur aijadalah determinan yang berasal darideterminan orde ke-n tadi dikurangi dengan bariske-i dan kolom ke-j. Dinotasikan dengan Mij
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3231
2221
13
3331
2321
12
3332
2322
11
131312121111
333231
232221
131211
det
aa
aaa
aa
aaa
aa
aaa
M a M a M aaaa
aaa
aaa
A
+−=
+−=
=
1
)102
A)@A%>@2A1>@2
A20A>@3@A)1>@2A01>@2
102
311
322
=++=
+−−−−=−−+−−−−−=
−−
−−= A
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KOFAKTOR MATRIKS
Kofaktor dari baris ke-i dan kolom ke-j dituliskandengan
Contoh :Kofaktor dari elemen a11
2323
32
23 A1@ M M c −=−= +
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TEOREMA LAPLACE
4747
Determinan dengan Ekspansi Kofaktor Pada BarisMisalkan ada sebuah matriks A berordo 3x3
Determinan Matriks A dengan metode ekspansikofaktor baris pertama|A|
=
333231
232221
131211
aaa
aaa
aaa
A
3231
2221
13
3331
2321
12
3332
2322
11
131312121111
131312121111
aa
aaa
aa
aaa
aa
aaa
M a M a M a
cacaca
+−=
+−=
++=
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TEOREMA LAPLACE
4848
Determinan Matriks A dengan metode ekspansi kofaktor baris
kedua|A|
Determinan Matriks A dengan metode ekspansi kofaktor barisketiga
|A|
3231
1211
23
3331
1311
22
3332
1312
21
232322222121
232322222121
aa
aaa
aa
aaa
aa
aaa
M a M a M a
cacaca
+−=
+−=
++=
2221
1211
33
2321
1311
32
2322
1312
31
333332323131
333332323131
aa
aaa
aa
aaa
aa
aaa
M a M a M a
cacaca
+−=
+−=
++=
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3231
2221
13
3331
2321
12
3332
2322
11
131312121111
333231
232221
131211
det
aa
aaa
aa
aaa
aa
aaa
M a M a M aaaa
aaa
aaa
A
+−=
+−=
=
−
−−−
=102
311
322
A
oba kerjakan dengan ko#ak%or baris ke:2 dan ke:
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DET MATRIKS SEGITIGA
5050
Jika A adalah matriks segitiga bujur sangkar berupasegitiga atas atau segitiga bawah maka nilai det(A)adalah hasil kali diagonal matriks tersebut
Contoh
dst aaa A ⋅⋅⋅⋅⋅⋅= 332211Adet@
125)45)A3@2Adet@ −=⋅⋅⋅−⋅= A
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TRANSPOSE MATRIKS
5151
Be*erapa i,at Matriks "ranspose :
T T
T T T
T T
T T T
kAkA
A B AB
A A
B A B A
=
=
=
+=+
A@
A@
A@
A@
=
314
131 A
=
31
13
41t A
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TRANSPOSE MATRIKS
5353
3em*uktian aturan no 2 :
=
232221
131211
aaa
aaa A
=
2313
2212
2111
aa
aaaa
AT
=
=232221
131211
2313
2212
2111
A@aaaaaa
aa
aa
aa
A
T
T T
,4"-)K,I
-uk%ikan si#a% ma%riks %rans'ose ke:
dan ke:> L
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INVERS MATRIKS
5555
Matriks invers dari suatu matriks A adalah matriks Byang apabila dikalikan dengan matriks A memberikansatuan I
AB = I
Notasi matriks invers :Sebuah matriks yang dikalikan matriks inversenya akanmenghasilkan matrik satuan
Jika
Maka
1− A
I A A =−1
=
d c
ba A
−
−−
=−
ac
bd
bcad A
11
AA@@Adet@
11 Aadjoin A
A =−
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INVERS MATRIX
5656
Langkah-langkah untuk mencari invers matriks M yang berordo 3x3 adalah :- Cari determinan dari M- Transpose matriks M sehingga menjadi
- Cari adjoin matriks- Gunakan rumus
T M
AA@@Adet@
11 M adjoin M
M =−
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INVERS MATRIX
5757
Contoh Soal :
- Cari Determinannya : det(M) = 1(0-24)-2(0-20)+3(0-5) = 1- Transpose matriks M
=
0)%
410
321
M
=
043
)12
%01T M
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INVERS MATRIX
5858
- Temukan matriks kofaktor dengan menghitung minor-minor matriksnya
- Hasilnya :
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14%
41%20
%124
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+−+
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14%
41%20
%124
8/15/2019 Lesson 01 Pendahuluan Inversi Geofisika
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INVERS MATRIX
5959
Hasil Adjoinnya :
Hasil akhir
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−=
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−=−
14%
41%20
%124
14%
41%20
%124
1
11 M
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−
14%
41%20
%124
8/15/2019 Lesson 01 Pendahuluan Inversi Geofisika
http://slidepdf.com/reader/full/lesson-01-pendahuluan-inversi-geofisika 60/61
Soal 1
Selesaikan 'ersamaan beriku% denganme%ode 4liminasi Gauss
Kerjakan dengan ara manual L
8/15/2019 Lesson 01 Pendahuluan Inversi Geofisika
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,)G$S I/DIMID)
• Kerjakan soal ma%rik dia%as
• -ua%lah essa( (ang menjelaskan PK; Sesar;embang 2016 mem'un(ai unsur 'emodelanmuka dan belakang berdasarkan me%ode:me%ode
(ang anda di'ergunakanL• Selesaikan salah sa%u me%ode dalam PK;
;embang menggunakan ma%rik me%ode 4liminasiGauss
• Di%ulis %angan• Ker%as $>
• Dikum'ulkan hari Senin sebelum jam 16.00 I-