laplace transf presentation by exar sept 2014 konsep dasar dan aplikasi

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Content

• Laplace Transforms (Lecture 01)

• Fourier Transforms (Lecture 02)

• Numerical Methods (Lecture 03)

• Statistical Analysis (Lecture 04)

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Lecture 01Lecture 01Section 01Introduction to Laplace Transforms

Pengantar LaplacePengantar LaplacePierre-Simon, marquis de Laplace

(23 March 1749 – 5 March 1827)French mathematicianAstronomer Development of mathematical astronomy

and statistics.

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BASIC CONCEPT

Frequency function

L

L - 1

Time Domain Frequency domain

Time function

internetinternet

BASIC CONCEPT

tteyy −=+''

?......)( =ty

22

)1(

1)1(

+=+s

Ys

)1()1(

122 ++

=ss

Y

transform

inverse transformtt etetty −− ++−=

2

1

2

1cos

2

1)(

HOW TO……?

Laplace Transformation

•La pl a c e T r a n s f o r m a t i on … … … … . ?• Tramsformasi fungsi dari kawasan waktu (t) ke

dalam kawasan frekwensi (s)• Simbol LL L

- 1L

- 1

II DD

Inverse Laplace Transformation

•I n v e r s e L a p la c e T r a n s f o r m a t i on … … … … . ?

• Tramsformasi fungsi dari kawasan frekwensi (s) ke dalam kawasan waktu (t)

• Simbol LL L- 1

L- 1

II DD

How to make a Laplace Transforms

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt)(tf )(tf

Time Domain Frequency domain

)( f(t)L

0∫∞

ste−. dt )(tf = )(sF

How to make a Laplace Transforms

)(sF=)(tf

Time Domain Frequency domain

)(sY=)(ty

)(sX=)(tx

(......)L f(t)

f(t)(......)L y(t)

f(t)(......)L x(t)

Contoh 1

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf 1

Time Domain Frequency domain

0......1 ≥tuntuk

0......0 <tuntuk

=)(sF 0

∞−

− se st

s

ee

−−=

−− ∞ 0

s

ee ss

−−=

−∞− 0..

s−−= 10

1

s=

)(tf

Konstanta di kawasan waktu akan dibagi oleh koefisien S di kawasan frekwensi

Contoh 2

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf (t)f

Time Domain Frequency domain

2......1 ≥tuntuk

2......0 <tuntuk

)(tf

2

0∫ (t)f ste−. dt

2∫∞

ste−. dt (t)f+

2

0∫ 0 ste−. dt

2∫∞

ste−. dt1+

{=)(tf2......1 ≥tuntuk

2......0 <tuntuk

contoh2

s

e s21 −

−∞e ee. Tidak ada yang punya pangkat tak hinggaee. Tidak ada yang punya pangkat tak hingga

0−e Tidak ada pangkat dekat ke …1

2

0∫ 0 ste−. dt

2∫∞

ste−. dt1+=)(sF

=)(sF∞−

−

2s

e st

=)(sFs

e s

−

∞− .

s

e s

−

− 2.

−s−

0

s

e s

−

−2

−

0s

e s

−

−2

−

=

= =

Tundaan sebesar 2 di

kawasan waktu akan

memunculkan

Di kawasan frekwensi

se 2−

Contoh 3

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf (t)f

Time Domain Frequency domain

2......2 ≥tuntuk

2......0 <tuntuk

)(tf

2

0∫ (t)f ste−. dt

2∫∞

ste−. dt (t)f+

2

0∫ 0 ste−. dt

2∫∞

ste−. dt2+

contoh3

s

e s22 −

−∞e ee. Tidak ada yang punya pangkat tak hinggaee. Tidak ada yang punya pangkat tak hingga

0−e Tidak ada pangkat dekat ke …1

2

0∫ 0 ste−. dt

2∫∞

ste−. dt2+=)(sF

=)(sF∞−

−

2s

e st

=)(sFs

e s

−

∞− .

s

e s

−

− 2..2−

s−0

s

e s

−

−2.2−

0s

e s

−

−2.2−

=

= =

Contoh 4

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf (t)f

Time Domain Frequency domain

3......1 ≥tuntuk

3......0 <tuntuk

)(tf

3

0∫ (t)f ste−. dt

2∫∞

ste−. dt (t)f+

3

0∫ 0 ste−. dt

2∫∞

ste−. dt1+

contoh4

s

e s31 −

−∞e ee. Tidak ada yang punya pangkat tak hinggaee. Tidak ada yang punya pangkat tak hingga

0−e Tidak ada pangkat dekat ke …1

3

0∫ 0 ste−. dt

2∫∞

ste−. dt1+=)(sF

=)(sF∞−

−

3s

e st

=)(sFs

e s

−

∞− .

s

e s

−

− 3.

−s−

0

s

e s

−

−3

−

0s

e s

−

−3

−

=

= =

Resume

{=)(tfQtuntukP ≥......

Qtuntuk <......0

=)(sFs

eP Qs−Konstanta P di kawasan waktu akan dibagi oleh koefisien S di

kawasan frekwensi

Tundaan sebesar Q di kawasan waktu akan memunculkan

Di kawasan frekwensi

se 2−

Contoh 5

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf (t)f

Time Domain Frequency domain

0...... ≥tuntukt

0......0 <tuntuk

)(tf=)(sF

t

Contoh5

=)(sF 0∫∞

ste−. dtt

=∫udv ∫− vduuv steu −=

dt

ed

dt

du st )( −

=

dtedu st−=

dvtdt =

dt

dvt =

tdt

dv =

dttv ∫=

ue st =−

2

2tv =

stedt

du −=

ste−

2

2tdte st−

2

2t

= −∫

Contoh5

=)(sF 0∫∞

ste−. dtt

=∫udv ∫− vduuv tu =

dt

td

dt

du )(=

dtdu =

dvdte st =−

dt

dve st =−

stedx

dv −=

dtev st∫ −=

ut =

s

ev

st

−=

−

1=dt

du

t

s

e st

−

−dt

s

e st

−

−

= −∫

s

te st

−=

−

2s

e st−

− C+

=)(sF 2s

e

s

te stst −−

−−[]∞0

Contoh5

=)(sF []∞2s

e

s

te stst −−

−− 2s

e

s

te stst −−

−−[]0−

ee. Tidak ada yang punya pangkat tak hinggaee. Tidak ada yang punya pangkat tak hingga

Jika sudah tidak ada pangkat , dekat ke …1

=)(sF

−

−∞ ∞−∞−

2

.

s

e

s

e ss

−

−

−−

2

00.0

s

e

s

e ss

−

=)(sF 2

1

snttf =)(

=)(sF 1

!+ns

n

n faktorial karena akan di integrasi

parsial sebanyak n

kali

N+1 karena setiap setiap integrasi akan

menghasilkan sebuah faktor pembagi s

dikawasan frekwensi

Contoh 7

)( f(t)L=)(sF

0∫∞

0),( >ttf

ste−. dt{=)(tf (t)f

Time Domain Frequency domain

0......2 ≥− tuntuke t

0......0 <tuntuk

)(tf

=)(sF

te 2−

=)(sF ste−. dtte 2−

∫∞

0

dt)2( ste −−∫∞

0∞−−

−−

0

)2(

)2( s

e st

0

)2()2(

)2()2(

−−

−

−−

−−∞−−

s

e

s

e stst

−−

−

−−

−−−−∞

)2()2(

)2(0)2(

s

e

s

e ss

−−

−

−− )2(

1

)2(

0

ss

−−

−

−− )2(

1

)2(

0

ss=)(sF

)2(

10

s−−−

)2(

1

+s=)(sF

tetf 2)( −=

Mengalikan fungsi dengan koefisien sebesar

pada kawasan waktu akan mengurangi nilai fungsi pada kawasan s

dengan faktor pembagi (s+2)

te 2−

=)(sF ste−. dtte2

∫∞

0

dt)2( ste −∫∞

0∞−

−

0

)2(

)2( s

e st

0

)2()2(

)2()2(

−

−

−

−∞−

s

e

s

e stst

−

−

−

−−∞

)2()2(

)2(0)2(

s

e

s

e ss

−

−

− )2(

1

)2(

0

ss

−

−

− )2(

1

)2(

0

ss

=)(sF

)2(

10

s−−

)2(

1

−s=)(sF

te2=)(tf

Mengalikan fungsi dengan koefisien sebesar

pada kawasan waktu akan menambah nilai fungsi pada kawasan s dengan faktor pembagi

(s-2)

te2

Lecture 01Lecture 01Section 02Laplace transformfor ODEs(algebraic equation)

Lecture 01Lecture 01Section 03Review of Complex Variabel

Lecture 01Lecture 01Section 04Laplace transformOf Trigonometric

Laplace transformLaplace transformOf Trigonometric Of Trigonometric Sin (at)Sinh (at)Cos (at)Cosh(at)

Teknik 1Teknik 2Teknik 3

30

Laplace Transform of Sin(at) 1Laplace Transform of Sin(at) 1stst Way Way

steu −=

dx

ed

dx

du st )( −

=

dxstsedu −=

tdtdv 4sin=

tdx

dv4sin=

dttv ∫= 4sin

∫ −− −−= dtsett

e stst .4

4cos

4

4cos

4

4cos tv −=

∫∫ −= vduuvudv

∫∫ −−− −−−−= dtsett

edtte ststst .4

4cos()

4

cos.4sin

( ) ?...4sin =tL

∫∞

−

0

4sin. tdte st

31

Laplace Transform of Sin(at) 1Laplace Transform of Sin(at) 1stst Way Way

∫∫ −−− −−= dtsett

edtte ststst .4

4cos

4

4cos4sin

2

∫∫ −−− −−= tdtsest

edtte ststst 4cos.44

4cos4sin

steu −=dtstsedu −=

tdtdv 4cos=dttv ∫= 4cos

4

4sin tv =

−−− ∫ −− dtse

tte

s stst .4

4sin

4

4sin

4

Laplace Transform of Sin(at) 1Laplace Transform of Sin(at) 1stst Way Way

−−−−= ∫∫ −−−− dtse

tte

stedtte stststst .

4

4sin

4

4sin

44

4cos4sin

+−−= ∫ −−− dtet

ste

ste ststst .4sin

44

4sin

44

4cos

∫ −−− −−−= dtetsst

est

e ststst .4sin4.44

4sin

44

4cos

∫ −−− −−−= dtets

test

e ststst .4sin16

4sin164

4cos 2

∫ −−− −−−= tdtes

test

e ststst 4sin.16

4sin164

4cos 2

test

etdtes

te stststst 4sin164

4cos4sin.

164sin.

2−−−− −−=+ ∫∫

test

es

te ststst 4sin164

4cos

1614sin.

2−−− −−=

+∫

test

es

te ststst 4sin164

4cos

1614sin.

2−−− −−=

+∫

Laplace Transform of Sin(at) 1Laplace Transform of Sin(at) 1stst Way Way

−−

+= −−−∫ te

ste

ste ststst 4sin

164

4cos

16

164sin.

2

−−

+= −− te

ste

sstst 4sin

164

4cos

16

162

[ ]∞

−−∞

−

+

+−=∫

0

20

4cos44sin16

44sin. ttse

s

ete st

stst

[ ] [ ]0

02

0

20cos40sin

16

44cos44sin

16

4

+

+−−

∞+∞

+−= −

−∞∞−

∞−s

ss

s

ses

ese

s

e

[ ] [ ]0

221.40.1

16

1cos4sin.0.

16

0

+

+−−

∞+∞

+−=

∞

ss

s

[ ]

+−−=

216

40

s 216

4

s+=

Laplace Transform of Sin(at)Laplace Transform of Sin(at)

2s

( )tt 4sin)( =f

{ } =)(tfL

2s

24....+

{ } ?.....)( =tfL

424+

4

Bentuk Semi Baku 7Bentuk Semi Baku 7

=∫ dxxe x 2sin4

−

2

2cos2sin

5

1 4 xxe x

=∫ dxxenx sin )cossin(12

xxnn

enx −+

( )xxe x 2cos22sin420

1 4 −=∫ dxxe x 2sin4

( ))cos()sin()(

122

qpxpqpxnenp

nx ±−±+

=±∫ dxqpxenx )sin(

Laplace Transform of Sinh(at)Laplace Transform of Sinh(at)

2s

( )tt 4sinh)( =f

{ } =)(tfL

2s

24....−

{ } ?.....)( =tfL

424−

4

Laplace Transform of Cos(at)Laplace Transform of Cos(at)

2s

( )tCost 4)( =f

{ } =)(tfL

2s

24....+

{ } ?.....)( =tfL

s24+

s

Laplace Transform of Cosh(at)Laplace Transform of Cosh(at)

2s

( )tCosht 4)( =f

{ } =)(tfL

2s

24....−

{ } ?.....)( =tfL

s24−

s

22ndnd Way Way

)sinIm(cossin titt +=iteIm=

33rdrd Way Way

)(2

1)sin( jatjat ee

jat −−=

Laplace Laplace TransformationTransformation 1

![ 1( )]

( )n at

n

AnAt e t

s a−

+=+

L

( ) cos( )1( )

cos( )1(

,

[ ])

f t A t t

A t tωω=

L 1( )2

j t j te eA t

ω ω− + =

L

1( ) 1( )2 2

j t j te eA t A t

ω ω− = +

L L

1 1

2 2

A A

s j s jω ω= +

− +( )

( ) ( )( )

2

s j s jA

s j s j

ω ωω ω

+ + −=

− +

2 2

As

s ω=

+

Lecture 01Lecture 01Section 05Laplace transform of a derivatif and integral

Laplace Laplace transform of a transform of a derivativederivative

( ) ( ) (0)d

L f t sF s fdt

= −

Primes and dots are often used as alternative notations for the derivative.

Dots are almost always used to denote time derivatives.

Primes might denote either time or space derivatives.

In problems with both time and space derivatives, primes are space derivatives and dots are time derivatives.

Note: Lower case f indicates function of time. Upper case F indicates function of s.

(Multiplication by s) = (differentiation wrt time)

Transform of DerivativesTransform of Derivatives

THEORM 1◦ Laplace of f(t) exists◦ f ’(t) exists and piecewise continuous for t>=0

THEOREM 2

)0()()'( ffsLfL −=

)0()0(')0()()( )1(21)( −−− −−−−= nnnnn ffsfsfLsfL

Differential EquationsDifferential Equations

)('" trbyayy =++

)]0(')0([ 2 ysyYs −−

basssQ

++=

2

1)(

)()( 1 YLty −=

1st step

2nd step

3rd step

1)0(' Ky =0)0( Ky =

)]0([ ysYa −+ bY+ )(sR=

)()()()]0(')0()[()( sQsRsQyyassY +++=

QuestionsQuestions

Does Laplace transform always exist?

When can Laplace transform be used to solve differential equations?

Advantages?◦ ◦

Transform of IntegralsTransform of Integrals

s

sF

s

fLdfL

t )()(})({

0

==∫ ττ

})(

{)( 1

0 s

sFLdf

t−=∫ ττ

Lecture 01Lecture 01Section 06Inverse Laplace transform

Frequency function

L

L - 1

Time Domain Frequency domain

Time function

internetinternet

BASIC CONCEPTBASIC CONCEPT

tteyy −=+''

?......)( =ty

22

)1(

1)1(

+=+s

Ys

)1()1(

122 ++

=ss

Y

transform

inverse transformtt etetty −− ++−=

2

1

2

1cos

2

1)(

HOW TO……?

INVERSE LAPLACE TRANSFORM INVERSE LAPLACE TRANSFORM METHODEMETHODE

Metode Partial Metode PartialContoh 1 Tentukan i(t) jika

)3)(4(

4)(

2 ++=

sssI

=++ )3)(4(

42 ss )3()4( 2 +

+++

s

C

s

BAs

Pangkat koefisien s dari

pembilang selalu lebih

rendah 1 pangkat

dibandingkan dengan

koefisien dari penyebut

)3)(4(

)4()3)((2

2

++++++

ss

CssBAs=

=++ )3)(4(

...................42 ss

CssBAs )4()3)(( 2 ++++

)3( +s

=4 CssBAs )4()3)(( 2 ++++

)4( 2 +s

Untuk mendapatkan nilai C, misalkan s=-3

sehingga suku pertama akan = 0

CBA )4)3(()33)()3(( 2 +−++−+−=4C)4)3(( 2 +−=

= C)13(

C =13

4

INVERSE LAPLACE TRANSFORM INVERSE LAPLACE TRANSFORM METHODEMETHODE=4 CssBAs )4()3)(( 2 ++++

433 22 CCsBBsAsAs +++++=43322 CBBsAsCsAs +++++43)3()( 2 CBsBAsCA +++++

==

Lihat koefisen s2

400 2 ++ ss

)( CA+ = 0A = C−

13

4−=

Lihat koefisen s

)3( BA+ = 0

B = A3−

=13

4.3 −− =

13

12

)3)(4(

4)(

2 ++=

sssI

)3()4( 2 ++

++

s

C

s

BAs= =)3(

134

)4(1312

134

2 ++

+

+−

ss

s

=

+

+++−

)3(

1

)4(

3

13

42 ss

s)(sI

INVERSE LAPLACE TRANSFORM INVERSE LAPLACE TRANSFORM METHODEMETHODE

)(ti = )(sIL - 1

= L - 1

+

+++−

)3(

1

)4(

3

13

42 ss

s

= L - 1

+

++

++

−)3(

1

)4(

3

)4(13

422 sss

s

= L - 1

+

++

++−

)3(

1

)2(

3

)2(13

42222 sss

s

= L - 1

++

++

+−

)3(

1

)2(

2.23

)2(13

42222 sss

s

= L - 1

+

++

++

−)3(

1

)2(

2

2

3

)2(13

42222 sss

s

=

++− − tett 32sin

2

32cos

13

4)(ti

Aplication Of Laplace IAplication Of Laplace IConsider the circuit when

the switch is closed at t = 0 with VC(0) = 1.0 V.

Solve for the current i(t) in the circuit?.

)()( tvtv Rc +

)()( tRitv RR =

dttiC

tv cc )(1

)( ∫==)(tv

+dttiC c )(1∫= )(tRiR

L

dttic )(10

16 ∫−= + )(103 tiR

)()()( tititi CR ==

5 dtti∫ )(

×5610−

610− + )(10 3 ti−= L

s

110.5 6− + )(10 3 sI−= [ ] 0.)(

1=∫ tdtti

s

dttiC

tv cc )(1

)( ∫=

=1

voltvc 1)0( =

[ ]0.6

)(10

1=− ∫ t

dtti

[ ] ==∫ 0.

)(t

dtti 610

s

110.5 6− + )(10 3 sI−= 610

1

s

s

110.5 6− + )(10 3 sI−= [ ] 0.)(

1=∫ tdtti

s

Aplication Of Laplace IAplication Of Laplace IConsider the circuit when

the switch is closed at t = 0 with VC(0) = 1.0 V.

Solve for the current i(t) in the circuit?.

2

)(

s

sI

s

610.5 −

+ 310)( −sI=

)(sIs610.5 − + 310)( −ssI=

×2s

)(sI 2610.5 s−+ 310)( −ssI= −0

)(sI2610.5 s− 310)( −ssI− =− 0

Aplication Of Laplace IIAplication Of Laplace II

)()()()( tvtvtvtv CLR ++=

dt

tdiL L )(+=)(tv )(tRiR dtti

C c )(1∫+

=0 dt

tdiR

)(+

dt

tdiL

L )(2

+ dttiC

)(1

=dt

tdiR

)(+

dt

tdiL

L )(2

+ dttiC

)(1 0 L L