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laplace transf presentation by exar sept 2014 konsep dasar dan aplikasi
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AZ. NASUTION
CEVEST
©2013
Content
• Laplace Transforms (Lecture 01)
• Fourier Transforms (Lecture 02)
• Numerical Methods (Lecture 03)
• Statistical Analysis (Lecture 04)
Advanced Search
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.
scan
Physical World Signal World
internetinternet
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