laplace transf presentation by exar sept 2014 konsep dasar dan aplikasi

1. AZ. NASUTION
CEVEST
©2013

2. Content
•
•
•
•
Laplace Transforms (Lecture 01)
Fourier Transforms (Lecture 02)
Numerical Methods (Lecture 03)
Statistical Analysis (Lecture 04)
Advanced
Search

3. Lecture 01
Section 01
Introduction to Laplace
Transforms

4. Pengantar Laplace
Pierre-Simon, marquis
de Laplace
(23 March 1749 – 5 March 1827)
French mathematician
Astronomer
Development of mathematical astronomy
and statistics.

5. BASIC CONCEPT
Mail
E-mail
scan
internet
internet
print
Physical World
Signal World

6. BASIC CONCEPT
Time function
y ' '+ y = te
HOW TO……? Frequency function
−t
transform
1
( s + 1)Y =
( s + 1) 2
2
L
internet
internet
y (t ) =...... ?
1
1
1
y (t ) = − cos t + te − t + e − t
2
2
2
Time Domain
L
-1
1
Y=
( s + 1) 2 ( s 2 + 1)
inverse transform
Frequency domain

7. Laplace Transformation
•Laplace Transformation………….?
• Tramsformasi fungsi dari kawasan waktu (t) ke
dalam kawasan frekwensi (s)
• Simbol
11
L
I
--
L
D

8. Inverse Laplace Transformation
•Inverse Laplace
Transformation………….?
• Tramsformasi fungsi dari kawasan frekwensi (s)
11
ke dalam kawasan waktu (t)
• Simbol
L
I
--
L
D

9. How to make a Laplace Transforms
Time Domain
Frequency domain
L ( f(t))
f (t )
L ( f(t))
∞
∫
0
∞
F (s ) =
∫
f (t ) .e −st dt
0
f (t ), t > 0
f (t ) .e −st dt = F (s )

10. How to make a Laplace Transforms
Time Domain
Frequency domain
L (......)
f(t)
f (t )
L (......)
y(t)
f(t)
y (t )
L (......)
x(t)
f(t)
x(t )
= F (s )
= Y (s )
= X (s )

11. Contoh 1
Konstanta di kawasan waktu
akan dibagi oleh koefisien S di kawasan frekwensi
Time Domain
Frequency domain
L ( f(t))
1....untuk ..t ≥ 0
F (s ) =
f (t ) = {
0....untuk ..t < 0
f (t )
∞
∫
.e −st dt
1
0
f (t ), t > 0
− st ∞
e
F (s ) =
−s
−∞
e −e
=
−s
=
0
−0
e
− s .∞
−e
−s
− s .0
0− 1 1
=
=
−s
s

12. Contoh 2
Time Domain
Frequency domain
L ( f(t))
1....untuk ..t ≥ 2
F (s ) =
f (t ) = {
0....untuk ..t < 2
f (t )
∞
∫
f (t) .e −st dt
0
2
∫
f (t) .e −st dt +
0
f (t ), t > 0
∞
∫
f (t) .e − st dt
2
2
∫
0
.e −st dt +
0
∞
∫
2
.e − st dt
1

13. contoh2
Tundaan
sebesar 2 di
kawasan
waktu
akan
memunculkan
e −2 s
Di kawasan
frekwensi
e
−∞
e
−0
ee. Tidak ada yang punya pangkat tak hingga
ee. Tidak ada yang punya pangkat tak hingga
Tidak ada pangkat dekat ke …1
2
F (s ) = ∫
0 .e −st dt +
0
F (s ) =
F (s ) =
∞
∫
.e − st dt
1
2
∞
e 


− s 2

− st
− s .∞
e
−s
− s .2
e
−
−s
e −2 s
=0 − − s
=
0
=
−s
−2 s
1e
s
e −2 s
−
−s

14. Contoh 3
Time Domain
Frequency domain
L ( f(t))
2....untuk ..t ≥ 2
F (s ) =
f (t ) = {
0....untuk ..t < 2
f (t )
∞
∫
f (t) .e −st dt
0
2
∫
f (t) .e −st dt +
0
f (t ), t > 0
∞
∫
f (t) .e − st dt
2
2
∫
0
.e −st dt +
0
∞
∫
2
.e − st dt
2

15. contoh3
e
−∞
e
−0
ee. Tidak ada yang punya pangkat tak hingga
ee. Tidak ada yang punya pangkat tak hingga
Tidak ada pangkat dekat ke …1
2
F (s ) = ∫
0 .e −st dt +
0
F (s ) =
F (s ) =
∞
∫
.e − st dt
2
2
∞
e 


− s 2

− st
− s .∞
e
−s
− s .2
2.e
−
−s
2.e −2 s
=0 − − s
=
0
=
−s
−2 s
2e
s
2.e −2 s
−
−s

16. Contoh 4
Time Domain
Frequency domain
L ( f(t))
1....untuk ..t ≥ 3
F (s ) =
f (t ) = {
0....untuk ..t < 3
f (t )
∞
∫
f (t) .e −st dt
0
3
∫
f (t) .e −st dt +
0
f (t ), t > 0
∞
∫
f (t) .e − st dt
2
3
∫
0
.e −st dt +
0
∞
∫
2
.e − st dt
1

17. contoh4
e
−∞
e
−0
ee. Tidak ada yang punya pangkat tak hingga
ee. Tidak ada yang punya pangkat tak hingga
Tidak ada pangkat dekat ke …1
3
F (s ) = ∫
0 .e −st dt +
0
F (s ) =
F (s ) =
∞
∫
.e − st dt
1
2
∞
e 


− s 3

− st
− s .∞
e
−s
− s .3
e
−
−s
e −3s
=0 − − s
=
0
=
−s
−3 s
1e
s
e −3s
−
−s

18. Resume
P....untuk ..t ≥ Q
f (t ) = { 0....untuk..t < Q
Konstanta P di kawasan waktu
akan dibagi oleh koefisien S di
kawasan frekwensi
Tundaan sebesar Q di kawasan
waktu akan memunculkan
e
−2 s
Di kawasan frekwensi
−Qs
Pe
F (s ) =
s

19. Contoh 5
Time Domain
Frequency domain
L ( f(t))
t....untuk ..t ≥ 0
F (s ) =
f (t ) = { 0....untuk..t < 0
f (t )
F (s ) =
∞
∫
0
f t(t) .e −st dt
f (t ), t > 0

20. Contoh5
∞
F (s ) = ∫ t .e
−st
dt
0
∫ udv = uv −∫vdu
=
−∫
tdt = dv
e − st = u
dv
e st
u = e −−st
t=
dt
du d (e − st )
=
dv
dt
dt
=t
dt
du
= e − st
v = ∫ tdt
dt
− st
tt22
du = e dt
v=
2
2

21. Contoh5
∞
F (s ) = ∫ t .e
−st
dt
0
∫ udv = uv −∫vdu
=
−∫
t =u
u = tt
e − st dt = dv
dv
− st
e =
dt
du d (t )
=
dv
dt
dt
= e − st
dx
du
=1
v = ∫ e − st dt
dt
e −−st
e st
du = dt
te − st e − st
+C
=
− 2
−s
s
v=
− st
− st
−s
−s
te
e
∞
− 2
F (s ) =
−s
s
0
[]

22. Contoh5
ee. Tidak ada yang punya pangkat tak hingga
ee. Tidak ada yang punya pangkat tak hingga
Jika sudah tidak ada pangkat , dekat ke …1
[
[] ]
F (s ) =
F (s ) =
F (s ) =
te − st e − st
− 2
−s
s
0
∞
te − st e − st
−
 ∞.e − s∞ e − s∞ 

 − s − s2  −



1
f (t ) = t n
s2
F (s ) =
−s
−
s2
 0.e − s 0 e − s 0 

 − s − s2 



n!
s n +1
n faktorial
karena akan
di integrasi
parsial
sebanyak n
kali
N+1 karena setiap
setiap integrasi akan
menghasilkan sebuah
faktor pembagi s
dikawasan frekwensi

23. Contoh 7
Time Domain
e
−2 t
Frequency domain
L ( f(t))
....untuk ..t ≥ 0
F (s ) =
f (t ) = { 0....untuk..t < 0
f (t )
F (s ) =
∞
∫
0
2t
ef−(t) .e −st dt
f (t ), t > 0

24. f (t ) = e
F (s ) =
−2 t
∞
∫
e −2t .e −st dt
F (s ) =
0−
0
∞
∫
e
t ( −2 − s )
dt
0

 

0
1

 −
 ( −2 − s )   ( −2 − s ) 


 

F (s ) =
1
( −2 − s )
1
( s + 2)
∞
 e t ( −2 −s ) 


(−2 − s ) 0

∞
 e t ( −2 −s ) 
 e t ( −2 −s ) 

 −

( −2 − s ) 

(−2 − s ) 0
 e ∞( −2 −s )   e 0 ( −2 −s ) 

 ( −2 − s )  −  ( −2 − s ) 
 


 


 

0
1

 −
 ( −2 − s )   ( −2 − s ) 


 

Mengalikan fungsi
dengan koefisien sebesar
e −2t
pada kawasan waktu
akan mengurangi nilai
fungsi pada kawasan s
dengan faktor pembagi
(s+2)

25. f (t ) = e 2t
F (s ) =
∞
F (s ) =
e 2t .e −st dt
∫
0−
0
∞
∫
 0   1 

 (2 − s)  − (2 − s ) 
 


 

 0   1 

 (2 − s)  − (2 − s ) 
 


 

et ( 2− s ) dt
0
∞
e



(2 − s ) 0

t ( 2 −s )
∞
e

 e t ( 2 −s ) 

 −

(2 − s) 

(2 − s ) 0
t ( 2 −s )
 e ∞( 2 −s )   e 0 ( 2 −s ) 

 (2 − s)  −  (2 − s ) 
 


 

F (s ) =
1
(2 − s )
1
( s − 2)
Mengalikan fungsi dengan
koefisien sebesar
e 2t waktu
pada kawasan
akan menambah nilai fungsi pada
kawasan s dengan faktor pembagi
(s-2)

26. Lecture 01
Section 02
Laplace transform
for ODEs
(algebraic equation)

27. Lecture 01
Section 03
Review of Complex Variabel

28. Lecture 01
Section 04
Laplace transform
Of Trigonometric

29. Laplace transform
Of Trigonometric
Sin
(at)
Sinh (at)
Cos (at)
Cosh(at)
Teknik
1
Teknik 2
Teknik 3

30. Laplace Transform of Sin(at) 1st Way
L ( sin 4t ) = ... ?
∞
∫e
− st
dv = sin 4tdt
du d (e − st ) dv
= sin 4t
=
dx
dx
dx
u = e − st
. sin 4tdt
0
∫ udv = uv − ∫ vdu
du = − se st dx
v = ∫ sin 4tdt
cos 4t
v=−
4
e −st sin 4tdt = e −st . −
∫
= −e
−st
cos t
cos 4t
) − (∫ −
. − se −st dt
4
4
cos 4t
cos 4t
−∫
.se −st dt
4
4
30

31. Laplace Transform of Sin(at) 1st Way
2
cos 4t
cos 4t − st
∫ e sin 4tdt = −e 4 − ∫ 4 .se dt
s
− st
− st cos 4t
e sin 4tdt = −e
− ∫ se − st . cos 4tdt
∫
4
4
− st
− st
dv = cos 4tdt
du = − se st dt v = ∫ cos 4tdt
u = e − st
v=
sin 4t
4
s  − st sin 4t
sin 4t

− st
− e
−∫
. − se dt 
4
4
4

31

32. Laplace Transform of Sin(at) 1st Way
e −st sin 4tdt = −e −st
∫
cos 4t s  −st sin 4t
sin 4t

− e
−∫
. − se −st dt 
4
4
4
4

= −e −st
= −e −st
= −e −st
= −e −st
∫e
−st
cos 4t
4
cos 4t
4
cos 4t
4
cos 4t
4
s  −st sin 4t s

e
+ ∫ sin 4t.e −st dt 
4
4
4


s
sin 4t s s
− e −st
− . ∫ sin 4t.e −st dt
4
4
4 4
s −st
s2
− e sin 4t − ∫ sin 4t.e −st dt
16
16
s −st
s 2 −st
− e sin 4t − ∫ e . sin 4tdt
16
16
−
s2
s −st
−st
−st cos 4t
. sin 4t + ∫ e . sin 4tdt = −e
− e sin 4t
16
4
16
∫e
−st
 s2 
s −st
−st cos 4t
. sin 4t 1 +  = −e
− e sin 4t
16 
4
16


33. Laplace Transform of Sin(at) 1st Way
∫e
−st
 s2 
cos 4t
s
. sin 4t 1 +  = −e −st
− e −st sin 4t
4
16
 16 
16  −st cos 4t
s

−e
− e −st sin 4t 
16 + s 2 
4
16


16  −st cos 4t
s

=
−e
− e −st sin 4t 
16 + s 2 
4
16


e −st . sin 4t =
∫
∞
∫e
0
−st
∞
 − 4e

−st
. sin 4t = 
se sin 4t + 4 cos 4t 
2
16 + s
0
−st
[
]
∞
 − 4e − s ∞

 − 4e − s 0

−s∞
=
se sin 4∞ + 4 cos 4∞  − 
se −s 0 sin 0 + 4 cos 0 
2
2
 16 + s

16 + s
0
∞
 −0
[ s.0. sin ∞ + 4 cos ∞] −  −1 2 [1.0 + 4.1]
=
2

16 + s

16 + s


0
[
 −4 
= [ 0] − 
2 
16 + s 
]
4
=
16 + s 2
[
]

34. Laplace Transform of Sin(at)
f (t ) = sin ( 4t )
L { f (t )} = ..... ?
.... + 4 2
s
2
L { f (t )} =
4
s
2
+4
4
2

35. Bentuk Semi Baku 7
e sin 2 xdx =
∫
4x
e sin xdx =
∫
nx
e sin 2 xdx =
∫
4x
∫e
nx
sin( px ± q )dx =
1 4x 
cos 2 x 
e  sin 2 x −

5
2 

nx
e
(n sin x − cos x)
2
n +1
1 4x
e ( 4 sin 2 x − 2 cos 2 x )
20
1
e nx ( n sin( px ± q ) − p cos( px ± q ) )
( p 2 + n2 )

36. Laplace Transform of Sinh(at)
f (t ) = sinh ( 4t )
L { f (t )} = ..... ?
.... − 4 2
s
2
L { f (t )} =
4
s
2
−4
4
2

37. Laplace Transform of Cos(at)
f (t ) = Cos ( 4t )
L { f (t )} = ..... ?
.... + 4 2
s
2
L { f (t )} =
s
s
2
+4
s
2

38. Laplace Transform of Cosh(at)
f (t ) = Cosh( 4t )
L { f (t )} = ..... ?
.... − 4 2
s
2
L { f (t )} =
s
s
2
−4
s
2

39. 2nd Way
sin t = Im(cos t + i sin t )
= Im e it

40. 3rd Way
1
sin( at ) =
(e jat − e − jat )
2j

41. Laplace
Transformation
An !
L [ At e 1(t )] =
n +1
(s + a)
n − at
f (t ) = A cos(ω t )1(t ),
jω t
− jω t
 e +e

1(t ) 
L [ A cos(ω t )1(t )] = L  A
2


 e jω t

 e − jω t

= L A
1(t )  + L  A
1(t ) 
2
 2



A 1
A 1
A ( s + jω ) + ( s − jω )
=
+
=
2 s − jω 2 s + jω 2 ( s − jω ) ( s + jω )
As
= 2
2
s +ω

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

43. Laplace
transform of a
derivative
d

L  f (t )  = sF ( s ) − f (0)
 dt

Note:
Lower case f indicates function
of time.
Upper case F indicates function
of s.
(Multiplication by s) =
(differentiation wrt time)
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.

44. Transform of Derivatives
 THEORM
1
◦ Laplace of f(t) exists
◦ f ’(t) exists and piecewise continuous for t>=0
L( f ' ) = sL( f ) − f (0)
 THEOREM
2
L( f ( n ) ) = s n L( f ) − s n−1 f (0) − s n− 2 f ' (0) −  − f ( n−1) (0)

45. Differential Equations
y"+ ay '+by = r (t )
1st step
y (0) = K 0
y ' ( 0) = K 1
[ s 2Y − sy (0) − y ' (0)] + a[ sY − y (0)] + bY = R (s )
2nd step
3rd step
1
Q(s) = 2
s + as + b
Y ( s ) = [( s + a ) y (0) + y ' (0)]Q( s ) + R ( s )Q( s )
y (t ) = L−1 (Y )

46. Questions
 Does
Laplace transform always exist?
 When
can Laplace transform be used to solve differential
equations?
 Advantages?
◦
◦

47. Transform of Integrals
t
L( f ) F ( s )
L{∫ f (τ )dτ } =
=
s
s
0
t
∫
0
F ( s)
f (τ )dτ = L {
}
s
−1

48. Lecture 01
Section 06
Inverse Laplace transform

49. BASIC CONCEPT
Time function
y ' '+ y = te
HOW TO……? Frequency function
−t
transform
1
( s + 1)Y =
( s + 1) 2
2
L
internet
internet
y (t ) =...... ?
1
1
1
y (t ) = − cos t + te − t + e − t
2
2
2
Time Domain
L
-1
1
Y=
( s + 1) 2 ( s 2 + 1)
inverse transform
Frequency domain

50. INVERSE LAPLACE TRANSFORM METHODE
Metode Partial
Pangkat
koefisien s dari
pembilang
selalu lebih
rendah 1
pangkat
dibandingkan
dengan
koefisien dari
penyebut
Metode Partial
Contoh 1
Tentukan i(t) jika
I ( s) =
4
( s 2 + 4)( s + 3)
4
As + B
C
= 2
+
( s 2 + 4)( s + 3)
( s + 4) ( s + 3)
4=
4=
=
=
Untuk mendapatkan
nilai C, misalkan s=-3
sehingga suku
pertama akan = 0
( As + B )( s + 3) + ( s 2 + 4)C
( A( −3) + B )(−3 + 3) + ((−3) 2 + 4)C
((−3) 2 + 4)C
(13)C
4
C =
13
( As + B )( s + 3) + ( s 2 + 4)C
=
( s 2 + 4)( s + 3)
4................... ( As + B )( s + 3) + ( s 2 + 4)C
=
2
( s + 4)( s + 3)
( s 2 + 4) ( s + 3)

51. INVERSE LAPLACE TRANSFORM METHODE
2
4 = ( As + B )( s + 3) + ( s + 4)C
Lihat koefisen s
= As + 3 As + Bs + 3B + Cs + C 4
2
2
= As + Cs + 3 As + Bs + 3B + C 4
2
0 s 2 + 0 s + 4 = ( A + C ) s + (3 A + B ) s + 3B + C 4
2
Lihat koefisen
s2
( A + C) = 0
A = −C
4
= −
13
2
(3 A + B ) = 0
B = − 3A
= − 3. −
4
12
=
13
13
4
12
4
− s+
4
As + B
C
13 + 13
= 2
= 13
I ( s) = 2
+
( s 2 + 4)
( s + 3)
( s + 4)( s + 3)
( s + 4) ( s + 3)
4  − s +3
1 
I (s ) =
 ( s 2 + 4) + ( s + 3) 
13 


52. INVERSE LAPLACE TRANSFORM METHODE
i (t ) =
L
I (s ) =
-1
L
-1
=
L
-1
=
L
-1
=
L
-1
=
L
-1
i (t ) =
4  − s +3
1 
 ( s 2 + 4) + ( s + 3) 
13 

4  −s
3
1 
 ( s 2 + 4) + ( s 2 + 4) + ( s + 3) 
13 

4  −s
3
1 
 ( s 2 + 2 2 ) + ( s 2 + 2 2 ) + ( s + 3) 
13 

3


.2
4 
s
1 
− 2
+ 22 2 +


13  ( s + 2 2 ) ( s + 2 ) ( s + 3) 


4 
s
3
2
1 
− ( s 2 + 2 2 ) + 2 ( s 2 + 2 2 ) + ( s + 3) 
13 

4 
3

− cos 2t + sin 2t + e −3t 
13 
2



53. Aplication Of Laplace I
 Consider
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?.
1
∫ i c (t )dt
C
v R (t ) = RiR (t )
v(t ) = vc (t ) + vR (t )
=
5 =
L
1
∫ i c (t )dt
C
+ RiR (t )
1
i (t )dt + 103 iR (t )
−6 ∫ c
10
5 10 −6 = L
5.10 −6
vc (t ) =
1 = 1
s
s
×10 −6
i (t )dt + 10 −3 i (t )
∫
[∫ i(t )dt ].
t =0
+10 −3 I ( s )
i R (t ) = iC (t ) = i (t )

54. vc (t ) =
1
∫ i c (t )dt
C
1
10 −6
1=
[∫i(t )dt ]
.t =0
5.10 −6
5.10 −6
vc (0) = 1volt
[∫i(t )dt ]
.t =0
= 10 6
[
]
−3
1 = 1
i (t )dt .t =0 +10 I ( s )
s
s ∫
1 = 1 6
10 + 10 −3 I ( s )
s
s

55. Aplication Of Laplace I
 Consider
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?.
I (s)
5.10 −6
=
+ I ( s )10 −3
s2
s
×
s2
−3
5.10 −6 s = I (s ) + sI ( s )10
0 = I (s ) + sI ( s )10 −3 − 5.10 −6 s 2
−3
5.10 −6 s 2 − sI ( s )10 − I (s ) = 0

56. Aplication Of Laplace II
v(t ) = vR (t ) + vL (t ) + vC (t )
di (t ) 1
v (t ) = RiR (t ) + L L
+ ∫ i c (t ) dt
C
dt
di 2 L (t )
di (t )
+ 1 i (t )dt
R
+L
0=
dt
C
dt
L
di 2 L (t )
di (t )
+ 1 i (t )dt = L
L
R
+
dt
C
dt
0