Jif 315 lesson 1 Laplace and fourier transform
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Transcript of Jif 315 lesson 1 Laplace and fourier transform
JIF 315Mathematical Methods
Lesson 1
Prepared by Ms. Phoong Seuk Wai
Outline
• Laplace Transform
– Basic Laplace Transform,
– Linearity of the Laplace Function,
– Inverse Laplace Transform,
– Initial Value Problem
• Fourier Analysis
– Fourier Transform,
– Linearity of Fourier Function
Laplace Transform
• Let f(t) is a function defined for t ≥ 0.
• Then the integral
F(s) = L(f) =
• is said to be the Laplace transform of f, provided the integral converges.
• Application: Spring/mass system or a series electrical circuit
0
ste f t dt
Example 1
• Let f(t) = 1 when t ≥ 0. Find F(s).
• Solution
0
stL f e f t dt
0
1 (1)stL e dt
0
ste
s
0
0e
s
10
s
1
s
Example 2
• Evaluate Find F(s).
• Solution
3 when 0.tL e t
3 3
0
t st tL e e e dt
0
stL f e f t dt
3
0
s te dt
( 3)
0( 3)
s te
s
1
3s
Example 3
• Find the Laplace Transform of the given function:
5 3 4( ) 6 5 9t tf t e e t
• Solution:
• Refer to the Laplace Transform Table
3 1
1 1 3! 1( ) 6 5 9
( 5) 3F s
s s s s
4
6 1 30 9
5 3s s s s
Example 4
• Find the Laplace Transform of the given function:
( ) 4cos(4 ) 9sin(4 ) 2cos(10 )g t t t t
• Solution:
• Refer to the Laplace Transform Table
2 2 2 2 2 2
4( ) 4 9 2
4 4 10
s sF s
s s s
2 2 2
4 36 2
16 16 100
s s
s s s
Linearity of the Laplace Function
• If L is a linear transform
or
0 0 0
st st ste af t bg t dt a e f t dt b e g t dt
L af t bg t aL f t bL g t
aF s bG s
Example 5
• Let f(t) = 3t - 5 sin 2t when t ≥ 0. Find F(s).
• Solution
• Refer to Laplace Transform Table
3 5sin 2 3 5 sin 2L t t L t L t
• Let f(t) = 3t - 5 sin 2t when t ≥ 0. Find F(s).
• Solution
2 2 2
1 23 5
2s s
2 2
3 10
4s s
2
2 2
12 7
4
s
s s
, 0s
3 5sin 2 3 5 sin 2L t t L t L t
Trigonometric Identity and Linearity
• Example 6
• Evaluate L{sin2t}
• Solution:
• Refer to Laplace Transform Table
2 1 cos2{sin }
2
tL t L
1 1
1 cos22 2
L L t
2 1 1{sin } 1 cos2
2 2L t L L t
2
1 1 1
2 2 4
s
s s
2
2
4s s
Transform of Piecewise Function
• Example 7
Show that the Laplace transform of the step function give
1 for 0 ≤ t <c
0 otherwise
is
( )f t
1
( )cse
L f t F ss
(shown)
• Solution
0
stL f e f t dt
0
(1) (0)
c
st st
c
e dt e dt
0
cste
s
1 cse
s
1cse
s s
Example 8
0, 0 t 3
• Evaluate L{f(t)} for f(t) =
2, t 3
Solution:
0
stL f e f t dt
3
0 3
(0) (2)st ste dt e dt
3
2 ste
s
32, 0
tes
s
Inverse Laplace Transform
• f(t) =
• Table of Laplace Transform
1L F s
• If is a linear transform,
• Evaluate
• Solution:
1L
1 1 1L aF s bG s aL F s bL G s
1
5
1L
s
1 1
5 4 1
1 1 4!
4!L L
s s
41
24t
Example 9
Example 10
• Evaluate
• Refer to Laplace Transform Table
1
2
1
64L
s
Example 10
• Evaluate
• Solution
1
2
1
64L
s
1 1
2 2 2
1 1 8
864 8L L
s s
1sin 8
8t
Example 11
• Evaluate
• Solution:
• Refer to Laplace Transform Table
1
2
3 5
7
sL
s
2 2 2
3 5 3 5
7 7 7
s s
s s s
1 1 1
2 2 2
3 5 5 73
7 7 77
s sL L L
s s s
53 cos 7 sin 7
7t t
Example 12
• Find the inverse Laplace transform for the following function.
6 1 4( )
8 3F S
s s s
• Solution:
• Refer to the Laplace Transform Table
1 1 1( ) 6 4
8 3F S
s s s
8 3( ) 6 1 4t tf t e e
8 36 4t te e
Example 13
• Evaluate
• Hint:
– Using Partial Fraction
1 1
1 2 4L
s s s
• Solution:
1
1 2 4 1 2 4
A B C
s s s s s s
2 4 1 4 1 2
1 2 4
A s s B s s C s s
s s s
1 2 4 1 4 1 2A s s B s s C s s
• By comparing the coefficients of the powers of s,
When s = -21 = A(0)(2)+B(-3)(2)+C(-3)(0)
= -6BB = -1/6
When s = 1A = 1/15
When s = -4C = 1/10
1 1 1 1
1 2 4 15 1 6 2 10 4s s s s s s
1 2 4 1 4 1 2A s s B s s C s s
1 11 1 1 1
1 2 4 15 1 6 2 10 4L L
s s s s s s
1 1 11 1 1 1 1 1
15 1 6 2 10 4L L L
s s s
2 41 1 1
15 6 10
t t te e e
Initial Value Problem
• Important Note
1 ( ) ( )L y s Y s
1 '( ) ( ) (0)L y s sY s y
1 2''( ) ( ) (0) '(0)L y s s Y s sy y
Example 14
• Use the Laplace Transform to solve the initial value problems.
• Solution
1, (0) 0dy
y ydt
' 1y y 1
( ) (0) ( )sY s y Y ss
1
1 ( )s Y ss
1
( )1
Y ss s
1
11
A B
s ss s
1 ( 1)A s Bs
When s = 11 = B
When s = 01 = A
1 1( )
1Y s
s s
( ) 1 ty s e
Example 15
• Solve
• Solution:
Knowing that
Then substitute inside will get
2' 3 , (0) 1ty y e y
1 ( ) ( )L y s Y s
1 '( ) ( ) (0)L y s sY s y
1( ) (0) 3 ( )
2sY s y Y s
s
1( ) 1 3 ( )
2sY s Y s
s
1
3 ( ) 12
s Y ss
1
3 ( )2
ss Y s
s
1
( )2 3
sY s
s s
• Carrying out the partial fraction decomposition,
When s = 2 When s = 3
2 – 1 = A (2 – 3) B = 2
A = -1
1
2 3 2 3
s A B
s s s s
1 ( 3) ( 2)s A s B s
1 1 2
( )2 3 2 3
sY s
s s s s
1 11 1
( ) 22 3
y t L Ls s
2 3( ) 2t ty t e e
Fourier Analysis
• General equation of the Fourier Transform is
• General equation of the Inverse Fourier Transform is
1
2
iwxf w f x e dx
1
2
iwxf w f x e dx
Example 16
• Find the Fourier Transform of
if x > 0 and f(x) = 0 if x < 0 with a >0.
Solution:
( )axF e ( ) axf x e
1
2
iwxf w f x e dx
0
0
1 1(0)
2 2
iwx ax iwxe dx e e dx
( )
0
1
2
ax iwxf w e dx
( )
0
1
2
a iw xe dx
( )
0
1
( )2
a iw xe
a iw
1 1
( )2 a iw
1
2 ( )a iw
Example 17
1
2
iwxf w f x e dx
1
2
a
x iwx
a
f w e e dx
11
2
aiw x
a
e dx
11
2 1
aiw x
a
e
iw
1 1
1
12
iw a iw ae e
f wiw
Example 18
• Find the Fourier Transform of f(x) = 1 if and f(x) = 0 otherwise.
• Solution
1x
1
2
iwxf w f x e dx
1
1
1(1)
2
iwxf w e dx
1
1
1
2
iwxef w
iw
1
2
iw iwe eiw
Express it using Euler Formula
• Euler Formula
• So from the example 18, we can express it as
and
cos sinixe x i x
cos siniwe w i w
cos siniwe w i w
• By subtraction
• Substitute in the previous equation will get
cos sin cos siniw iwe e w i w w i w
2 sini w
sin
2
wf w
w