Jif 315 lesson 1 Laplace and fourier transform

JIF 315
Mathematical 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
st
e f t dt




Example 1
• Let f(t) = 1 when t ≥ 0. Find F(s).
• Solution
   
0
st
L f e f t dt


 
 
0
1 (1)st
L e dt


 
0
st
e
s


 
  
 
0
0
e
s

 
   
 
1
0
s
 
1
s


Example 2
• Evaluate Find F(s).
• Solution
 3
when 0.t
L e t

 3 3
0
t st t
L e e e dt

  
 
   
0
st
L f e f t dt


 
 3
0
s t
e dt

 
 
( 3)
0
( 3)
s t
e
s

 
 
  
  
1
3s



Example 3
• Find the Laplace Transform of the given
function:
5 3 4
( ) 6 5 9t t
f t e e t
   

• Solution:
• Refer to the Laplace Transform Table
3 1
1 1 3! 1
( ) 6 5 9
( 5) 3
F 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:
2 2 2 2 2 2
4
( ) 4 9 2
4 4 10
s s
F 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 st
e 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 2
3 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:
2 1 cos2
{sin }
2
t
L t L
 
  
 
   1 1
1 cos2
2 2
L L t 

   2 1 1
{sin } 1 cos2
2 2
L 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
( )
cs
e
L f t F s
s

   
 

(shown)
• Solution
   
0
st
L f e f t dt


 
0
(1) (0)
c
st st
c
e dt e dt

 
  
0
c
st
e
s
 
  
 
1 cs
e
s



1cs
e
s s

 
    
 

Example 8
0, 0 t 3
• Evaluate L{f(t)} for f(t) =
2, t 3
Solution:
   
0
st
L f e f t dt


 
3
0 3
(0) (2)st st
e dt e dt

 
  
3
2 st
e
s

 
  
 
3
2
, 0
t
e
s
s

 

Inverse Laplace Transform
• f(t) =
• Table of Laplace Transform
  1
L F s

• If is a linear transform,
• Evaluate
• Solution:
1
L
       1 1 1
L aF s bG s aL F s bL G s          
     
1
5
1
L
s
  
 
 
1 1
5 4 1
1 1 4!
4!
L L
s s
 

   
   
   
41
24
t
Example 9

Example 10
• Evaluate
1
2
1
64
L
s
  
 
 

Example 10
• Evaluate
• Solution
1
2
1
64
L
s
  
 
 
1 1
2 2 2
1 1 8
864 8
L L
s s
    
   
    
1
sin 8
8
t

Example 11
• Evaluate
• Solution:
1
2
3 5
7
s
L
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 7
3
7 7 77
s s
L L L
s s s
  
      
      
        
5
3cos 7 sin 7
7
t t 

Example 12
• Find the inverse Laplace transform for the
following function.
6 1 4
( )
8 3
F S
s s s
  
 

• Solution:
1 1 1
( ) 6 4
8 3
F S
s s s
   
     
    
  8 3
( ) 6 1 4t t
f t e e  
8 3
6 4t t
e e  

Example 13
• Evaluate
• Hint:
– Using Partial Fraction
   
1 1
1 2 4
L
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 = -2
1 = A(0)(2)+B(-3)(2)+C(-3)(0)
= -6B
B = -1/6
When s = 1
A = 1/15
When s = -4
C = 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 4
L L
s s s s s s
     
       
       
       
1 1 11 1 1 1 1 1
15 1 6 2 10 4
L L L
s s s
 
  2 41 1 1
15 6 10
t t t
e 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) 0
dy
y y
dt
  
' 1y y 
1
( ) (0) ( )sY s y Y s
s
  
  1
1 ( )s Y s
s
 

 
1
( )
1
Y s
s s


 
1
11
A B
s ss s
 

1 ( 1)A s Bs  
When s = 1
1 = B
When s = 0
1 = A
1 1
( )
1
Y s
s s
 

( ) 1 t
y s e 

Example 15
• Solve
• Solution:
Knowing that
Then substitute inside will get
2
' 3 , (0) 1t
y y e y  
 1
( ) ( )L y s Y s

 1
'( ) ( ) (0)L y s sY s y
 

1
( ) (0) 3 ( )
2
sY s y Y s
s
  

1
( ) 1 3 ( )
2
sY s Y s
s
  

  1
3 ( ) 1
2
s Y s
s
  

  1
3 ( )
2
s
s Y s
s

 

  
1
( )
2 3
s
Y 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
s
Y s
s s s s
 
  
   
   
1 11 1
( ) 2
2 3
y t L L
s s
 
      
     
       
2 3
( ) 2t t
y t e e  

Fourier Analysis
• General equation of the Fourier Transform is
• General equation of the Inverse Fourier
Transform is
   1
2
iwx
f w f x e dx



 
   1
2
iwx
f 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:
( )ax
F e
( ) ax
f x e

   1
2
iwx
f w f x e dx




 
0
0
1 1
(0)
2 2
iwx ax iwx
e dx e e dx
 

  

  

  ( )
0
1
2
ax iwx
f w e dx


  
 
( )
0
1
2
a iw x
e dx


 
 
( )
0
1
( )2
a iw x
e
a iw

 
 
  
  
1 1
( )2 a iw
 
  
 
1
2 ( )a iw



Example 17
   1
2
iwx
f w f x e dx




 
  1
2
a
x iwx
a
f w e e dx



 
 11
2
a
iw x
a
e dx



 
 



 
   
1
1
2 1
aiw x
a
e
iw

 
   1 1
1
12
iw a iw a
e e
f w
iw
  
 
  

Example 18
• Find the Fourier Transform of f(x) = 1 if
and f(x) = 0 otherwise.
• Solution
1x 
   1
2
iwx
f w f x e dx




 
 
1
1
1
(1)
2
iwx
f w e dx



 
 
1
1
1
2
iwx
e
f w
iw


 
  
 
 1
2
iw iw
e e
iw 

 

Express it using Euler Formula
• Euler Formula
• So from the example 18, we can express it as
and
cos sinix
e x i x 
cos siniw
e w i w 
cos siniw
e w i w
 

• By subtraction
• Substitute in the previous equation will get
   cos sin cos siniw iw
e e w i w w i w
    
2 sini w
  sin
2
w
f w
w



Jif 315 lesson 1 Laplace and fourier transform

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