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:


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:


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


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:


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:


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

Jif 315 lesson 1 Laplace and fourier transform

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