C HAPTER 6. L APLACE T RANSFORMS

C ^HAPTER 6. L ^APLACE T ^RANSFORMS

2018.6

서울대학교

조선해양공학과

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)



Laplace Transform (라플라스 변환):

- 적분 과정을 통해 주어진 함수를 새로운 함수로 변환하는 것



Inverse Transform (역 변환):

     

0

F s f e st f t dt



 ^L   

   

1 F f t





L

 Ex.1 Let f (t) = 1 when t ≥ 0. Find F(s).

 Ex.2 Let f(t) = e ^at

when t ≥ 0, where a is a constant. Find L( f ).

     

0 0

1 1

1 ^{st st} 0

f e dt e s

s s

 

 

      

L L

  ^{ }

0 0

1 s a t 1

at st at

e e e dt e

a s s a

 

  

  

 



L

 Theorem 1 Linearity of the Laplace Transform

The Laplace transform is a linear operation;

that is, for any functions f(t) and g(t) whose transforms exist and any constants a and b, the transform of af(t) + bg(t) exists, and

   

 ^{af t  bg t}  ^{ a}  ^{f t  }  ^{ b}  ^{g t  } 

L L L

 Ex.3

Find the transform of cosh at and sinh at.

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)

       

      2 2

1 1 1 1

cosh , inh

2 2

1 1 1 1

cosh

at at at at at at

at at

at e e at e e e , e

s a s a

at e e s

  



      

 

 

 

              

L L

L L L

s

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)

 Theorem 2 First Shifting Theorem (제 1 이동 정리), s-shifting (S-이동)

If f (t) has the transform F(s) (where s > k for some k),

then e

^at f (t) has the transform F(s ‒ a) (where s – a > k). In formulas,

or, if we take the inverse on both sides, .

^{e f t}

^at

  ^{ L}

^-

¹  ^{F s a}  ^  

Proof) We obtain F(s‒a) by replacing s with s ‒ a in the integral, so that

 ^{e f t at}   

^

^{F s a }

^

^

L

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)



 ^{  }

^

^{ }

^





0 0

)

(

( ) [ ( )] { ( )}

)

(

s a e f t dt e e f t dt e f t

F ^{s a t st at}

L

^at

  ¹    

at -

e f t  L F s  a

 Ex. 5 Formulas 11 and 12 in Table 6.1,

For instance, use these formulas to find the inverse of the transform.



^at

cos  _ s _

2

a

2

, 

^at

sin  _{ }

2 2

e t e t

s a s a

  

 

  

   

L L

  ₂ ^{3 137}

2 401 f s

s s

 

  L

 

       

1 1 1

2 2 2

3 1 140 1 20

3 7 3cos 20 7 sin 20

1 400 1 400 1 400

s s

t

f e t t

s s s

            

                            

L L L

 ^{e f t}

^at

   ^{ F s a   }

L

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)

(a = -1,  = 20)

 Theorem 3 Existence Theorem for Laplace Transform

If f (t) is defined and piecewise continuous on every finite interval on the semi- axis t ≥ 0

and satisfies | f(t) | ≤ Me

^kt

for all t ≥ 0 and some constants M and k, that is, f (t) does not grow too fast (“growth restriction (증가제한)”), then the Laplace transform

L

( f ) exists for all s > k.

6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)

0 0

0

| ( ) |

^st

( ) ( )

^st

kt st

f e f t dt f t e dt Me e dt M

s k

   

 

 

 



 



L

Proof)

Piecewise continuous → e

^-st f (t) is

integrable over any finite interval on t-axis.

 Theorem 1 Laplace Transform of Derivatives (도함수의 라플라스 변환)

6.2 Transform of Derivatives and Integrals. ODEs

     

  ²      

' 0

'' 0 ' 0 .

f s f f ,

f s f sf f

 

  

L L

L L

0 0 0

{ ( )} ( ) ( ) ( )

(0) { ( )}

st st st

f t e f t dt e f t s e f t dt

f s f t

     

    

  

 

L

L

 

{ f t  ( )}  s f  f (0)

L L

 

0 0 0

{ ( )} ( ) ( ) ( )

(0) { ( )}

[ (0)] (0)

st st st

f t e f t dt e f t s e f t dt

f s f t

s s f f f

     

      

 

  

   

 

L

L L

 

{ f  ( )} t  s 2 f  sf (0)  f  (0)

L L

 

3 2

{ f  ( )} t  s f  s f (0)  sf  (0)  f  (0)

L L

Proof)

 









 

 

 

 

 

 







 



 





 



 

0 0

0

0 0

0

) ( )

( )

(

) ( )

( )

(

) ( )

( )

(

dt t f e dt

t f se t

f e

dt t f e dt

t f se t

f e

t f e t f se t

f e

st st

st

st st

st

st st

st



 ^{g ( t ) f  ( t ) dt  g ( t ) f ( t )  g  ( t ) f ( t ) dt}

(integration by parts)

6.2 Transform of Derivatives and Integrals. ODEs

 Ex. 1

Let f (t) = tsinωt. Find the transform of f (t).

 

   

 

     

2

2 2

0 0,

' sin cos , ' 0 0,

'' 2 cos sin

'' 2 (0) (0)

f

f t t t t f

f t t t t

f s f s f sf f

  

   

 





  

 

      

L  L L

 Theorem 2 Laplace Transform of Derivatives

Let be continuous for all t ≥ 0 and satisfy the growth restriction

| f(t) | ≤ Me

^kt

.

Furthermore, let f

⁽ⁿ⁾

be continuous on every finite interval on the semi-axis t

≥ 0.

^L   ^f  

ⁿ

^{ s}

ⁿ

^{L   f  s}

ⁿ

^{ 1 f   0  s}

ⁿ

^{ 2 f ' 0     f }

ⁿ

^{ 1    0}

) 1

,...., (

, f  f

ⁿ

^

f

 Theorem 3 Laplace Transform of Integral (적분의 라플라스 변환)

Let F(s) denote the transform of a function f(t) which is piecewise continuous for all t ≥ 0

and satisfies a growth restriction | f(t) | ≤ Me

^kt

. Then, for s > 0, s > k, and t >

0,

           

¹

 

0 0

1 1

t t

f t F s f d F s , f d

-

F s

s s

   

   

             

L L L

6.2 Transform of Derivatives and Integrals. ODEs





^t f d t

g

( )

0

(



)

 Proof)

→ g(t) also satisfies a growth restriction.

g'(t) = f (t) , except at points at which f(t) is discontinuous.

g'(t) is piecewise continuous on each finite interval.

g(0) = 0.

)}

( { )

0 ( )}

( { )}

( ' { )}

(

{ f t L g t s L g t g s L g t

L    

0 0 0

| ( ) |

^t

( )

^t

( )

^{t k} M

(

^kt

1)

M ^kt

( 0)

g t f d f d M e d e e k

k k

     













   

{ ( )} g t 1 { ( )} f t

 L  L

From Theorem 1

 Theorem 3 Laplace Transform of Integral

 Ex. 3 Find the inverse of and .

 ^{2 1 2}  ²  ^{2 1 2} 

s s   s s  

1

2 2

1 1

sin t

s 

 

         L

    ^{       }

¹

^{ }

0 0

1 1

t t

f t F s f d F s , f d

-

F s

s s

   

   

             

L L L

6.2 Transform of Derivatives and Integrals. ODEs

2

} 2

{sin 

 

  t s L

  ^{ }

1

2 2 2

0

1 sin 1

1 cos

t

d t

s s   

 



  

 

   

  

  

L

2 2

{cos } s

t s

 ^  

L



Differential Equations, Initial Value Problems:

Step 1

Setting up the subsidiary equation (보조방정식).

Step 2

Solution of the subsidiary equation by algebra.

Transfer Function (전달함수):

Solution of the transfer function:

Step 3

Inversion of Y to obtain y.

y(t) = L ^-1

(Y).

    0   1

'' ' , 0 , ' 0

y  ay  by  r t y  K y  K

   

 ^{Y  L y , R  L r} 

          ^{       }

2 2

0 ' 0 0 0 ' 0

s Y sy y a sY y bY R s s as b Y s a y y R s

                  

 

 

_{2 2}

2

1 1

1 1

2 4

Q s s as b

s a b a

 

          

      ^{0 ' 0}        

Y s    s a y   y   Q s  R s Q s

6.2 Transform of Derivatives and Integrals. ODEs

 Ex. 4

Solve

Step 1

Step 2 Step 3

   

'' , 0 1, ' 0 1

y   y t y  y 

     

2 2

2 2

1 1

0 ' 0 1 1

s Y sy y Y s Y s

s s

        

   

2 2 2 2 2 2 2

1 1 1 1 1 1 1

1

1 1 1 1 1

Q Y s Q Q s

s s s s s s s s

  

                  

 

¹

 

¹

¹

¹ ₂

¹

¹

¹

₂

^sinh

1 1

y t Y e

t

t t

s s s

 

 



 



 

 L  L       L       L       

6.2 Transform of Derivatives and Integrals. ODEs

 Ex. 5

Solve

Step 1

Step 2 Step 3

6.2 Transform of Derivatives and Integrals. ODEs

0 ) 0 ( 16

. 0 ) 0 ( .

0

9    

 

  y y y y

y

4 ) 35 2 / 1 (

08 . 0 ) 2 / 1 (

16 . 0 9

) 1 (

16 . 0

2 2







 





 

s s s

s Y s

 

/ 2

/ 2

35 0.08 35

( ) ( ) 0.16 cos sin

2 35 / 2 2

0.16 cos 2.96 0.027 sin 2.96

t

t

y t Y e t t

e t t





 

    

 

L

^-1

) 1 (

16 . 0 )

9 (

. 0 9

16 . 0 16

.

0 ²

2 Y  s  sY   Y  s  s  Y  s 

s

(a = -1/2,  =

³⁵₄

)

6.2 Transform of Derivatives and Integrals. ODEs

7 12 21 3 ^t , (0) 3.5, (0) 10

y   y   y  e y  y   

Q? Solve

6.2 Transform of Derivatives and Integrals. ODEs

[ Relation of Time domain and s-Domain ] Time domain analysis

(differential equation)

Problem Answer

Laplace Transform

S-domain analysis (algebraic equation)

Inverse Transform



The process of solution consists of three steps.

Step 1

The given ODE is transformed into an algebraic equation (“subsidiary equation”).

Step 2

The subsidiary equation is solved by purely algebraic manipulations.

Step 3

The solution in Step 2 is transformed back, resulting in the solution of the given problem.

IVP Initial Value

Problem (초기값 문제)

AP Algebraic

Problem (보조방정식)

① Solving

AP by Algebra

② Solution

of the IVP

③

Solving an IVP by Laplace transforms

6.2 Transform of Derivatives and Integrals. ODEs

6.2 Transform of Derivatives and Integrals. ODEs



Advantages of the Laplace Method

1.

Solving a nonhomogeneous ODE does not require first solving the homogeneous ODE.

2.

Initial values are automatically taken care of.

3.

Complicated inputs can be handled very efficiently.

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

 Unit Step Function (단위 계단 함수)

•

Unit Step Function (Heaviside function):

•

Laplace transform of unit step function:

   

 

0 1

t a u t a

t a

 

    

 

s e s

dt e e

dt a t u e a

t u

as

a t st a

st st

 



  

       



 ⁾  ^{( )}  ¹

( 0

L

“on off function”

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

 Unit Step Function

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

 Theorem 1 Second Shifting Theorem (제 2이동 정리); Time Shifting

 ^{f t}    ^{ F s   }  ^{f t   a u t    a }  ^{ e}

^as

^{F s   , f t   a u t    a  }

^-1

 ^e

^as

^{F s  } 

L L L

dt d

a t t

a    

  

 , thus ,

a t

a t a

t a f

t u a t f t

f 



 



 





 if

if ) (

) 0 (

) (

) ˆ (



 ^{    }



  

0

) (

0 ( ) ( )

)

( s e e ^ f  d  e ^ f  d  F

e ^{as as s s a}

( ) ( )

( ) ( ) ˆ ( )

as st

a

st st

e F s e f t a dt

e f t a u t a dt e f t dt

  

   

 

   



 

Proof)

 

s e s

dt e e dt a t u e a t u

as

a t st a

st st

 



  

 

     



 ⁾  ^{( )}  ¹

(

0

L

     

0

F s f e

st

f t dt



 ^L  



 Ex. 1 Write the following function using unit step functions and find its

transform.

Step 1 Using unit step functions:

 

 

 

 

2

2 0 1

2 1 2

cos 2

t

f t t t

t t





  

 

   

  



  ^{2 1}   ¹   ^{1 2  1  1  cos  1}

2 2 2

f t   u t   t ^   u t   u t ^     ^   ^    t u t ^     ^  

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

𝜋

2

 Ex. 1 Write the following function using unit step functions and find its

transform.

Step 1 Using unit step functions:

Step 2

  ^{2 1}   ¹   ¹

²

 ¹  ¹  ^cos  ¹

2 2 2

f t   u t   t ^   u t   u t ^     ^   ^    t u t ^     ^  

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

     ²   

2

3 2

1 1 1 1 1 1

1 1 1 1

2 2 2 2

t u t t t u t e s

s s s

  

               

       

       

L L

2 2 2

2 2

3 2

1 1 1 1 1 1 1

2 2 2 2 2 2 8 2 2 8

t u t t t u t e

s

s s s

   



 ^{ }   ^  ^{  }

^

                       

                       

         

L L

 ^cos  ^{1 sin 1 1} ₂ ^{1 2}

2 2 2 1

t u t t u t e s

s

 ^   ^{ } 

                

                  

     

L L

   

 ^{f t  a u t  a}  ^{ e}

^as

^{F s  }

L

 

1

n

!

n

t n

s

^



L

 Ex. 2

Find the inverse transform f(t) of

   

2 3

2

2 2 2 2

2

s s s

e e e

F s s  s  s

  

  

  

  ¹     ^{  1}  ^{ }  ^{   }

²

^{ }

³

^{ }

f t sin  t 1 u t 1 sin  t 2 u t 2 t 3 e

^t

u t 3

 

        

 



1

2 2

1 sin t

s



 

         L

 

1 1 2

2 2

1 1

2

t te

t

s s

    

      

   

    

L L

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

 

   

 

 

²

^{ }

³

 

0 0 t 1 sin 1 t 2

0 2 t 3 3

^t

t 3

t

t e





 

 

 

   

  

  

  



     

 

   

1

2 2

2 1

2 2

1 sin 1 1

1 sin 2 2

s

s

e t u t

s

e t u t

s

  

  

 

 

 

        

 

        

L

L

   

 ^{f t  a u t  a}  ^{ e}

^as

^{F s  }

L

2

} 2

{sin



 

  t s L

 Time shifting

 ^{e f t}

^at

   ^{ F s a   }

L

 S-shifting

    ^{ }  ^{ }  ^{ }

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (t-Shifting)

3 2 1 if 0 1 and 0 1

(0) 0, (0) 0

y y y t if t

y y

       

  

Q? Solve using Laplace transform.

6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions



Dirac delta function or the unit impulse function:



Definition of the Dirac delta function



Laplace transform of the Dirac delta function

   

 

0 otherwise t a t a

    ^{  }



   

     

0

1

lim 0 otherwise

k k

k

a t a k

f t a k



t a f t a



   

       



   

0 0

1 1 1

a k

k

a

f t a dt dt t a dt

k 

  

     

  

  ¹      

f

k

t a u t a u t a k

k  

       

 

  ^{1   1}  ^{ } 

ks a k s

as as as

k

f t a e e e e t a e

ks ^{      } ks ^  ^

 L        L  

1 1

0

1 1

lim = lim = = ( ) 1

ks ks s ks

e e e de

s

   

 

        

 

  

 Ex. 1 Mass Spring System Under a Square Wave Step 1

Step 2 Step 3

6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions

0 ) 0 ( 0 ) 0 ( ).

2 ( ) 1 ( ) ( 2

3         



y y r t u t u t y y

y

t

t

e

e F

t

f ²

2 1 2

) 1 ( )

(  L ^{- 1}   ^  ^

) )(

2 3 (

) 1 ( ) 1(

2

3

² ₂ ²

2 s s s s

e s e

s s s

Y e

s e Y sY

Y

s ^{  }



^



 









2 2 / 1 ) 1 (

1 2

/ 1 ) 2 )(

1 (

1 )

2 3 (

) 1

(

₂

 

 

 

 



 

s s

s s

s s s

s s s

F

 

 

















































) 2 (

) 2 1

(

) 1 0

( 2

/ 1 2

/ 1

2 / 1 2

/ 1

0

) 2 ( 2 )

1 ( 2 )

2 ( ) 1 (

) 1 ( 2 )

1 (

t t

t

e e

e e

e e

t t

t t

t t

) 2 ( ) 2 ( ) 1 ( ) 1 (

) ) ( )

(

( ²















 ^{ }

t u t f t

u t f

e s F e

s F

y L ^-1

^{s s}

   

  ^{ }

 Time shifting

 ^{e f t}

^at

   ^{ F s a   }

L

 S-shifting

   

 ^{f t  a u t  a}  ^{ e}

^as

^{F s}

^{ }

L

 Time shifting

 Ex. 2 Find the response of the system in Example 1 with the square wave

replaced by a unit impulse at time t = 1.

Initial value problem:

Subsidiary equation:

     

'' 3 ' 2 1 , 0 0, ' 0 0 y  y  y   t  y  y 

2 3 2 ^s

s Y  sY  Y  e ^

     ^{1 1}

1 2 1 2

s

e s

Y s e

s s s s

   

           

6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions

  ^{e at 1 f t ( )= e t , ( )= g t e 2 t}

s a

 

 

L 

   

 ^{f t  a u t  a}  ^{ e}

^as

^{F s  }

L

     

     

1

1 2 1

0 0 t 1 t 1

t t

y t Y

e e



   

  

   

 

L 

( 1) ( 1) ( 1)( 1)

f t u t g t u

     

( 1) 2( 1)

( 1) ( 1)

t t

e ^{ } u t e ^{ } u

   

 

 ^{ t}

^

^a 

^

^{e  as}

L

 Dirac’s Delta

 Time shifting

6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions

9 ( / 2), (0) 2, (0) 0

y ^  y   t   y  y ^ 

Q? Solve



Transform of a product is generally different from the product of the transforms of the factors.



Convolution (합성곱):

6.5 Convolution. Integral Equations

      

0 t

f  g t   f  g t    d

) ( )

( )

( f g L f L g

L    L ( fg )  L ( f ) L ( g )

 Ex. Transform of a Convolution

Evaluate

Solution)

f ( t )  e ^t , g ( t )  sin t

 ⁰ 

2 2

sin( ) ( ) ( ) { } {sin }

1 1 1

1 1 ( 1)( 1)

t e τ t dτ F s G s e t t

s s s s



  

  

   



L L L

1

2 2

2 2

{ } ! { } 1

{sin } {cos }

n

n

at

t n

s

e s a

t s

t s s

 



 

 

 

 

  L

L L

 ^{ 0 t e  sin( t    ) d}  ^{  e t *sin t } L

L L

 ^{f g}      ^{f g}

 L   L L



Convolution:

 Theorem 1 Convolution Theorem

If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order, then

6.5 Convolution. Integral Equations

      

0 t

f  g t   f  g t    d

 ^{f  g}  ^     ^{f g}

L L L

 Ex. 1 Let . Find h(t). ^{H s}     ^{ s  1 a s}

   

1 1 1 1 1

, 1 1 1 1

t

at at a at

e h t e e ^ d  e

                        

L L

( ) h  H s ( )

L g s

a f s

s a g s

f g

f

h 1

) ( 1 ,

) ( 1 ,

) 1 ( ) ( )

* ( )

( 

 

 



 L L L L L

L

6.5 Convolution. Integral Equations

 Ex. 2 Convolution

Evaluate

¹

2 2 2

1

( )

-

s 

 

  

 

L

Solution)

) (

) 1 ( )

(

Let _{2 2}



 

 G s s s

F

1

2 2

1 1

( ) ( ) sin

f t g t t

s

 

  

  

  L      

1

2 2 2 2 0

1 1

sin sin ( )

( )

- t

t d

s

   

 

 

 

  

  

L























t t

d t t

d t

t

2 0 2 0

)]

cos(

) 2

[cos(

2 1

)]

cos(

) [cos(

2 1 1







 













 

 

  

 t t  t





 ^cos

sin 2

1

2

2

0

1 1

sin (2 ) cos

2 2

t

t t

   

 

 

      

 

 , find ( )

) (

) 1

( _{2 2 2} h t

s s

H 

1

2 2

2 2

{ } ! { } 1

{sin } {cos }

n

n

at

t n s

e s a

t s t s

s

 



 





 

 

 

L L L L

      

0 t

f  g t   f  g t    d

6.5 Convolution. Integral Equations

 Ex. 4 Repeated Complex Factors. Resonance – Undamped mass-spring system

0 ) 0 ( 0 ) 0 ( sin

₀

2

0

   



y K t y y

y  

2 2 0 0

0 2 2 2 2 2

0 0

( ) ( )

K K

s Y Y Y s

s s

 

  

   

 

The subsidiary equation

2 2

2 )

( ) 1

(   

s s

H  

  

 t t t

t

h 





 ^cos

sin 2

) 1

( ₂

In Ex. 2, we found

The solution



Property of convolution



Commutative Law (교환법칙):



Distributive Law (분배법칙):



Associative Law (결합법칙):



Unusual Properties of Convolution

^:

6.5 Convolution. Integral Equations

f    g g f

 1 2  1 2

f  g  g     f g f g

 ^{f  g}  ^{   v f}  ^{g v } 

0 0

0   

 f

f

1 f   f

 Ex. 3 Unusual Properties of Convolution

t t

d

t   0 ^t   2 ²  1 1

1

*  

 Application to Nonhomogeneous Linear ODEs by convolution

6.5 Convolution. Integral Equations

If

    0   1

'' ' , 0 , ' 0

y  ay  by  r t y  K y  K

       

  ^{       }

2

2

0 ' 0 0

0 ' 0

s Y sy y a sY y bY R s

s as b Y s a y y R s

           

 

      

   

 ^{Y  L y , R  L r} 

  _{2 2}

2

1 1

1 1

2 4

Q s s as b

s a b a

 

      

0 ) 0 ( 0 ) 0

( 

y

 

y Y  RQ

 ^

 ^t q t r d

t

y ( ) 0 (  ) (  ) 

      

0 t

f  g t   f  g t    d

f    g g f

 Theorem 1 Convolution Theorem

If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order, then

6.5 Convolution. Integral Equations

 ^{f  g}  ^     ^{f g}

L L L

Proof)

) , [ ,

,    



 p  p t  t 

t

0 0

( ) ^s ( ) ( ) ^sp ( )

F s   ^ e ^{ } f   d and G s   ^ e ^ g p dp



 ^{       }

 





 g t  dt e e g t  dt

e s

G ( ) ^{s ( t )} ( ) ^{s st} ( )











  



 f e e g t dt d f e g t dt d

e s

G s

F ^{( ) ( ) }  0 ^{  s ( ) s}  ^{  st (  ) }  0 ^{ ( )}  ^{  st (  )}

) ( )

( )

( )

( ) ( )

( )

( s G s 0 e 0 f g t d dt 0 e h t dt h H s F   ^{  st}  ^{t }  ^{ }   ^{  st}  ^L 

  0 ^{ } ( ) 

 f t dt d

  0 ^ 0 ^t f (  ) d  dt

 ^{f  g}  ^{ ( ) h  H s ( )}

L L

 Ex. 5 Response of a Damped Vibrating System to a Single Square Wave

6.5 Convolution. Integral Equations

0 ) 0 ( 0 ) 0 ( ).

2 ( ) 1 ( ) ( 2

3         



y y r t u t u t y y

y

t

t e

e t

q ( )  ^  ^{ 2}

2 1 1

1 )

2 3 (

) 1 ( )

1( 2

3

² ₂

2

 

 



 











^{ }

s s

s s s

Q e

s e Y sY Y

s ^{s s}

0 ,

1 For t  y 

( ) 2( )

1 1

( ) 2( ) ( 1) 2( 1)

1

For 1 2, ( ) 1

1 1 1

2 2 2

t t

t t

t

t t t t

t y q t d e e d

e e e e

 

 

  ^{   } 

       

      

 

        

 

2 2

( ) 2( )

0 1 1

For 2  t y ,  

^t

q t (    ) 1 d    q t (    ) 1 d    e ^{ }

^t

^  e ^{ }

^t

^ d  ( ) 1, only for 1 2

r t    t



Integral Equation: Equation in which the unknown function appears in an integral.

 Ex. 6 Solve the Volterra integral equation of the second kind.

Equation written as a convolution:

Apply the convolution theorem:

     

0

sin

t

y t



 y  t



  d



t

sin

y   y t  t

    ₂ ^{1 1} ₂

Y s Y s 1

s s

 



  ² ₄ ^{1 1} ₂ ¹ ₄   ³

6

s t

Y s y t t

s s s

      

6.5 Convolution. Integral Equations (적분방정식)

) ( )

( y  Y t L

Unknown

 ^



 g t ^t f h t d

t

f ( ) ( ) 0 (  ) (  ) 

: Volterra integral equation for f(t)

1

{ } ⁿ n _n !

t  s ^

L

6.5 Convolution. Integral Equations

Solve

^{f t  t  e  t }  0 ^{t f e t  d for f t}

2 ( ) ( )

3 )

(  ^ 

Solution)

1 2! 1 3! 1 1 1 1

( ) 3 2

f t  L ^      L ^      L ^      L ^    

 Example An Integral Equation

 ^{f t ( )}  ^{ 3}     ^{t 2  e  t }  ^{ 0 t f ( )  e t   d } 

L L L L

1 ) 1

1 ( 1

! 3 2 )

( ₃

 

 



 F s s

s s s

F

1 2 1

6 ) 6

( _{3 4}

 





 s s s s

s F

      

  ⁰    

1

2 2

2 2

{ } ! { } 1

{sin } {cos }

t

n

n

at

f g t f g t d

f g f g

t n

s

e s a

t s

t s s

  

 



 



  

 



 

 

 



L L L

L

L

L

L

6.5 Convolution. Integral Equations

( ) 2 ^t 0 ^t ( ) ^t y t  e  y  e d ^{ }   te Q? Solve

Q? find if equals: ^{f t}   ^L  ^{f t}   

    ^{9 ,}   ^?

( 3)

f t f t

 s s 

L 

6.6 Differentiation and Integration of Transforms.

ODEs with Variable Coefficients

 Differentiation of Transforms

L (f ) f (t)

 Ex.1 We shall derive the following three formulas.

 ^s

²

^{ 1 }

²



²

^{2 1 }

³

^{ sin  t   t cos  t }

 ^{s 2  s  2}  ² 2 ^t _ ^{sin  t}

 

s 2

 sin  ^t  ^t cos  ^t 

1 

     

0

F s f e st f t dt



 ^L    ^{     }

0

dF

^st

F s e tf t dt tf

ds







       ^L

    ^{  1}  ^{ }  ^{ }

tf t F s  , ^ F s  tf t

 L   L  