Laplace Transform (1)

Laplace Transform (1)

– If f(t) is defined for all t ≥ 0, its Laplace transform is

– Integral transform with kernel

– Inverse transform

– Example: f(t) = 1

     

0

  

F s L f e f s dt

     

0^

,

 

F s k s t f s dt k s t   ,  e

  

 

f t L

F

   

0 0

1 1

1

   

  

 



f e dt e

s s

L L  s  0 

Laplace Transform (2)

– Example: f(t) = e^at

– Linearity of Laplace transform

– Application: hyperbolic function

 

0

1

1

   

  

 



at st at

e e e dt e

a s s a

L  s   a 0 

   

 af t  bg t   a  f t     b   g t  

L L L

       

       

2 2

2 2

1 1 1 1

cosh ,

2 2

1 1 1 1

sinh 2 2





 

           

 

           

at at

at at

at e e s

s a s a s a

at e e a

s a s a s a

L L L

L L L

Laplace Transform (3)

– Application: trigonometric function

– By substitution

0 0

0

0 0

0

cos cos sin 1 ,

sin sin cos

 

  

 

  

 

   

 

   

    



   



 

 

st

st st

c s

st

st st

s c

L e tdt e t e tdt L

s s s s

L e tdt e t e tdt L

s s s

2 2

2 2

1 , ,

1 ,

 



  



 

       

 

       

c c c

s c s

L L L s

s s s s

L L L

s s s s

Laplace Transform (4)

– Or by complex method

– Basic functions and their transforms

    



1   

     

 

    

     

i t

s i s i s

e i

s i s i s i s s s

L

  e

  cos  t  i sin  t    cos  t   i  sin  t 

L L L L

 

f t L   f

t t

1 s 1 s

  f

  L f t 1

2! s

 

, 0,1,

t

n  n s !

 

, : positive

t

a    a 1  s

e

1  s  a 

cos  t s s 

 



sin  t   s

 



Laplace Transform (5)

– Basic functions and their transforms (cont’d)

– Proofs

 

f t L   f f t   L   f

cosh at

s

s  a sinh at

a

s  a

at

cos

e  t  

s a

s a 



 

at

sin

e  t  s a 



  

 

0

1 1

n st n st n

n

t e t dt e t e t dt

s s

  



 

 

  

L

 





1

1 ! 1 !

n n

n n n n

s t s s

s

  

 L  

s-Shifting (1)

– Proofs --- by setting st = x

– s-Shifting --- If the transform of f(t) exists for some s greater than some k, then the transform of e^atf(t) exists for s-a>k.

– Proof

 

 

1 1

1

a

a st a x x n

a a

x dx

t e t dt e e x dx a

s s s s

     

 

              

L

 e f t

    F s   a 

L

 

   

e f t

 L

F s  a

 

 

  

  

F s  a  

e

f t dt  

e

  e f t   dt  L e f t

Existence/Uniqueness of Laplace Transform

– Existence --- If f(t) is defined and piecewise continuous on the semi- axis t≥0, and satisfies the relation below for all t≥0 and some

constants M and k, then its Laplace transform exists for all s>k.

– Proof

– Uniqueness --- If two continuous functions have the same transform, then they are completely identical.

 

f t  Me

     

0 0 0

st st kt st

M

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

s k

     

   

   

L

Transform of Derivatives (1)

– Proof

– For any order of derivative

  f   s   f  f   0

L L

  f   s

  f  sf   0  f    0

L L

       

0 0 0

st st st

f   

e

f  t dt    e

f t  

 s 

e

f t dt L

           

     

2

0 0 0

0 0

f s f f s s f f f

s f sf f

            

   

L L L

L

  f

 s

  f  s

f   0  s

f    0   f

  0

L L

Transform of Derivatives (2)

– Example 1

– Example 2

   

   

 

sin , 0 0,

sin cos , 0 0,

2 cos sin

f t t t f

f t t t t f

f t t t t



  

   

 

    

  

  f 2

s

  f s

  f

 s 

    

L  L L

   



2 s



f f

s



   L L 

  cos ,   0 1,   0 0,  

cos

f t   t f  f   f  t     t

  f s

  f s

    f , f

s

 s

      

L L L L 

Solution of DE by Laplace Transform (1)

– Initial Value Problem

– Step 1: Apply Laplace transform (Subsidiary eqn.)

– Step 2: Solve algebraically for Y (Transfer function)

Solution

    , 0

,   0

y   ay   by  r t y  K y   K

       

2

0 0 0

s Y sy y  a sY y bY R s

           

 

 s Y

 as b Y      s a y    0  y    0  R s  

 

1

Q s  s as b

 

      0     0    

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

Solution of DE by Laplace Transform (2)

– If

– Step 3: Reduce Y to a sum of terms (partial fractions) and apply inverse transform

– Example: IVP – Step 1:

  0 0,   0 0

y  y  

   

   

 

output input Q s Y s

 R s  L L

   

, 0 1, 0 1

y    y t y  y  

Transfer Function

   

2 2

0 0 1 ,

s Y  sy  y    Y s

 s

 1  Y    s 1 1 s

Solution of DE by Laplace Transform (3)

– Step 2:

– Step 3:

– Advantages of the Laplace transform method for solving DE

1) No need to determine a general sol. for a homogeneous eqn.

2) No need to determine an arbitrary constants in a general sol.

 

1

Q s 1

 s



 1  1

s

1 1

 1

1  1 1

1 1 1

Y s Q Q

s s s s s s s

  

               

 

 

1

1

1

1 1

t

sinh

y t Y

s s s

e t t

 

 

 

 

                 

  

- - - -

L L L L

Transform of Integrals (1)

– Proof

 0t f     d   1 s F s  

L  s  0, s  k 

 

 

0

1

t

f d F s

  

  s  

 

 L

   

0

g t  

f   d

 

 

M 

1  M

g t f d M e d e e

k k

 



        k  0 

 f t      g t      s   g t    g   0

L L L  s  k 

Transform of Integrals (2)

– Example 1

– Example 2: Shifted data problem

  f  s s 

1  

 , f t    ?

L

1

2 2

1 1

sin t

s 

 



        L

 

1

2 2 0 2

1 1 1 1

sin 1 cos

t

d t

s s   

  



                 L

1 1 1

2 , , 2 2

4 2 4

y    y t y          y          

Transform of Integrals (3)

– Set

– Step 1:

– Step 2:

0

1 4 , 1 4

t   t   t 

   

1 1

2 , 0 , 0 2 2

4 2

y    y    t      y   y   

   

2

2

2 2

0 0 ,

s Y sy y Y

s s

 

    



2 1 



2 1    0

s 1   0

1 1

Y y y

s s

s s s s

 

   

 

 

  2  sin  1 2  1 cos  1 2 cos  2 2 sin 

y t  t  t    t   t   t

Unit Step Function (1)

– Set back to

– Unit step function (Heaviside function)

– u(t) u(t-a)

1 4 t   t 

  2 sin cos

y t   t t  t

  0 if

1 if

t a u t a

t a

  

        a  0 

Unit Step Function (2)

– Usages

Time Shifting Theorem (1)

– “Shifted function”

– Proof

– Set

       

0 if

if

t a f t f t a u t a

f t a t a

  

         

   

 f t  a u t  a   e

F s  

L

   



  

f t  a u t  a  L

e

F s

   

 

0 0

s a

as as s

e

F s  e



e

f   d  

e

f   d

, ,

a t t a d dt

       

   

as as st

e

F s  e



e

f t  a dt

Time Shifting Theorem (2)

– RHS is equivalent to

       

0 0

as as st st

e

F s  e



e

f t  a u t  a dt  

e

f t dt

– Example 1: Find the transform.

– Step 1:

– Step 2: Arrange each term in order for t-shifting theorem

 

2 if 0 1

2 if 1 2

cos if 2

t

f t t t

t t





   

 

    

  

 

  2 1   1   1

 1  1

2 2

f t   u t   t    u t   u t           

    

  

1

1 1

1 1 1 1

2 t u t  2 t t 2 u t 

          

     

     

L L

Application of Shifting Theorems (1)

 cos  1

t u t  2  

     

Application of Shifting Theorems (2)

3 2

1 1 1

2

e

s s s

 

      

2 2

1

1 1 1 1 1

2 t u t 2    2 t 2   2 t 2   8  u t 2  

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

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

       

L L

2

2

3 2

1

2 8

e

s s s

 

 

    

 

 cos  1 sin 1 1

2 2 2

t u t   t  u t  

              

                 

     

L L

2 2

1 1

e

s



 



Application of Shifting Theorems (3)

– Add together

– Or other convenient form of t-shifting theorem

– Proof

  2 2 1

1

1 1

2 2 8

s s s

f e e e

s s s s s s s s

 



 

 

                L

2 2

1 1

e

s







   

 f t u t  a   e

 f t   a  

L L

      ,  

f t  a  g t f t  g t  a

Application of Shifting Theorems (4)

– Then

   

2 2

1 1 1 1

1 1

2 2 2 2

s s

t u t e

t e

t t

           

     

     

L L L

3 2

1 1 1

2

e

s s s

 

      

 cos  1

cos 1

 sin 

2 2

s s

t u t  e

t  e

t

            

           

   

L L L

2 2

1 1 e

s



 



Application of Shifting Theorems (5)

Application of Shifting Theorems (6)

– Example 2: Find the inverse transform.

– Without the exponential functions in the numerator

– By the t-shifting theorem

– The second and third terms cancel each other when t>2

   

2 3

2 2 2 2 2

2

s s s

e e e

F s s  s  s

  

  

  

1 1

sin  t , sin  t te ,

 



             

 

 

1 1

sin 1 1 sin 2 2

3

3

f t t u t t u t

t e u t

 

 

 

      

 

Application of Shifting Theorems (7)

– Finally

   

 

0 if 0 1

sin if 1 2

0 if 2 3

if 3

3

t

t t

f t t

t e t

 

 

   

    

 

      

   

 

Application of Shifting Theorems (8)

– Example 3: Response of RC-circuit to a single rectangular wave.

– Input:

– Governing DE

         

   

0

1

Ri t q t Ri t i d v t V u t a u t b

C C  

            

   

V u t

   a  u t  b  

Application of Shifting Theorems (9)

– Subsidiary eqn.

– Solve algebraically for I(s)

– By the t-shifting theorem

    I s V

RI s e e

sC s

 

 

    

      ,  

  ,

 

1

as bs

V R V

I s F s e e F s F e

s RC R

   

   

 L

         

   

 

 

1 1

0

as bs

t a RC t b RC

i t I e F s e F s

V e u t a e u t b

R

   

   

  

 

     

L L

Application of Shifting Theorems (10)

– Finally

 

 

1

1 2

if if

t RC

t RC

K e a t b

i t K K e t b





   

 

        

   

1 0 ^{a RC}

,

K  V e R K  V e R