LAPLACE TRANSFORMS · 2016-04-05 · LAPLACE TRANSFORMS. 1-2 1. LaplaceTransform. Inverse Transform. Linearity. Shifting 2. Transforms of Derivatives and Integrals. Differential Equations

1-1 Chosun University

조조 선선 대대 학학 교교 기계공학과기계공학과

담당교수담당교수 정정 상상 화화

LAPLACELAPLACETRANSFORMSTRANSFORMS

1-2

1. Laplace Transform. Inverse Transform.Linearity. Shifting

2. Transforms of Derivatives and Integrals.Differential Equations

3. Unit Step Function. Second Shifting Theorem.Dirac`s Delta Function

4. Differentiation and Integration of Transforms

5. Convolution. Integral Equations

6. Partial Fractions. Differential Equations

7. Systems of Differential Equations

8. Laplace Transform : General Formulas

9. Table of Laplace Transforms

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목차

1-3

5.1 Laplace Transform. Inverse Transform

Linearity. Shifting

Let be a given function that is defined for all we multiply by

and integrate with respect to from zero to infinity. Then, if the resulting integral exists, it is a function of say, ;

This function of the variable is called the Laplace transform of the

Original function , and will be denoted by .Thus

( )f t

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0t ≥ ( )f tste− t

s ( )F s

0( ) ( )stF s e f t dt

∞ −= ∫( )F s s

( )f t ( )L f

0( ) ( ) ( )stF s L f e f t dt

∞ −= = ∫

(1)


The original function is called the inverse transform or inverse of

And will be denoted by

THEOREM 1 (Linearity of the Laplace transform)

The Laplace transform is a linear operation; that is, for any

functions and whose Laplace transforms exist and

any constants and ,

( )f t ( )F s1 ( )L F−

1( ) ( )f t L F−=

( )f t ( )g t

a b

{ ( ) ( )} { ( )} { ( )}L a f t bg t aL f t bL g t+ = +


THEOREM 2 (First Shifting theorem)

If has the transform (where ), then has

the transform (where ). In formulas,

or, if we take the inverse on both sides,

()f t ( )F s s k> ( )ate f t( )F s a− s a k− >

{ ( )} ( )atL e f t F s a= −

1( ) { ( )}ate f t L F s a−= −

1-6

A Short List of Important Transforms

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THEOREM 3 (Existence theorem for Laplace transforms)

Let be a function that is piecewise continuous on every

finite interval in the range and satisfies

for all

and for some constants and . Then the Laplace transform

of exists for all .

()f t0t ≥

(2) ( ) ktf t Me−≤ 0t ≥

k M()f t s k>


EXAMPLE 1 Laplace transform

Let when .Find

Solution. From (1) we obtain by integration

The interval of integration in (1) is infinite. Such an integral is called an improper integral and, by definition, is evaluated according to the rule

() 1f t = 0t ≥ ( )F s

0

0

1( ) (1)1st stL f L e dt

se

s

∞

∞ − −= = = =−∫

0 0( ) lim ( )

Tst st

Te f t dt e f t dt

∞ − −

→∞=∫ ∫

( 0)s >


Hence our convenient notation means

0

00

1 1 1 1lim limT

st st sT

T Te dt e e e

s s s s∞ − − −

→∞ →∞

⎡ ⎤ ⎡ ⎤= − = − + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ( 0)s >

1-10

The Laplace transform is a method of solving differential equations.

The crucial idea is that the Laplace transform replaces operations of calculus

by operation of algebra on transforms. Roughly, differentiation of is

Replaced by multiplication of by . Integration of is replaced by

division of by .

THEOREM 1 [Laplace transform of the derivative of ]

Suppose that is continuous for all , satisfies (2),

Sec.5.1, for some and , and has a derivative that is

piecewise continuous on every finite interval in the range .

Then the Laplace transform of the derivative exists when

, and

( )f t

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5.2 Transforms of Derivatives and integrals

Differential Equations

( )L s S ( )f t( )L s S

( )f t

( )f t 0t ≥k M ' ( )f t

0t ≥' ( )f t

s k>

1-11

(1)

THEOREM 2 (Laplace transform of the derivative of any order )

Let and its derivatives

be continuous functions for all , satisfying (2), Sec.5.1,

for some and , and let the derivative be piecewise

continuous on every finite interval in the range .

Then the Laplace transform of exists when and is

given by

(4)

'( ) ( ) (0)L f sL f f= −

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n

( )f t ' '' ( 1 )( ) , ( ) , , ( )nf t f t f t−i i i0t ≥

k M ( ) ( )nf t

0t ≥( ) ( )nf t s k>

( ) 1 2 ' ( 1)( ) ( ) (0) (0) (0).n n n n nL f s L f s f s f f− − −= − − − −i i i

1-12

Differential Equations, Initial Value Problems

We begin with an initial value problem

(5)

With constant and . Here is the input applied to the mechanical

system and is the output. In Laplace’s method we do three steps:

1st Step. We transform (5) by means of (1) and (2), writing and

. This gives

This is called the subsidiary equation. Collecting Y-terms, we have

'' ' '0 1( ), (0) (0)y a y b y r t y K y K+ + = = =

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a b ( )r t( )y t

( )Y L y=( )R L r=

2 '[ (0) (0)] [ (0)] ( ).s Y sy y a sY y bY R s− − + − + =

2 '( ) ( ) (0) (0) ( ).s as b Y s a y y R s+ + = + + +


2nd step. We solve the subsidiary equation algebraically for Y. Division by

and use of the so-called transfer function

(6)

gives the solution

(7)

If this is simply ; thus is the quotient

2s a s b+ +

2

1( )Q ss as b

=+ +

'( ) [( ) (0) (0)] ( ) ( ) ( ).Y s s a y y Q s R s Q s= + + +

'(0) (0) 0y y= = Y RQ= Q

( )( )

Y L outputQR L input

= =


and this explains the name of .

THEOREM 3 [Integration of ]

Let be the Laplace transform of . If is piecewise

continuous and satisfies an inequality of the form (2),Sec.5.1 then

(8) (s > 0, s > k)

or, if we take the inverse transform on both sides of (8)

(9)

Q

( )f t

( )F s

{ }0

1( ) ( )t

L f d F ss

τ τ =∫

( )f t ( )f t

1

0

1( ) ( ) .t

f d L F ss

τ τ − ⎧ ⎫= ⎨ ⎬⎩ ⎭∫


EXAMPLE 1. Let . Derive from

Solution. Since and

we obtain from (2)

hence

EXAMPLE 2. Initial value problem: Explanation of the basic steps

Solve

Solution.

1st Step. From (2) and Table 5.1 we get the subsidiary equation

thus .

2nd step. The transfer function is , and (7) becomes

2( )f t t= ( )L f (1)L

' ''(0) 0, (0) 0, ( ) 2f f f t= = = 2(2) 2 (1) ,L Ls

= =

'' 22( ) (2) ( ),L f L s L fs

= = = 23

2( )L ts

=

'' '(0) 1, (0) 1y y t y y− = = =

2 ' 2(0) (0) 1/ ,s Y s y y Y s− − − = 2 2( 1) 1 1/s Y s s− = + +21/( 1)Q s= −


3rd step. From this expression for and Table 5.1 we obtain the solution

EXAMPLE 3. An application of Theorem 3

Let Find .

Solution. From Table 5.1 in Sec. 5.1 we have

2 2 2 2

1 1 1( 1)1 ( 1)

sY s Q Qs s s s

+= + + = +

− −

2 2

1 1 11 ( 1)s s s

⎛ ⎞= + −⎜ ⎟− −⎝ ⎠

1 1 12 2

1 1 1( ) sinh .1 1

ty t L L L e t ts s s

− − −⎧ ⎫ ⎧ ⎫ ⎧ ⎫= + − = + −⎨ ⎬ ⎨ ⎬ ⎨ ⎬− − ⎩ ⎭⎩ ⎭ ⎩ ⎭

2 2

1( )( )

L fs s ω

=+

( )f t

12 2

1 1 sin .L ts

ωω ω

− ⎛ ⎞=⎜ ⎟+⎝ ⎠


From this and Theorem 3 we obtain the answer

12 2 20

1 1 1 1sin (1 cos )t

L d ts s

ωτ τ ωω ω ω

− ⎧ ⎫⎛ ⎞⎪ ⎪ = = −⎨ ⎬⎜ ⎟+⎪ ⎪⎝ ⎠⎩ ⎭∫


5.3 Unit Step Function

Second Shifting Theorem

Dirac’s Delta Function

Unit Step Function

By definition, is 0 for , has a jump of size 1 at

and is 1 for :

(1)

( )u t a−

( )u t a− t a=t a<

t a>

0( )

1if t a

u t aif t a

<⎧− = ⎨ >⎩

( 0)a ≥


Figure 110 shows the special case , which has the jump at zero, and

Fig. 111 the general case for an arbitrary positive .

The unit step function is also called the Heaviside function.

( )u t a− a( )u t


- Shifting: Replacing by in

THEOREM 1 (Second shifting theorem; t-shifting)

If has the transform , then the “shifted function”

(2)

has the transform . That is,

(3)

or, if we take the inverse on both sides, we can write

t t t a− ( )f t

( )f t ( )F s

0( ) ( ) ( )

( )if t a

f t f t a u t af t a if t a

<⎧= − − = ⎨ − >⎩

( )ase F s−

{ }( ) ( ) ( )asL f t a u t a e F s−− − =

{ }1( ) ( ) ( ) .asf t a u t a L e F s− −− − =


(4) { }( ) .aseL u t a

s

−

− = ( 0)s >


Short Impulses. Dirac’s Delta Function

we consider the function

Its impulse is 1, since the integral evidently gives the area of the rectangle

in Fig. 117:

1/( )

0k

k if a t a kf t a

otherwise≤ ≤ +⎧

− = ⎨⎩


We can represent in terms of two unit step functions,

From (4) we obtain the Laplace transform

The limit of as is denoted by ,

is called the Dirac delta function.

0

1( ) 1.a k

k k aI f t a dt dt

k∞ +

= − = =∫ ∫( )kf t a−

1( ) [ ( ) ( ( ))].kf t a u t a u t a kk

− = − − − +

{ } ( )1 1( ) [ ]ks

as a k s ask

eL f t a e e eks ks

−− − + − −

− = − =

( )kf t a− 0 ( 0)k k→ > ( )t aδ −

0( ) lim ( ).kkt a f t aδ

→− = −

( )t aδ −


The Laplace transform of ( )t aδ −

{ }( ) .asL t a eδ −− =

1-25

5.4 Differentiation and Integration of Transform

• -Differentiation of Transforms-

consequently, if , then

differentiation of the transform of a function corresponds to the multiplication of the function by .

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0( ) ( ) ( )stF s L f e f t dt

∞ −= = ∫

( ) ( )L f F s=0

( ) [ ( ) ]stF s e t f t dt∞ −′ = −∫

{ ( )} ( )L t f t F s′= −

t−1{ ( )} ( )L F s t f t− ′ = −

1-26

• -Integration of Transforms-

If satisfies the conditions of the existence and the limit of ,

as approaches 0 from the right, exists, then

integration of the transform of a function corresponds to the division of by .

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( ) ( )s

f tL F s d st

∞⎧ ⎫ =⎨ ⎬⎩ ⎭ ∫

t

( )f t

{ }1 ( )( )s

f tL F s d st

∞− =∫

( )f t ( )f t t

( )f tt

1-27

EXAMPLE 1 Integration of transforms

Find the inverse transform of the function

Sol.>>

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2

2ln 1sω⎛ ⎞

+⎜ ⎟⎝ ⎠

2 2 2

22 3 2 2 2 2

2

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

d sds s s s s s s

s

ω ω ωω ω ω

⎛ ⎞− + = − ⋅ − = = −⎜ ⎟ + +⎝ ⎠ +

1 12 2

2( ) ( ) 2 2 2cossf t L F L ts s

ωω

− − ⎧ ⎫= = − = −⎨ ⎬+⎩ ⎭

21 1

2

( )ln 1 ( )s

f tL L F s d ss tω ∞

− −⎧ ⎫ ⎧ ⎫⎛ ⎞⎪ ⎪+ = =⎨ ⎬ ⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠ ⎩ ⎭⎩ ⎭

∫

21

2

2ln 1 (1 cos )L ts tω ω− ⎧ ⎫⎛ ⎞⎪ ⎪+ = −⎨ ⎬⎜ ⎟

⎪ ⎪⎝ ⎠⎩ ⎭

1-28

5.5 Convolution. Integral Equations

THEOREM 1 (CONVOLUTION THEOREM)

Let and satisfy the hypothesis of the existence theorem.

Then the product of their transforms and

is the transform of the convolution of and ,

which is denoted by and defined by

• The convolution has the properties

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( ) ( )F s L f=

0( ) ( ) ( ) ( ) ( )

th t f g t f g t dτ τ τ= ∗ = −∫

( )f t ( )g t( ) ( )G s L g=

( ) ( )H s L h= ( )f t ( )g t( )h t( )( )f g t∗

( ) ( )f g g f∗ = ∗

1 2 1 2( )f g g f g f g∗ + = ∗ + ∗

( )f g∗

(commutative law)

(distributive law)

1-29

EXAMPLE 1 Convolution

Using convolution, find the inverse of

Sol>>

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2 2 2 2

1 1 1( )( 1) 1 1

H ss s s

= = ⋅+ + +

( ) ( )f g v f g v∗ ∗ = ∗ ∗

0 0 0f f∗ = ∗ =

(associative law)

( )h t

1

0

0 0

( ) ( ) sin sin

sin sin ( )

1 1cos cos(2 )2 2

1 1cos sin2 2

t

t t

h t L H t t

t d

t d t d

t t t

τ τ τ

τ τ τ

− = ∗

= −

= − + −

= − +

∫

∫ ∫

1-30

• -Differential Equations-

has the subsidiary equation

The solution of the latter is

we have

Obtain from the convolution theorem the integral representation.

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0( ) ( ) ( )

ty t g t r dτ τ τ= −∫

( )y ay by r t′′ ′+ + =

2( ) ( ) (0) (0) ( )s as b Y s a y y L r′+ + = + + +

2

( ) [ ( ) (0) (0) ] ( ) ( ) ( )( ) ( )( ) 1 ( )

Y s s a y y Q s R s Q swith R s L r

Q s s as b

′= + + +=

= + +

(0) (0) 0y y′= = Y RQ=


5.6 Partial Fractions

Differential Equations

The solution of a subsidiary equation of a differential equation usually

Comes out as a quotient of two polynomials,

Hence we can often determine its inverse by writing as a sum of

Partial fractions and obtain the inverse of the latter from a table and the first

Shifting theorem (Sec.5.1)

The form of the partial fractions depends on the kind of factors in the product

form of .

( )Y s

( )( )( )

F sY sG s

=

( )Y s

( )G s


Case 1. Unrepeated Factor

Case 2. Repeated Factor

Repeated factors , etc., require partial fractions

(1) , etc., respectively.

Case 3. Unrepeated Complex Factors

Such factors occur, for instance, in connection with vibrations. If

with complex is a factor of , so is with

the conjugate. To there

corresponds the partial fraction.

s a−

( )ms a−

2 3( ) , ( )s a s a− −

2 12( )

A As a s a

+− −

3 2 13 2( ) ( )

A A As a s a s a

+ +− − −

( ) ( )s a s a− −

s a−

a iα β= + ( )G s s a−

a iα β= − 2 2( ) ( ) ( )s a s a s α β− − = − +


(3) or

Case 4. Repeated Complex Factors

In this case the partial fractions are of the form

(4)

This case is important, for instance, in connection with resonance.

( )( )As B

s a s a+

− − 2 2( )As B

s α β+

− +

2[( ) ( )]s a s a− −

2[( )( )] ( )( )As B Ms N

s a s a s a s a+ +

+− − − −


5.7 Systems of Differential Equations

For a first-order linear system

Writing , we obtain

From (1) in Sec. 5.2 the subsidiary equation

Or, by collecting the and terms,

'1 11 1 12 2 1

'2 21 1 22 2 2

( )

( )

y a y a y g t

y a y a y g t

= + +

= + +(1)

1 1 11 1 12 2 1

2 2 21 1 22 2 2

(0) ( )

(0) ( )

sY y a Y a Y G s

sY y a Y a Y G s

− = + +

− = + +

11 1 12 2 1 1

21 2 22 2 2 2

( ) (0) ( )

( ) (0) ( )

a s Y a Y y G s

a Y a s Y y G s

− + =− −

+ − =− −

1Y − 2Y −

1 1 2 2 1 1 2 2( ), ( ), ( ), ( )Y L y Y L y G L g G L g= = = =

(2)


This must be solved algebraically for and .

The solution of the given system is then obtained if we take the inverse

.

1( )Y s 2( )Y s

1 11 1 2 2( ), ( )y L Y y L Y− −= =


5.8 Laplace Transform: General Formulas



5.9 Table of Laplace Transforms



Documents

LAPLACE TRANSFORMS · 2016-04-05 · LAPLACE TRANSFORMS. 1-2 1. LaplaceTransform. Inverse Transform. Linearity. Shifting 2. Transforms of Derivatives and Integrals. Differential Equations