LAPLACE TRANSFORMS · 2016-04-05 · LAPLACE TRANSFORMS. 1-2 1. LaplaceTransform. Inverse...
Transcript of LAPLACE TRANSFORMS · 2016-04-05 · LAPLACE TRANSFORMS. 1-2 1. LaplaceTransform. Inverse...
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조조 선선 대대 학학 교교 기계공학과기계공학과
담당교수담당교수 정정 상상 화화
LAPLACELAPLACETRANSFORMSTRANSFORMS
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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
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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)
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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+ = +
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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−= −
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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>
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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 >
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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 >
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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>
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(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
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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+ + = + + +
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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
= =
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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 − ⎧ ⎫= ⎨ ⎬⎩ ⎭∫
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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= −
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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
− ⎛ ⎞=⎜ ⎟+⎝ ⎠
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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
− ⎧ ⎫⎛ ⎞⎪ ⎪ = = −⎨ ⎬⎜ ⎟+⎪ ⎪⎝ ⎠⎩ ⎭∫
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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 ≥
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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
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- 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− −− − =
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(4) { }( ) .aseL u t a
s
−
− = ( 0)s >
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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≤ ≤ +⎧
− = ⎨⎩
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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 −
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The Laplace transform of ( )t aδ −
{ }( ) .asL t a eδ −− =
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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− ′ = −
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• -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
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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− ⎧ ⎫⎛ ⎞⎪ ⎪+ = −⎨ ⎬⎜ ⎟
⎪ ⎪⎝ ⎠⎩ ⎭
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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)
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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
− = ∗
= −
= − + −
= − +
∫
∫ ∫
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• -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=
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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
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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− − = − +
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(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+ +
+− − − −
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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)
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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− −= =
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5.8 Laplace Transform: General Formulas
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5.9 Table of Laplace Transforms
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