Ch. 6 Laplace Transforms
2
Ch. 6 Laplace Transforms
• The Laplace transform method is a powerful method for
solving linear ODEs and corresponding initial value
problems as well as systems of ODEs arising in
engineering.
• Advantage of the Laplace transform
▫ Problems are solved more directly, initial value problems without
first determining a general solution, and nonhomogeneous ODEs
without first solving the corresponding homogeneous ODE.
▫ The use of the unit step function and Dirac’s delta make the
method particularly powerful for problems with inputs that have
discontinuities or represent short impulses or complicated
periodic functions.
3
Ch. 6 Laplace Transforms
• 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
①
AP
Algebraic
Problem
②
Solving
AP
by Algebra
< Solving an IVP by Laplace transforms >
③
Solution
of the
IVP
4
Ch. 6 Laplace Transforms
s-domain analysis
(algebraic equation)
Laplace
Transform
Problem
Inverse
Transform
Time domain analysis
(differential equation)
< Relation of Time domain and s-Domain >
Answer
5
6.1 Laplace Transform. Linearity.
First Shifting Theorem (s-Shifting)

• Laplace Transform : F ( s ) = L ( f ) =  e− st f ( t )dt
• Inverse Transform : L
−1
( F ) = f (t )
0
▫ Ex.1) Let f ( t ) = 1 when t  0 . Find F ( s ) .


1
1
L ( f ) = L (1) =  e− st dt = − e− st =
s
s
0
0
( s  0)
▫ Ex.2) Let f ( t ) = eat when t  0 ,where a is a constant. Find L ( f ).


1 − ( s − a )t
1
L ( eat ) =  e− st eat dt =
e
=
a−s
s−a
0
0
6
6.1 Laplace Transform. Linearity.
First Shifting Theorem (s-Shifting)
• 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
L ( af ( t ) + bg ( t ) ) = aL ( f ( t ) ) + bL ( g ( t ) )
▫ Ex.3) Find the transform of cosh at and sinh at.
cosh at =
1 at
1
e + e − at ) , sinh
inh at = ( e at − e − at )
(
2
2


L ( e at ) =
1
1
, L ( e − at ) =
s−a
s+a
L ( cosh at ) =
1
1 1
1 
s
 L ( e at ) + L ( e − at )  = 
+
=
 2  s − a s + a  s 2 − a 2
2
L ( sinh at ) =
1
1 1
1 
a
 L ( e at ) − L ( e − at )  = 
−
=
 2  s − a s + a  s 2 − a 2
2
7
6.1 Laplace Transform. Linearity.
First Shifting Theorem (s-Shifting)
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6.1 Laplace Transform. Linearity.
First Shifting Theorem (s-Shifting)
• Theorem 2 First Shifting Theorem, s-shifting
If f(t) has the transform F(s) ( where s > k for some k ), then
eat f ( t ) has the transform F(s - a) (where s – a > k). In
formulas, L eat f ( t ) = F ( s − a ) or, if we take the inverse on both
sides, eat f ( t ) = L -1 F ( s − a ) .
▫ Ex.5) From Example 4 and the first shifting theorem we
immediately obtain formulas 11 and 12 in Table 6.1,
L ( eat cos t ) =
s−a
at
,
L
e
sin t ) =
(
2

(s − a) + 
(s − a) + 2
▫ For instance, use these formulas to find the inverse of the
3s − 137
transform L ( f ) = 2
.
2
2
s + 2s + 401
 3 ( s + 1) − 140 



 −t
s +1
20
−1
−1
f =L 
 = 3L 
 − 7L 
 = e ( 3cos 20t − 7sin 20t )
 ( s + 1)2 + 400 
 ( s + 1)2 + 400 
 ( s + 1)2 + 400 






−1
9
6.1 Laplace Transform. Linearity.
First Shifting Theorem (s-Shifting)
• 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 )  Mekt
for all t  0 and some constants M and k, then the Laplace
transform L ( f ) exists for all s > k.
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6.2 Transform of Derivatives and Integrals.
ODEs
• Theorem 1 Laplace Transform of Derivatives
L ( f ') = sL ( f ) − f ( 0 ) ,
L ( f '') = s 2L ( f ) − sf ( 0 ) − f ' ( 0 ) .
• Theorem 2 Laplace Transform of Derivatives
L ( f ( n) ) = s n L ( f ) − s n−1 f ( 0 ) − s n−2 f ' ( 0 ) − − f ( n−1) ( 0 )
▫ Ex.1) Let f ( t ) = t sin t . Find the transform of f (t).
f ( 0 ) = 0,
f ' ( t ) = sin t + t cos t ,
f ' ( 0 ) = 0,
f '' ( t ) = 2 cos t −  2t sin t

s
2
2
L ( f '' ) = 2 2
−

L
f
=
s
L (f)
(
)
s + 2

∴ L ( f ) = L ( t sin t ) =
2 s
(s + )
2
2 2
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6.2 Transform of Derivatives and Integrals.
ODEs
• Theorem 3 Laplace Transform of Integral
L ( f (t )) = F ( s )

t
 1
L   f ( ) d  = F ( xs ) ,
0
 s
▫ Ex.3) Find the inverse of
 1  1
L  2
= sin t
2 
s
+


 
−1



-1  1
s
f

d

=
L
F
x
(
)
(
)


0
s

t
1
1
and
.
s ( s2 +  2 )
s2 ( s2 +  2 )

 t sin 
1
1

L 
=
d

=
(1 − cos t )
 s ( s 2 +  2 )  0 
2



 1 t
1
t sin t
−1

L  2 2
=
1
−
cos

d

=
−
(
)
 s ( s +  2 )   2 0
2
3


−1
12
6.2 Transform of Derivatives and Integrals.
ODEs
• Differential Equations, Initial Value Problems :
y ''+ ay '+ by = r ( t ) , y ( 0 ) = K 0 , y ' ( 0 ) = K1
Step 1 Setting up the subsidiary equation. (Y = L ( y ) ,
 s 2Y − sy ( 0 ) − y ' ( 0 ) + a  sY − y ( 0 ) + bY = R ( s ) 
R = L ( r ))
( s + as + b )Y = ( s + a ) y ( 0) + y ' ( 0) + R ( s )
2
Step 2 Solution of the subsidiary equation by algebra.
1
Transfer Function : Q ( s ) = 2 1
=
2
s + as + b
Solution :
Y ( s ) = ( s + a ) y ( 0 ) + y ' ( 0 ) Q ( s ) + R ( s ) Q ( s )
Step 3 Inversion of Y to obtain y.
1 
1 2

s
+
a
+
b
−
a


2 
4

13
6.2 Transform of Derivatives and Integrals.
ODEs
▫ Ex.4) Solve y ''− y = t , y ( 0 ) = 1, y ' ( 0 ) = 1 .
1
s
Step 1 s 2Y − sy ( 0 ) − y ' ( 0 ) − Y = 2

( s 2 − 1) Y = s + 1 +
1
s2
Step 2
Q=
1
s2 −1
Step 3

Y = ( s + 1) Q +
1
s +1
1
1
1
 1
Q
=
+
=
+
−


s2
s 2 − 1 s 2 ( s 2 − 1) s − 1  s 2 − 1 s 2 
1 
 1 
−1 
−1  1 
t
y ( t ) = L −1 (Y ) = L −1 
+
L
−
L
=
e
+ sinh t − t

 2 
 2
 s −1 
 s −1 
s 
14
6.2 Transform of Derivatives and Integrals.
ODEs
• Advantages of the Laplace Method
1.1. Solving a nonhomogeneous ODE does not require first
solving the homogeneous ODE.
2.2. Initial values are automatically taken care of.
3.3. Complicated inputs can be handled very efficiently.
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Thank you
• Reading assignment
▫ ~ Section 6.4