Advanced Engineering Mathematics
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.
Advanced Engineering Mathematics
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
Advanced Engineering Mathematics
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
Advanced Engineering Mathematics
6.1 Laplace Transform. Linearity.
First Shifting Theorem(s-Shifting)
 Laplace Transform : =
F ( s ) L=
(f)
 Inverse Transform : L −1 ( F ) = f ( t )
∞
∫e
− st
f ( t )dt
0
 Ex.1 Let
=
f ( t ) 1 when t ≥ 0. Find F ( s ) .
∞
∞
1
1
L (f )=
L (1) =
− e − st =
( s > 0)
∫0 e dt =
s
s
0
− st
 Ex.2 =
Let f ( t ) e at when t ≥ 0, where a is a constant. Find L ( f ) .
∞
∞
1 − ( s − a )t
1
=
L ( e ) ∫=
e e dt
=
e
a−s
s−a
0
0
at
− st at
Advanced Engineering Mathematics
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 a f(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
1
1
cosh at = ( e at + e − at ) , inh at = ( e at − e − at )
2
2
⇒
⇒
1
1
, L ( e − at ) =
L ( e at ) =
s−a
s+a
1
1 1
1 
s
 L ( e at ) + L ( e − at )  =
+

= 2


2
2  s − a s + a  s − a2
1
1 1
1 
a
 L ( e at ) − L ( e − at )  =
sinh at )
L (=
−

= 2


2
2  s − a s + a  s − a2
L ( cosh at )=
Advanced Engineering Mathematics
6.1 Laplace Transform. Linearity.
First Shifting Theorem(s-Shifting)
Advanced Engineering Mathematics
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 e at f ( t ) has the transform F(s - a)
(where s – a > k). In formulas, L {e at f (=
t )} F ( s − a )
or, if we take the inverse on both sides, =
e at f ( t ) L -1 { F ( s − a )}
 Ex.5 From Example 4 and the first theorem we immediately obtain formulas 11 and 12 in Table 6.1,
=
L ( e at cos ωt )
s−a
, L ( e at sin ωt )
=
2
2
(s − a) + ω
ω
(s − a) + ω2
2
For instance, use these formulas to find the inverse of the transform
L (f )=
3s − 137
s + 2 s + 401
2
 3 ( s + 1) − 140 



 −t
s +1
20
f =
L −1 
3L −1 
e ( 3cos 20t − 7 sin 20t )
=
 − 7L −1 
=
2
2
2
 ( s + 1) + 400 
 ( s + 1) + 400 
 ( s + 1) + 400 






Advanced Engineering Mathematics
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 ) ≤ Me kt for all t ≥ 0 and some constants M and k, then the Laplace transform
L ( f ) exists for all s > k.
Advanced Engineering Mathematics
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=
s n L ( f ) − s n −1 f ( 0 ) − s n − 2 f ' ( 0 ) −  − f (
n
n −1)
( 0)
 Ex.1 Let f ( t ) = t sin ωt . Find the transform of f (t).
f ( 0 ) = 0,
f '(t ) =
sin ωt + ωt cos ωt ,
f ' ( 0) =
0,
f '' ( t ) 2ω cos ωt − ω 2t sin ωt
=
⇒
L ( f '')= 2ω
s
− ω 2L ( f )= s 2L ( f )
2
s +ω
2
⇒
 L ( f )= L ( t sin ωt )=
2ω s
(s +ω )
2
2 2
Advanced Engineering Mathematics
6.2 Transform of Derivatives and Integrals.
ODEs
 Theorem 3 Laplace Transform of Integral
L ( f ( t ) ) =⇒
F (s)
 Ex.3 Find the inverse of
t
 1
L  ∫ f (τ ) dτ  =
F ( x) ,
0
 s
t
1
-1
0
1
1
and 2 2
.
2
s(s +ω )
s (s + ω2 )
2
 1  1
sin ωt ⇒
=
L −1  2
2 
 s +ω  ω
⇒

L  F ( x) 
∫ f (τ ) dτ =
s


 t sin ωτ
1
1

dτ =
=
L −1 
(1 − cos ωt )
∫
2
2
 s(s +ω )  0 ω
ω2



 1 t
1
t sin ωt

L −1  2 2
=
−
(1 − cos ωτ )dτ =
 s ( s + ω 2 )  ω 2 ∫0
ω2
ω3


Advanced Engineering Mathematics
6.2 Transform of Derivatives and Integrals.
ODEs
 Differential Equations, Initial Value Problems : y ''+ =
ay '+ by r ( t ) , =
y ( 0) K0 , =
y ' ( 0 ) K1
Step 1 Setting up the subsidiary equation.
=
( y ) , R L ( r ))
(Y L=
 s 2Y − sy ( 0 ) − y ' ( 0 )  + a  sY − y ( 0 )  + bY = R ( s ) ⇒
( s + as + b ) Y = ( s + a ) y ( 0 ) + y ' ( 0 ) + R ( s )
2
Step 2 Solution of the subsidiary equation by algebra.
Transfer Function=
: Q (s)
1
1
=
2
2
s + as + b 
1 
1 2
s + a +b − a
2 
4

Solution of the transfer function : Y ( s ) =+
( s a ) y ( 0 ) + y ' ( 0 )  Q ( s ) + R ( s ) Q ( s )
Step 3 Inversion of Y to obtain y.
Advanced Engineering Mathematics
6.2 Transform of Derivatives and Integrals.
ODEs
 Ex. 4 Solve =
y ''− y t , =
y ( 0 ) 1,
Step 1 s Y − sy ( 0 ) − y ' ( 0 ) − Y =
2
1
Step 2 Q = 2
s −1
⇒
=
y ' ( 0) 1
1
s2
⇒
Y =( s + 1) Q +
( s − 1) Y = s + 1 + s1
2
2
1
s +1
1
1
1
 1
+ 2 2
= + 2
− 2
Q =2
2
s
s − 1 s ( s − 1) s − 1  s − 1 s 
1 
 1 
−1 
−1  1 
L −1 (Y ) =
L −1 
et sinh t − t
Step 3 y ( t ) =
 + L  2  − L  2  =+
 s −1 
 s −1 
s 
Advanced Engineering Mathematics
6.2 Transform of Derivatives and Integrals.
ODEs

1.
2.
3.
Advantages of the Laplace Method
Solving a nonhomogeneous ODE does not require first solving the homogeneous ODE.
Initial values are automatically taken care of.
Complicated inputs can be handled very efficiently.
Advanced Engineering Mathematics
6.3 Unit Step Function(Heaviside Function).
Second Shifting Theorem(t-Shifting)
 Unit Step Function
0
1
• Unit Step Function (Heaviside function) : u ( t − a ) =

(t < a )
(t > a )
e − as
• Laplace transform of unit step function : L {u ( t − a )} =
s
 Theorem 1 Second Shifting Theorem; Time Shifting
L ( f ( t ) )= F ( s )
⇒
L ( f ( t − a ) u ( t − a ) )= e − as F ( s ) ,
f ( t − a ) u ( t − a )= L -1 {e − as F ( s )}
Advanced Engineering Mathematics
6.3 Unit Step Function(Heaviside Function).
Second Shifting Theorem(t-Shifting)
 Ex.1 Write the following function using unit step functions and find its transform.

 2
 2
f (t )  t
=
2

 cos t

( 0 < t < 1)
(1 < t < π 2 )
(t > π 2 )
1 
 1 
 1 
Step 1 Using unit step functions : f ( t ) = 2 (1 − u ( t − 1) ) + t 2  u ( t − 1) − u  t − π   + ( cos t ) u  t − π 
2 
 2 
 2 
1
 1
 1 1 1 
2
1

Step 2 L  t 2u ( t − 1)  = L  ( t − 1) + ( t − 1) +  u ( t − 1)  =  3 + 2 +  e − s
2
2s 
2

 2
 s s
 1  1  2 π  1  π 2   1    1
 1 2  1 
π π 2  −π s 2
u  t − π  =
+
+
L  t u  t − π  =
L   t − π  +  t − π  +

 3
e
2


8s 
 2  2 
 2  2  2  2  8   2    s 2 s
−π s
   1    1 
1

 1 
L ( cos t ) u  t − π   =
e 2
− 2
L −  sin  t − π   u  t − π   =
s +0
 2 

   2    2 
⇒
−π s
1
2 2
π π 2  −π s 2
1
1 1 1 
e 2
∴ L ( f ) = − e− s +  3 + 2 +  e− s −  3 + 2 +
− 2
e
s s
2s 
8s 
s +0
s s
 s 2s
Advanced Engineering Mathematics
6.3 Unit Step Function(Heaviside Function).
Second Shifting Theorem(t-Shifting)
 Ex. 2 Find the inverse transform f (t) of
F (s) =
e− s
e −2 s
e −3 s
+
+
s 2 + π 2 s 2 + π 2 ( s + 2 )2
 1  sin π t
=
L −1  2
2 
π
 s +π 
 1 
1
t
⇒ L −1 
te −2t
L −1  2  =
=
( The first shifting theorem )
2


s 
s
2
+
(
)


f (t )
⇒ =
1
π
sin (π ( t − 1) ) u ( t − 1) +
0

− ( sin π t )
π
=
0

−2( t −3)
 ( t − 3) e
1
π
sin (π ( t − 2 ) ) u ( t − 2 ) + ( t − 3) e −2(t −3)u ( t − 3)
( 0 < t < 1)
(1 < t < 2 )
( 2 < t < 3)
( t > 3)
Advanced Engineering Mathematics
6.4 Short Impulses. Dirac’s Delta Function.
Partial Fractions
∞
 0

 Dirac delta function or the unit impulse function : δ ( t − a ) =

a)
(t =
( otherwise )
• Definition of the Dirac delta function
 1
f k ( t=
− a)  k
 0
∞
(a ≤ t ≤ a + k )
( otherwise )
a+k
• ∫ f k ( t − a ) dt = ∫ 1 dt = 1
k
a
0
⇒
δ ( t=
− a ) lim f k ( t − a )
k → 0
∞
⇒
∫ δ ( t − a )dt = 1
0
 Laplace transform of the Dirac delta function
1
u ( t − a ) − u ( t − ( a + k ) ) 
k
− ks
1  − as −( a + k ) s 
− as 1 − e
⇒ L ( f k ( t − a=
e −e
=
))
 e
ks 
ks
fk (t − =
a)
⇒
L {δ ( t − a=
)} e− as
Advanced Engineering Mathematics
6.4 Short Impulses. Dirac’s Delta Function.
Partial Fractions
 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 : y ''+ 3 y '+ 2 y = δ ( t − 1) ,
y ( 0 ) = 0,
Subsidiary equation : s 2Y + 3sY + 2Y =
e− s
e− s
1  −s
 1
=
−
e
( s + 1)( s + 2 )  s + 1 s + 2 
⇒ Y (s) =
0
∴ y ( t=
) L −1 (Y=)  −(t −1) −2(t −1)
−e
 e
( 0 < t < 1)
( t > 1)
y ' ( 0) = 0
Advanced Engineering Mathematics
6.5 Convolution. Integral Equations
t)
 Convolution : ( f ∗ g )(=
t
∫ f (τ )g ( t − τ ) dτ
0
 Property of convolution
• Commutative Law : f ∗ g = g ∗ f
• Distributive Law : f ∗ ( g1 + g 2 ) = f ∗ g1 + f ∗ g 2
• Associative Law : ( f ∗ g ) ∗ v = f ∗ ( g ∗ v )
• f ∗0 = 0∗ f = 0
• Unusual Properties of Convolution : f ∗1 ≠ f
 Theorem 1 Convolution Theorem
L ( f ∗ g) =
L ( f )L ( g )
 Ex. 1 Let H ( s ) = 1
( s − a ) s 
 1  − at
L −1 
= e ,
 s−a
. Find h(t).
1
L −1  = 1
s
⇒
h ( t )= e at ∗1=
t
∫ e •1dτ = a ( e − 1)
0
aτ
1
at
Advanced Engineering Mathematics
6.5 Convolution. Integral Equations
 Integral Equation : Equation in which the unknown function appears in an integral.
 Ex.6 Solve the Volterra integral equation of the second kind
t
y ( t ) − ∫ y (τ ) sin ( t − τ ) dτ =
t
0
Equation written as a convolution : y − y ∗ sin t =
t
1
1
Apply the convolution theorem : Y ( s ) − Y ( s ) 2
=
s + 1 s2
s2 + 1 1 1
Y (s) =
=
+
s4
s2 s4
⇒
t3
y (t ) =
t+
6
Advanced Engineering Mathematics
6.6 Differentiation and Integration of
Transforms.
ODEs with Variable Coefficients
 Differentiation of Transforms
∞
∞
L (f )=
F (s) =
( t ) ∫ e f ( t ) dt
− st
⇒
0
⇒
dF
=
− ∫ e − st tf ( t ) dt =
−L ( tf )
F '( s) =
ds
0
L ( tf ( t ) ) =
− F ' ( s ) , L −1 { F ' ( s )} =
−tf ( t )
 Ex.1 We shall derive the following three formulas.
L( f )
1
(s + β )
2 2
2
s
(s + β )
2 2
2
s2
(s + β )
2
2 2
f(t)
1
2β 3
( sin β t − β t cos β t )
t
sin βt
2β
1
(sin βt + βt cos βt )
2β
Advanced Engineering Mathematics
6.6 Differentiation and Integration of
Transforms.
ODEs with Variable Coefficients
L ( sin β t ) =
L ( cos β t ) =
β
s +β
2
By differentiation
2
s
2
s +β2


1
L  t cos β t ± sin β t =

β


L ( t sin β t ) =
By differentiation
s2 − β 2
1
± 2
=
2
2
2 2
(s + β ) s + β
2β s
(s + β )
2
2 2
 t

s
sin β t  =
L
2
 2β
 ( s2 + β 2 )
( s 2 + β 2 ) − 2s 2 =
s2 − β 2
L ( t cos β t ) =
−
2
2
( s2 + β 2 )
( s2 + β 2 )
(s − β ) ± (s + β )
(s + β )
2
2
2
2
2 2
2
Advanced Engineering Mathematics
6.6 Differentiation and Integration of
Transforms.
ODEs with Variable Coefficients
 Integration of Transforms : L 
⇒
∞
 f (t )
L  ∫ F ( s )ds  =
t
s

 ω2 
s2 + ω 2
 Ex.2 Find the inverse transform of ln 1 + 2  =
ln
s 
s2

Derivative
 ω2 
s2 + ω 2
ln 1 + 2  =
ln
s 
s2

2s
2s
d
2
ln ( s 2 + ω 2 ) − ln s=
− 2
2
2
ds
s +ω
s
(
)
 f (t )  ∞
=
∫ F ( s )ds
 t  s
−1
Case 1 Differentiation of transform
 ω2 
F ( s ) =L ( f ) =ln 1 + 2 
s 

2 cos ωt − 2
∴ f (t ) =
−t
⇒
2s 
 2s
− 2  =2 cos ωt − 2 =−tf ( t )
L −1 ( F ' ( s ) ) =L −1  2
2
s 
 s +ω
Case 2 Integration of transform
2s
2
G ( s )= 2
−
⇒
g ( t ) = L −1 ( G ) = 2 ( cos ωt − 1)
2
s +ω
s
  ω2 
∞

g (t ) 2
∴ L −1  ln 1 + 2   =
−
=
L −1  ∫ G ( s ) ds  =
(1 − cos ωt )
s 
t
t
s

 
Advanced Engineering Mathematics
6.6 Differentiation and Integration of
Transforms.
ODEs with Variable Coefficients
 Special Linear ODEs with Variable Coefficients
d
dY
 sY − y ( 0 )  =−Y − s
ds
ds
d
dY
L ( ty '') =
−  s 2Y − sy ( 0 ) − y ' ( 0 )  =
−2 sY − s 2
+ y ( 0)
ds
ds
L ( ty ') =−
 Ex.3 Laguerre’s Equation, Laguerre Polynomials
ny 0 =
Laguerre’s ODE is ty ''+ (1 − t ) y '+=
( n 0,1, 2,)
dY 



2 dY
 −2 sY − s ds + y ( 0 )  + sY − y ( 0 ) −  −Y − s ds  + nY= 0
⇒
dY
n +1− s
n +1 
 n
ds =
=
−
−

 ds
2
Y
s-s
s 
 s −1
1,

 t n
∴ ln ( t =
) L (Y =)  e d n −t
(t e ) ,

 n ! dt n
⇒
n=0
−1
n = 1, 2, 
⇒
( s − 1)
Y=
s n +1
n
+ ( n + 1 − s ) Y= 0
( s - s ) dY
ds
2
Advanced Engineering Mathematics
6.7 Systems of ODEs
 Ex.3 Model of Two Masses on Springs
The mechanical system consists of two bodies of mass 1 on three springs of the same spring constant k and of
negligibly small masses of the springs. Also damping is assumed to be practically zero.
Model of the
−ky1 + k ( y2 − y1 )
physical system : y1 '' =
y2 '' =
−k ( y2 − y1 ) − ky2
Initial
conditions :
=
y1 ( 0 ) 1,=
y2 ( 0 ) 1,
y1 ' ( 0 ) =
Laplace transform
s 2Y1 − s − 3k =−kY1 + k (Y2 − Y1 )
s 2Y2 − s + 3k =−k (Y2 − Y1 ) − kY2
3k , y2 ( 0 ) = − 3k
Cramer’s rule
or Elimination
3k )
( s + 3k ) ( s + 2k ) + k ( s − =
Inverse =
Y
2
−1
y1 ( t ) L=
=
(Y1 ) cos kt + sin 3kt
−1
=
y2 ( t ) L=
(Y2 ) cos kt − sin 3kt
transform
3k
s
+ 2
s + k s + 3k
( s + 2k ) − k
3k )
( s − 3k ) ( s + 2k ) + k ( s − =
3k
s
=
Y
−
s + k s + 3k
( s + 2k ) − k
1
2
2
2
2
2
2
2
2
2
2
2