Transcript Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe

Group 2
Bhadouria, Arjun Singh
Glave, Theodore Dean
Han, Zhe
Chapter 5. Laplace Transform
Chapter 19. Wave Equation
Laplace Transform
Chapter 5
5. Laplace Transform
•
•
•
•
5.1. Introduction & Definition
5.2. Calculation of the Transform
5.3. Properties of the Transform
5.4. Application to the Solution of Differential
Equations
• 5.5. Discontinuous Forcing Functions; Heaviside
Step Function
• 5.6. Impulsive Forcing Functions; Dirac Impulse
Function
• 5.7. Additional Properties
5.1. Introduction & Definition
• The Laplace transform is a widely used integral transform. Denoted , it is a
linear operator of a function f(t) with a real argument t (t ≥ 0) that
transforms it to a function F(s) with a complex argument s. The Laplace
transform has the useful property that many relationships and operations
over the originals f(t) correspond to simpler relationships and operations
over the images F(s). The Laplace transform has many important
applications throughout the sciences.
• The Laplace transform is used for solving differential and integral
equations. In physics and engineering, it is used for analysis of linear timeinvariant systems. In this analysis, the Laplace transform is often
interpreted as a transformation from the time-domain, in which inputs
and outputs are functions of time, to the frequency-domain, where the
same inputs and outputs are functions of complex angular frequency, in
radians per unit time. Given a simple mathematical or functional
description of an input or output to a system, the Laplace transform
provides an alternative functional description that often simplifies the
process of analyzing the behavior of the system, or in synthesizing a new
system based on a set of specifications.
http://en.wikipedia.org/wiki/Laplace_transform
Basic idea
Laplace
Transformation
Time domain
unknown f(t), d/dt, Diff Eqs
Solve
Differential
Equations
Time domain
known f(t)
Frequency domain
unknown F(s), Alg Eqs
Solve
Algebraic
Equations
Frequency domain
known F(s)
Inverse
Laplace
Transform
http://faculty.mercer.edu/olivier_pd/documents/Ch2LaplaceTransforms.ppt
• Given a known function K(t,s), an integral transform
of a function f is a relation of the form
b
F ( s)   K (t , s) f (t )dt
a
• The Laplace Transform of f is defined as

L f (t )  F (s)   est f (t )dt
0
where the kernel function is K(s,t) = e-st , a=0, b= .
F (s) is the symbol for the Laplace transform,
L is the Laplace transform operator,
and f(t) is some function of time, t.
How to solve problems
Time Domain
y
2 
t   y t   yt   xt 
Laplace transform
Frequency Domain
1
H(s )  2
s  3s  2
1
Xt  
s
1
x t   1
Solve algebraic equation
Inverse Laplace transform
1 t 1  2 t
yt    e  e
2
2
1
1
s s 2  3s  2
Douglas Wilhelm Harder, University of Waterloo.
5.2. Calculation of the Transform
Since the Laplace Transform is defined by
an improper integral, thus it must be checked
whether the transform F(s) of a given function
f(t) exists, that is
whether the integral converges.

F (s)   est f (t )dt
0
EXAMPLE 2.1.
Consider the following improper integral.


0
est dt
We can evaluate this integral as follows:


0
st b
e
e dt  lim  e dt  lim
b  0
b  s
st
b
st
0


1
 lim e sb  1
s b 
Note that if s = 0, then est = 1. Thus the following two
cases hold:


0
1
e dt   , if s  0; and
s
st


0
e st dt diverges , if s  0.
EXAMPLE 2.2.
Consider the following improper integral.


0
st costdt
We can evaluate this integral using integration by parts:


0
b
st costdt  lim  st costdt
b  0
b
b

 lim st sin t 0   s sin tdt

0
b  


 lim st sin t 0  s cost
b 
b
b
0

 lim sb sin b  s cosb  1
b 
Since this limit diverges, so does the original integral.
Exponential order
Suppose that f is a function for which the following hold:
(1) f is piecewise continuous on [0, b] for all b > 0.
(2) | f(t) |  Kect when t  T, for constants c, K, M, with K, M > 0.
A function f that satisfies the conditions specified above is said
to to have exponential order as t  
EXAMPLE 2.3. Is f(t)=exp(4t)cost of exponential order?
Solution: Yes.
f (t )  e 4t cos t  e 4t  10e 4t
Therefore, K=10, c=4, and T=10 for instance
EXAMPLE 2.4. Is f(t)=ln(t) of exponential order?
Solution: Yes.
lim
t 
ln(t )
1 1
 lim
0
ct
ct
e
t  t ce
Therefore, K=100, c=1, and T=10 for instance
Piecewise Continuous
If an interval [a, b] can be partitioned by a finite number of
points a = t0 < t1 < … < tn = b such that
(1) f is continuouson each (t k , t k 1 )
(2) lim f (t )  , k  0, , n  1
t t k
(3) lim f (t )  , k  1,, n
t t k 1
Then the function f is piecewise continuous
Or we can say f is piecewise continuous on [a, b] if it is
continuous there except for a finite number of jump
discontinuities.
Picture from Paul's Online Math Notes
EXAMPLE 2.5. Piecewise continuous function
Consider the following piecewise-defined function f.
• (a)
t 2 ,
0  t 1

f (t )  3  t , 1  t  2
t  1 2  t  3

From this definition of f, and from the graph of f below, we see that f is
piecewise continuous on [0, 3].
• (b)
t 2  1,
0  t 1

1
f (t )  2  t  , 1  t  2
4,
2t 3

From this definition of f, and from the graph of f below, we see that f is
NOT piecewise continuous on [0, 3].
http://ebookbrowse.com/ch06-laplace-transform-ppt-d116265990
THEOREM 2.1
Existence of the Laplace Transform
Let f(t) satisfy these conditions:
(i) f(t) is piecewise continuous on 0  t  A for every A>0,
(ii) f(t) is of exponential order as t  
That is to say
(1) f is piecewise continuous on [0, b] for all b > 0.
(2) | f(t) |  Kect when t  T, for constants c, K, M, with K, M > 0.
Then the Laplace Transform of f exists for s > c.

L f (t )  F (s)   est f (t )dt finite
0
Inverse Laplace transform operator
By definition, the inverse Laplace transform operator,
L-1, converts an s-domain function back to the
corresponding time domain function:
1  i
st
f  t   L {F  s } 
F
(
s
)
e
ds

2 i  i
1
Importantly, both L and L-1 are linear operators.
Thus,
L{au  t   bv  t }  aL{u  t }  bL{v  t }  aU  s   bV  s 
L1{aU  s   bV  s }  au  t   bv  t 
Examples of Calculation
2.6.
2.7 f (t )  e  at ;
2.8

a
F  s    ae st dt   e  st
0
s
f(t)  a  constant;


0
0
F ( s)   e at e  st dt   e ( s  a )t dt 

0
1
sa
f (t )  sin at

F ( s )  L sin(at)    e
0
 st
b
sin atdt  lim  e  st sin atdt
b  0
b
s b  st


 st
 lim  (e cos at) / a   e cos at
0
b 
a 0


b
1 s
s b  st
  st

  lim (e sin at) / a   e sin at
0
a a b  
a 0

1 s2
a
  2 F ( s)  F ( s)  2
, s0
2
a a
s a
 a a
 0  
 s s
Laplace transform table
5.3. Properties of the Transform
• A various types of problems that can be
treated with the Laplace transform include
ordinary and partial differential equations as
well as integral equations.
THEOREM 3.0
The transform of an expression that is multiplied by a
constant is the constant multiplied by the transform.
That is:
Lkf (t )  kL f (t ) and L1 kF (s)  kL1 F (s)
THEOREM 3.1
Linearity of the Transform
• Suppose u and v are functions whose Laplace
transforms exist for s > a1 and s > a2, respectively.
• Then, for s greater than the maximum of a1 and a2,
the Laplace transform of au(t) + bv(t) exists. That is,
L au(t )  bv(t )    est au(t )  bv(t )dt is finite

0
With

L au(t )  bv(t )   a  est u(t )dt
0
Therefore

 b est v(t )dt  aLu(t )  bLv(t )
0
Lau(t )  bv(t )  aLu(t ) bLv(t )
EXAMPLE 3.1.
f (t) = 5e-2t - 3sin(4t) for t  0.
by linearity of the Laplace transform, and using results
of Laplace transform table, the Laplace transform F(s)
of f is:


F ( s)  L{ f (t )}  L 5e 2t  3 sin(4t )
5
12
 2t
 5L e
 3L sin(4t )  
 2
, s0
s  2 s  16


THEOREM 3.2
Linearity of the Inverse Transform
For any U(s) and V(s) such that the inverse transforms
L1{U  s } u  t  and L1{V  s } v  t  exist.
L1{aU  s   bV  s }  au  t   bv  t 
For any constants a,b.
Basic idea :
Consider a general expression,
F s 
N s
Ds

N s
n
  s  bi 
i 1
Expand a complex expression for F(s) into simpler terms, each
of which appears in the Laplace Transform table. Then you can
take the L-1 of both sides of the equation to obtain f(t).
EXAMPLE 3.2.
F s 
s5
 s  1 s  4 
Perform a partial fraction expansion (PFE)
1
s5
F (s) 
 s  1 s  4 


2
s 1 s  4
where coefficients 1 and  2 have to be determined.
To find 1 : Multiply both sides by s + 1 and let s = -1
To find  2 : Multiply both sides by s + 4 and let s = -4
 1 
Therefore,
s5
s4

s 1
4
3
2 
s 5
s 1

s 4
4 1
1 1

}
3 s 1 3 s  4
4 1 1
1 1 1
4 t 1 4t
 L {
} L {
} e  e
3
s 1 3
s4
3
3
f (t )  L1{F ( s )}  L1{
1
3
THEOREM 3.3
Transform of the Derivative
Let f(t) be continuous and f’(t) be piecewise continuous on
0≤t≤t。For every finite t。, and let f(t) be of exponential
order as t   so that there are constants K, c, T such that |
f(t) |  Kect when t  T. Then L{f’(t)} exists for all s>c.
L{ f '(t )}  sL{ f (t )}  f  0 
• This is a very important transform because derivatives appear
in the ODEs we wish to solve.
• Similarly, for higher order derivatives:
1
n2
n1
L{ f ( n) (t )}  s n F  s   s n1 f  0   s n2 f    0   ...  sf    0   f    0 
where:
n
f    0 
dn f
dt n
t 0
Proof
Deriving the Laplace transform of f (t) often requires integration by parts.
However, this process can sometimes be avoided if the transform of the
derivative is known:
For example, if f (t ) = t then f ‘(t ) = 1 and f (0) = 0 so that, since:
L f (t )  sL{ f (t )}  f (0) then L{1}  sL{t}  0
That is:
1
 sL{t} therefore
s
L{t} 
1
s2
It has already been established that if:
F (s)  L{ f (t )} and G(s)  L{g (t )}
then:
L{ f (t )}  F ( s )  f (0) and L{g (t )}  G ( s )  g (0)
Now let
g (t )  f (t ) so g (0)  f (0) and G ( s)  sF ( s )  f (0)
so that:
Therefore:
L{g(t )}  L{ f (t )}  sG(s)  g (0)  s  sF (s)  f (0)  f (0)
L{ f (t )}  s 2 F ( s)  sf (0)  f (0)
For example, if:
F ( s)  L{ f (t )}
then:
L{ f (t )}  F ( s )  f (0)
and
L{ f (t )}  s 2 F ( s)  sf (0)  f (0)
L{ f (t )}  s3 F ( s)  s 2 f (0)  sf (0)  f (0)
Similarly:
And so the pattern continues.
EXAMPLE 3.3
f (t )  sin kt so that f (t )  k cos kt and f (t )  k 2 sin kt
Then substituting in:
yields
So:
L{ f (t )}  s 2 F ( s)  sf (0)  f (0)
L{k 2 sin kt}  k 2 L{sin kt}  s 2 L{sin kt}  s.0  k
L{sin kt} 
k
s2  k 2
EXAMPLE 3.4
d
s2
s2
s 2  ( s 2  1)
1

L{ cos(t )}  2
 f (0 )  2
1 
 2
 L{ sin(t )}
2
dt
s 1
s 1
s 1
s 1
*Additional Section


f
(
0
),
f
(
0
) & f (0)
•COMMENT: Difference in
The values are only different if f(t) is not continuous at t=0
Example of discontinuous function: u(t)

f (0 )  lim u (t )  0
t 0

f (0 )  lim u (t )  1
t 0
f (0)  u (0)  1
EXAMPLE 3.5
Try to solve the differential equation:
f (t )  f (t )  1 where f (0)  0
take the Laplace transform of both sides of the differential equation to yield:
L f (t )  f (t )  L1 so that L f (t )}  L{ f (t )  L 1
 sF (s)  f (0)  F (s) 
1
1
which means that ( s  1) F ( s) 
s
s
1
s ( s  1)
The right-hand side can be separated into its partial fractions to give:
1
1
F ( s)  
s s 1
From the table of transforms it is then seen that:
Resulting in:
F (s) 
1 
1
1  1 
1  1 
t
f (t )  L1  
 L   L 
  1 e
 s s  1
s
 s  1
Thus,
f (t )  1  et
THEOREM 3.4
Laplace Convolution Theorem
L{ f  t }  F  s  and L{g  t } G  s 
If
both exist for s>c, then
L {F  s  G  s }  f ( )g (t   )d
t
1
0
or
define
L{ f ( )g (t   )d } F  s  G  s  for s  c.
t
0
t
( f  g )(t )  f ( )g (t   )d
0
As Laplace convolution of f and g.
L{ f  t }  F  s 
Proof:
We have,

F ( s)   e
 su
0





0


0

t

f (u )du][  e  sv g (v)dv]
0
e  st f (u )g (t  u )dudt
  e [
 st
t 0
 su
0
e  s ( u  v ) f (u )g (v)dudv
t 0 u 0

G(s)   e sv g (v)dv
f (u )du
F ( s )G ( s )  [  e
0

t
u 0
f (u )g (t  u )du]dt
t
 L{ f (u )g (t  u )du}
0
therefore
L {F  s  G  s }  f ( )g (t   )d
1
t
0
EXAMPLE 3.6
If f(t)=exp(t), g(t)=t, then
t

 t
f * g (t )   e (t   )d  te  (e  e )  et  t  1
0
0
0
t


EXAMPLE 3.7
1
L{e * e } 
( s  a)(s  b)
at
bt
at
bt
t
1
e

e
L1{
}  e at * ebt   e a eb (t  ) d 
0
( s  a)(s  b)
a b
(a  b)
Schiff. Joel L. The Laplace transform: theory and applications. P92
5.4. Application to the Solution of
Differential Equations
• Laplace transforms play a key role in important engineering
concepts and techniques.
Examples:
• Transfer functions
• Frequency response
• Control system design
• Stability analysis
•…
Solution of ODEs by Laplace Transforms
Procedure:
1. Take the L of both sides of the ODE.
2. Rearrange the resulting algebraic equation in the
s domain to solve for the L of the output variable,
e.g., F(s).
3. Perform a partial fraction expansion.
4. Use the L-1 to find f(t) from the expression for
F(s).
EXAMPLE 4.1.
Solve
f '' f  1, f (0)  f '(0)  0
Equation with initial conditions
Laplace transform is linear
Apply derivative formula
Rearrange
Take the inverse
f '' f  1, f (0)  f '(0)  0
L{ f ''}  L{ f }  L{1}
s 2 F ( s )  sf (0)  f '(0)  F ( s) 
F ( s) 
1
1
s
 
s ( s 2  1) s s 2  1
f (t )  1  cos t
1
s
EXAMPLE 4.2.
f ''' 6 f '' 11 f ' 6 f  1, f (0)  0
resulted in the expression
F s 

1
s s3  6s 2  11s  6

3
1  2
4
1
F s 
 


s  s  1 s  2  s  3 s s  1 s  2 s  3
1/ 6 1/ 2 1/ 2 1/ 6




s
s 1 s  2 s  3
Take L-1 of both sides:
1/ 6
1/ 2
1/ 2
1/ 6
L1{F ( s )}  L1{
}  L1{
}  L1{
}  L1{
}
s
s 1
s2
s3
1 1 t 1 2t 1 3t
f t    e  e  e
6 2
2
6
EXAMPLE 4.3.
3 f ' 2 f  4e  x  2, f (0)  0
Taking Laplace transforms of both sides of this equation gives:
4
2 6s  2
3[ sF ( s)  f (0)]  2 F ( s) 
 
s  1 s s( s  1)
6s  2
27 1
1 4 1
F ( s) 
 (
)  (
)
s ( s  1)(3s  2) 5 3s  2 s 5 s  1
27
1
1 4 1
 (
)  (
)
15 s  2 / 3 s 5 s  1
9
4
f (t )  e2 x /3  e x  1
5
5
K.A. Stroud. Engineering Mathematics. P1107
EXAMPLE 4.4.
A mass m is suspended from the end of a vertical
spring of constant k (force required to produce
unit stretch). An external force f(t) acts on the
mass as well as a resistive force proportional to
the instantaneous velocity. Assuming that x is the
displacement of the mass at time t and that the
mass starts from rest at x=0,
(a) Set up a differential equation for the motion
(b) Find x at any time t
Solution:
(a)The resistive force is given by –βdx/dt. The restoring force is
given by –kx. Then by Newton’s law,
m
d 2x
dt
or
where
2
 
dx
 kx  F (t )
dt
d 2x
dx
m

 kx  F (t )
2
dt
dt
x  0, x '(0)  0
(1)
(1)
(2)
EXAMPLE 4.4.
(b) Taking the Laplace transform of (1), using
we obtain
L{ f  t }  F  s  , L( x)  X
m[s 2 X  sx(0)  x '(0)]   [sX  x(0)]  kx  F (s)
So that on using (2)
F ( s)
F ( s)
X

2
ms   s  k m[( s   / 2m) 2  R]
Where
k 2
R 
m 4m 2
There are three cases to be considered.
(3)
EXAMPLE 4.4.
• Case 1, R>0. In this case let
We have
1
1
L {
( s   / 2 m)  
2
2
} e
R  2
  t /2m sin t

Then we find from (3)
1 t
  (t u )/2m
x
F
(
u
)
e
sin (t  u )du

0
m
• Case 2, R=0. In this case
1
L {
thus
1
( s   / 2 m) 2
}  te  t /2m
1 t
x   F (u )(t  u )e  (t u )/ m du
m 0
EXAMPLE 4.4.
• Case 3. R<0, In this case let
We have
1
1
L {
( s   / 2 m)  
2
2
} e
R   2
  t /2m sinh  t

1 t
  (t u )/2m
x
F
(
u
)
e
sinh (t  u )du

m 0
Schaum’s Outline of Theory and Problems of Advanced
Mathematics for Engineers and Scientists. P115, Problem 4.46
5.5. Discontinuous Forcing Functions;
Heaviside Step Function
The unit step function is widely used. It is defined
as:
0, t  0
H (t )  
1, t  0
Because the step function is a special case of a
“constant”, it follows
L{H (t )}  
t 
t 0
H (t )e dt  
 st
t 
t 0
1  st t   1
e dt  [ e ]

t 0 s
s
 st
More generally,
0, t  a
H (t  a)  
1, t  a
It is possible to express various discontinuous functions in
terms of the unit step function.
a
L{H (t  a)}   e (0)dt  
 st
0
 as
e
Therefore, L{H (t  a)} 
s

a
e st  e as
e (1)dt  0 

if s  0
s a
s
 st
s  0 or
L{H (t  a) f (t  a)}  eas F (s)
 as
e
L1{
}  H (t  a)
s
s0
EXAMPLE 5.1
0

 1
1

H (t ) 
( H a (t  a)  H b (t  b))  
ba
b  a
0


ta
at b
t b
e as  ebs
L{H (t )} 
s(b  a)
Schiff. Joel L. The Laplace transform: theory and applications. P92
EXAMPLE 5.2
Determine L{g(t)} for
0  t 1
 0,
g (t )  
2
(
t

1)
,
t 1

2 2e s
L{g (t )}  L{H (t  1)(t  1) }  e L{t }  e 3  3
s
s
2
s
2
s
Schiff. Joel L. The Laplace transform: theory and applications. P92
5.6. Impulsive Forcing Functions;
Dirac Impulse Function
Pictorially, the unit impulse appears as follows:
Mathematically:
1 / 
 (t  t0 )  
 0

 (t  t0 ;  )  
lim
 0
0
0t 
t 
0t 
t 
t 0 
 (t  t )dt  1
0
 0
t0 
Picture from web.utk.edu
We can prove that

1
g (t ) (t  t )dt  lim g (t )  lim g (t )  g (t )
lim





0
0 0
Where
0
1
0
1
0
t0  t1  t0  
t1  t0  t 2
 g (t0 )
t g (t ) (t  t0 )dt  0
t0  t1 , t0  t 2
1
t2
further

 g (t ) (t )dt g (0)
thus
0
 (t )
is known as Dirac delta function, or unit impulse function
The Laplace transform of a unit impulse:
if we let f(t) = (t) and take the Laplace

L{ (t )}    (t )e dt  e
 st
0
* Rectangular Pulse Function
*Additional Section
0 for t  0

f  t   h for 0  t  t w
0 for t  t
w

F s 

h
1  e t w s
s

0 s
1
EXAMPLE 6.1
Some useful properties of
Dirac Impulse Function
sa
L {
}   (t )  2ae  at
sa


du 1
  (ax)dx    (u) a  a
1
 (ax) 



 ( x)
a
f (t ) (t  T )dt  f (T )
5.7. Additional Properties
THEOREM 7.1 S-plane (frequency) shift
If L{ f  t }  F  s  exists for s>s0, then for any real constant a,
L{eat f (t )}  F (s  a)
for s+a>s0, or equivalently,
L1{F (s  a)}  eat f (t )
Proof :
L{e
 at

f (t )}   e
 at
0
EXAMPLE
7.1.
EXAMPLE
7.2.

f (t )e dt   f (t )e ( s  a )t dt F ( s  a)
 st
L{e  at sin(t )} 
0

( s  a) 2   2
2s  1
2s  1
1
1 2( s  1)  1
}

L
{
}

L
{
}
2
2
2
s  2s  5
( s  1)  4
( s  1)  4
( s  1)
1
1
t
t sin 2t
 2 L1{
}

L
{
}

2
e
cos
2
t

e
( s  1) 2  4
( s  1) 2  4
2
L1{
THEOREM 7.2 Time Shift
If L{ f  t }  F  s  exists for s>s0, then for any real constant
a>0,
 as
L{H (t  a) f (t  a)}  e F (s)
for s>s0, or, equivalently,
L1{eas F (s)}  H (t  a) f (t  a)
Proof :

0
a
L{H (t  a) f (t  a)}   H (t  a) f (t  a)e  st dt   f (t  a)e  st dt
u  t  a, t  u  a
set
a

0
EXAMPLE
7.3.

f (u )e
 s (u  a )
du e
 as


f (u )e  su du e  as F ( s)
0
10 s
e
L{H (t  10)e a (t 10 ) } 
sa
THEOREM 7.3 Multiplication by 1/s (Integrals)
If L{ f  t }  F  s  exists for s>s0, then
F ( s)
L{ f ( )d }  L{D f (t )} 
0
s
for s>max{0,s0}, or, equivalently,
t
1 F ( s )
L {
}   f ( )d
0
s
t
Proof :
1
0
t
1
dv  e  st dt, v   e  st
s
1
1  st
 st 
 [ g (t )e ]0   e d ( g (t ))
s
s
set u  g (t )  D f (t )   f ( )d , du  f (t )dt,
1
0
0


1  st
L{ f ( )d }  L{g (t )}   g (t )e dt   g (t )( )de
0
s
0
0
t
 st
1
1
F ( s)
 [ g (t )e  st ]0   f (t )e  st dt 
s
s
s
Notice:


If t  0, g(t)  0, for (t  )   f (t )e dt  , so  f (t )dt  g (t )  , slower than e  st  0
 st
0
EXAMPLE
7.4.
0
1
s
1
L{ f ( )d }  L{ cos( )d }  ( )( 2 )  2
 L{sin( t )}
0
0
s s 1 s 1
t
t
COMMENT:
Another way to prove Time Integration Property
 t
  t
L   f τ dτ     f τ dτe  st dt
0
 0 0



f τ e  st dtdτ
0 τ



0

f τ  e  st dtdτ
τ

1
  f τ e  st dτ
s 0
F s 

s
From Douglas Wilhelm Harder, University of Waterloo.
THEOREM 7.4 Differentiation with Respect to s
(Multiplication by tn)
If L{ f  t }  F  s  exists for s>s0, then
dn
L{t f (t )}  (1)
F (s)
n
ds
n
for s>s0, or, equivalently,
Proof :
n
d
F ( s)
n n
L1{
}

(

1
)
t f (t )
n
ds



0
0
0
L{t n f (t )}   t n f (t )e  st dt   f (t )t n e  st dt (1) n 
n
 (1)
s n
n
EXAMPLE
7.5.
n


0
n
f (t )e dt (1)
F ( s)
n
s
 st
n
n
d
1
n!
L{t nu (t )}  L{t n1}  (1) n n ( )  n 1
ds s
s
 n  st
f (t ) n e dt
s
EXAMPLE
7.6.
Firstly 
f (t )  t sint 
1



t
sin
t
e
dt

t sin t e  st

s
0

 st
0

1
sin t   t cost e  st dt
s
0




1 
 st
 st
   sin t e dt   t cost e dt 

s  0
0


where
1


t
cos
t
e
dt


t cost e  st

s
0

 st
0

 
0
1
cost   t sin t e  st dt
s



1 
 st
 st

cost e dt    t sin t e dt 


s  0
0

substitute






1
1
 st

st

st

st
  t sint e dt   s   sint e dt  s   cost e dt    t sint e dt  
0
0
0

0

Douglas Wilhelm Harder, MMath
[email protected]
EXAMPLE
7.6.
f (t )  t sint 
substitute






1
1
 st
 st

st

st

 cost e dt   t sin t e dt  





t
sin
t
e
dt


sin
t
e
dt



 

 
s
s
0
0
0

0

therefore
1 1
1 s

L t sin t     2
  2
 L t sin t  
s  s 1 s  s 1

1
1
L t sin t 
 2
 2

s s 1 s s 1
s2

 

 s2 1
2


 L t sin t  2    2
s s 1
 s 
2s
d 1
L t sin t   

2
2
2
ds
s
1
s 1




THEOREM 7.5 Integration with Respect to s
If there is a real number s0 such that L{ f  t }  F  s 
exists for s>s0, and limt 0 f  t  / t exists, then

f (t )
L{
}   F ( s ' )ds '
s
t
for s>s0, or, equivalently,

L1{ F ( s ' )ds '} 
s
f (t )
t
sa
L {ln
}
s
s
d
sa
1
1
sa
s1  a
1
1
ln


ln
 ln
 (
 )ds'
s
1 s 'a
ds
s
sa s
s
s1
s'
EXAMPLE 7.7. To evaluate
1
1
1
f (t )  L {
 }  e at  1
sa s
1
s  a 1  e at
L {ln
}
s
t
1

sa
1
1
ln
  (
 )ds '
s
s
s 'a s '
THEOREM 7.6 Large s Behavior of F(s)
Let f(t) be piecewise continuous on 0  t  t0 for each finite t0
and of exponential order as , t   then
(i) F ( s )  0 as s  
(ii) sF ( s ) is bounded as s  
Proof : Since f(t) is of exponential order as t  
then there exist real constants K and c, with K  0 such that
| f(t) |  Kect for all t  T.
Since f(t) is piecewise continuous on 0  t  t0
There must be a finite constant M such that | f(t) |  M, on 0  t  t0
For all s>c:
F ( s) 


0

t0

f (t )e dt   f (t ) e dt   f (t ) e dt   f (t ) e  st dt
 st
0
t0

0
t0
  Me st dt  
 st
0
 st
t0
 st 0
( s c ) t
 M
1

e
e
K
( s c ) t
Ke
dt  M
K


s
 ( s  c) t0
s sc
THEOREM 7.7 Initial-Value Theorem
Let f be continuous and f’ be piecewise continuous on 0  t  t0
for each finite t0 , and let f and f’ be of exponential order ast  
then
[ sF ( s)]  f (0)
lim
s 
NOTE:
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.
Proof : with the stated assumptions on f and f’, it follows that
L{ f '(t )}  sL( f (t ))  f (0)  sF (s)  f (0)
Since f’ satisfies the conditions of THEOREM 4.7, it follows that
L{ f '(t )}  0 as s  
thus
lim[sF (s)]  f (0)
s 
EXAMPLE
7.8.
Given
F(s) 
(s  2 )
(s  1 )2  52
Find f(0)
f( 0 )  lim [ sF(s)]  lim
s 
 lim
s 
s 
(s  2 )
[s
]  lim
2
2
(s  1 )  5
s 
s2 s2  2 s s2
1
2
2
2
2
s s  2s s  ( 26 s )


s 2  2s
 2

 s  2 s  1  25
THEOREM 7.8 Final Value Theorem
*Additional Section
Let f be continuous and f’ be piecewise continuous on 0  t  
, and let f and f’ be of exponential order as t  
then
[ sF ( s)] 
f (t )
lim
s 0
lim
t 
NOTE:
It can be used to find the steady-state value of a closed loop system (providing
that a steady-state value exists.
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.
EXAMPLE
7.9.
Suppose
then
F s 
5s  2
s  5s  4 
 5s  2 
f     lim f  t   lim 
 0.5

t 
s 0  5s  4 
EXAMPLE 7.10.
Given:
(s  2 )2  32
F(s)
(s  2 )2  32 


note F 1(s)  te 2t cos3t
find f ()
( s  2) 2  32
f ()  lim [ sF ( s)]  lim [ s
]0
.
2
2
( s  2)  3 
s 0
s 0


THEOREM 7.9 Transform of Periodic Function
If f is periodic with period T on 0  t   and piecewise
continuous one period, then
T
1
 st
for s  0
L{ f (t )} 
f
(
t
)
e
dt

1  e sT 0
EXAMPLE 7.11.
(a)cos(t) is repeated with period 2 (b)cos(t) is repeated with period 
cost 
Lcost 
f t 
L f t 
π
From Douglas Wilhelm Harder,
University of Waterloo.
 sπ
se
  s 
 cost e dt
2
s
1
0
L f t  

1  e  sπ
1  e  sπ
 st
EXAMPLE 7.12.
Consider f(t) below:
(a)
(b)
ut 
f t 
1
Lu t  
s
L f t 
Lu t   

0
1 1
e dt  

s
s
 st
From Douglas Wilhelm Harder,
University of Waterloo.
1
s
e
 e dt  1 1  es
s 
L f t   0  s 2 
1 e
1  e 2 s
s 1  e 2 s
 st


THEOREM 7.10 Scaling in Time
*Additional Section
If L{ f  t }  F  s  exists for s>s0, then for any real constant
1
s
a>0,
L{ f (at )} 
for s>s0
Proof :
F( )
a a
u
1
let u  at, t  , dt  du
a
a

L{ f (at)}  
0
EXAMPLE 7.13.

a
s
( )u
1
1
s
 st
f (at)e dt   f (u )e a du  F ( )
a0
a a
2

L{sin(t )}  (
 1)  ( 2
) 2
2
2
 (s )
 s 
s 2

1
s /
s
Lcosωt  

ω ( s /  ) 2  1 s 2  ω2
1
1
1
THEOREM 7.11 Time delay
*Additional Section
Time delays occur due to fluid flow, time required to do an
analysis (e.g., gas chromatograph). The delayed signal can be
f  t  θ  θ  time delay
represented as
then
L{ f  t  θ }  eθs F  s 
EXAMPLE 7.14.
L{ f  t  θ }  L{cos  (t  θ)}  e
θs
s
s2   2
Common Transform Properties Table
f(t)
F(s)
e  at f (t )
F ( s  a)
H (t  a ) f (t  a ), a  0
t
 f ( )d
0
e  asF ( s )
n
t f (t )
d n f (t )
dtn
f (t )
t
f (t )
f (at)
f (t   )
F ( s)
s
d n F (s)
(1)
dsn
n
s n F ( s )  s n 1 f (0)  s n  2 f ' (0)  ...  s 0 f n 1 f (0)


s
F ( s ' )ds'
T
1
 st
f
(
t
)
e
dt
 sT 0
1 e
1 s
F( )
a a
e s F ( s )
EXAMPLE 7.15
Solve the ODE,
df
5 4f 2
dt
f 0  1
First, take L of both sides of the equation,
2
5  sF  s   1  4 F  s  
s
Rearrange,
5s  2
F s 
s  5s  4 
Take L-1,
From Table,
References
• Schiff. Joel L. The Laplace transform: theory and applications.
• Murray. R. SPIEGEL. Schaum’s Outline of Theory and Problems
of Advanced Mathematics for Engineers and Scientists.
• K.A. Stroud. Engineering Mathematics.
• http://en.wikipedia.org/wiki/Laplace_transform
• Douglas Wilhelm Harder, Math. University of Waterloo
• http://faculty.mercer.edu/olivier_pd/documents/Ch2LaplaceT
ransforms.ppt
• http://ebookbrowse.com/ch06-laplace-transform-pptd116265990