CHAPTER 4
The Laplace Transform
Chapter Contents
4.1 Definition of the Laplace Transform
4.2 The Inverse Transform and Transforms of
Derivatives
4.3 Translation Theorems
4.4 Additional Operational Properties
4.5 The Dirac Delta Function
4.6 Systems of Linear Differential Equations
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Ch4_2
4.1 Definition of Laplace Transform
Basic Definition
If f(t) is defined for t  0, then


0
b
K ( s, t ) f (t )dt  lim  K ( s, t ) f (t )dt
b 0
(1)
Definition 4.1.1 Laplace Transform
If f(t) is defined for t  0, then

L { f (t )}   e
0
 st
f (t )dt
(2)
is said to be the Laplace Transform of f.
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Ch4_3
Example 1 Using Definition 4.1.1
Evaluate L {1}
Solution:
Here we keep that the bounds of integral are 0 and  in
mind.
From the definition

b
L (1)   e (1)dt  lim  e st dt
 st
b 0
0
 st b
e
 lim
b
s
0
 e sb  1 1
 lim

b
s
s
Since e-st  0 as t , for s > 0.
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Ch4_4
Example 2 Using Definition 4.1.1
Evaluate L {t}
Solution:
 te
L {t} 
s
 st 
0
1  st
  e dt
s 0
1
11 1
 L {1}     2
s
ss s
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Ch4_5
Example 3 Using Definition 4.1.1
Evaluate L {e-3t}
Solution:


L {e }   e e d t   e ( s3) t dt
3 t
 st 3t
0
0
( s 3 ) t 
e

s3

0
1
, s  3
s3
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Ch4_6
Example 4 Using Definition 4.1.1
Evaluate L {sin 2t}
Solution:

L {sin2 t}   e st sin 2t dt
0

e
 st

sin 2t
2  st
  e cos 2t dt
s
s 0
0
2  st
  e cos 2t dt , s  0
s 0
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Ch4_7
Example 4 (2)
lim e st cos 2t  0 , s  0
t 

Laplace transform of sin 2t
↓

 st


2  e cos 2t
2  st
 
  e sin 2t dt 
s 
s
s 0

0
2 4
 2  2 L {sin 2t}
s s
2
L {sin 2t}  2
,s0
s 4
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Ch4_8
L is a Linear Tramsform
We can easily verify that
L { f (t )   g (t )}
 L { f (t )}  L {g (t )}
  F ( s)  G ( s)
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(3)
Ch4_9
Theorem 4.1.1 Transform of Some Basic
Functions
(a) L {1} 
(b) L {t n }  nn!1 , n  1, 2, 3, 
s
(d) L {sin kt} 
k
s2  k 2
k
(f) L {sin kt )  2 2
s k
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1
s
(c) L {e at }  1
sa
(e) L {cos kt} 
s
s2  k 2
(g) L {cosh kt} 
s
s2  k 2
Ch4_10
Fig 4.1.1 Piecewise-continuous function
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Ch4_11
Definition 4.1.2 Exponential Order
A function f(t) is said to be of exponential order,
if there exists constants c, M > 0, and T > 0, such that
|f(t)|  Mect for all t > T. See Fig 4.1, 4.2.
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Ch4_12
Fig 4.1.2 Function f is of exponential order
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Ch4_13
Fig 4.1.3 Functions with blue graphs are of
exponential order
|t |e
t
t
|e |e
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t
2 cos t  2et
Ch4_14
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Ch4_15
Theorem 4.1.2 Sufficient Conditions for
Existence
If f(t) is piecewise continuous on [0, ) and of
exponential order c, then L {f(t)} exists for s > c.
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Ch4_16
Example 5
Find L {f(t)} for
0 , 0  t  3
f (t )  
2 , t  3
Solution:
3

L { f (t )}   e 0dt   e st 2dt
 st
0
2e st

s
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3

3
2e 3 s

,s0
s
Ch4_17
Fig 4.1.5 Piecewise-continuous function in Ex
5
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Ch4_18
4.2 The Inverse Transform and Transform of
Derivatives
Theorem 4.2.1 Some Inverse Transform
(a) 1  L
(b)
tn  L
1
 n!  , n  1, 2, 3, 
 n1 
s 
(d) sin kt  L
(f) sinh kt  L
1
 k 
 2
2
s

k


 k 
 2
2
s

k


1
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1 
 
s 
1
(c) e  L
at
1
(e) cos kt  L
 1 


s  a 
1
(g) cosh kt  L
 s 
 2
2
s

k


1
 s 
 2
2
s  k 
Ch4_19
Example 1 Applying Theorem 4.2.1
Find the inverse transform of
(a) L
1
 1  (b) L
 5
s 
1
 1 
 2

s  7
Solution:
(a) L 1 15   1 L 1 45!   1 t 4
 s  4!
 s  24
1  1
7  1
1
1
(b) L  2   L  2   sin 7t
7
7
s  7
s  7
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Ch4_20
L -1 is also linear
We can easily verify that
L 1{F ( s )  G ( s )}
 L
{F ( s )}  L
1
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1
{G ( s )}
(1)
Ch4_21
Example 2 Termwise Division and Linearity
Find L
1
  2s  6 
 2

s

4


Solution:
 2s  6 

L  2
 L
 s 4 
1
  2s  6 
 2

2
s

4
s

4


s  6
1
 2L  2
 L
s  4 2
 2 cos 2t  3 sin 2t
1
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1
 2 
 2

s

4


(2)
Ch4_22
Example 3 Partial Fractions and Linearity
Find L
1


s 2  6s  9


( s  1)(s  2)(s  4) 
Solution:
Using partial fractions
s 2  6s  9
A
B
C



( s  1)(s  2)(s  4) s  1 s  2 s  4
Then
s 2  6s  9
 A( s  2)(s  4)  B( s  1)(s  4)  C ( s  1)(s  2)
(3)
If we set s = 1, 2, −4, then
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Ch4_23
Example 3 (2)
A  16/5, B  25/6, c  1/30
(4)
Thus
L
1


s 2  6s  9


( s  1)(s  2)(s  4) 
16
 L
5

1
 1   25 L


 s  1 6
16 t 25 2t 1 4t
e  e  e
5
6
30
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 1  1 L


 s  2  30
1
1
 1 


s  4
(5)
Ch4_24
Transform of Derivatives
 L { f (t )}

 e
 st
0
f (t )dt e
 st


f (t ) 0  s  e st f (t )dt
0
  f (0)  sL { f (t )}
L { f (t )}  sF ( s)  f (0)
(6)
 L { f (t )}

 e
0
 st
f (t )dt  e
 st


f (t ) 0  s  e st f (t )dt
0
  f (0)  sL { f (t )}
 s[ sF ( s )  f (0)]  f (0)
L { f (t )}  s 2 F ( s)  sf (0)  f (0)
L { f (t )}  s 3 F ( s)  s 2 f (0)  sf (0)  f (0)
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(7)
(8)
Ch4_25
Theorem 4.2.2 Transform of a Derivative
( n1)

f
,
f
,

,
f
If
are continuous on [0, ) and are of
exponential order and if f(n)(t) is piecewise-continuous
On [0, ), then
L { f ( n ) (t )}
 s n F ( s )  s n1 f (0)  s n2 f (0)    f ( n1) (0)
where F (s)  L { f (t )}.
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Ch4_26
Solving Linear ODEs
n
n1
d
y
d
 an n  an1 n1y    a0 y  g (t )
dt
dt
y (0)  y0 , y(0)  y1 , y ( n1) (0)  yn1
Then
d n y 
 d n1 y 
anL  n   an1L  n1     a0L { y}  L {g (t )}
 dt 
 dt 
(9)
an [ s nY ( s)  s n1 y (0)    y ( n1) (0)]
 an1[ s n1Y ( s )  s n2 y (0)    y ( n2 ) (0)]
   a0Y ( s )
(10)
 G( s)
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Ch4_27
We have
P( s)Y ( s)  Q( s)  G( s)
Q( s ) G ( s )
Y ( s) 

P( s ) P( s)
(11)
where P( s)  an s n  an1s n1    a0
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Ch4_28
Find unknown
y (t ) that satisfies
a DE and Initial
Apply Laplace
transform L
Transformed DE
becomes an
algebraic equation
In Y (s )
Apply Inverse
transform L 1
Solve transformed
equation for Y (s )
condition
Solution y (t ) of
original IVP
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Ch4_29
Example 4 Solving a First-Order IVP
dy
Solve  3 y  13 sin 2t , y (0)  6
dt
Solution:
dy
L    3L { y}  13L {sin 2t}
 dt 
26
sY ( s )  6  3Y ( s )  2
s 4
(12)
26
( s  3)Y ( s )  6  2
s 4
6
26
6s 2  50
Y ( s) 


2
s  3 ( s  3)(s  4) ( s  3)(s 2  4)
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(13)
Ch4_30
Example 4 (2)
6s 2  50
A
Bs  C

 2
2
( s  3)(s  4) s  3 s  4
6s 2  50  A( s 2  4)  ( Bs  C )(s  3)
We can find A = 8, B = −2, C = 6
Thus
6s 2  50
8
 2s  6
Y ( s) 

 2
2
( s  3)(s  4) s  3 s  4
y (t )  8L
1
 1   2L


 s  3
1
 s   3L
 2

s  4
1
 2 
 2

s  4
y(t )  8e 3t  2 cos 2t  3 sin 2t
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Ch4_31
Example 5 Solving a Second-Order IVP
4 t
y
"

3
y
'

2
y

e
, y(0)  1, y' (0)  5
Solve
Solution:
d 2 y 
L  2   3L
 dt 
 dy 
4 t

2
L
{
y
}

L
{
e
}
 
 dt 
1
s Y ( s)  sy (0)  y(0)  3[ sY ( s)  y (0)]  2Y ( s) 
s4
1
2
( s  3s  2)Y ( s)  s  2 
s4
s2
1
s 2  6s  9
(14)
Y ( s)  2
 2

s  3s  2 ( s  3s  2)(s  4) ( s  1)(s  2)(s  4)
2
Thus
16 t 25 2t 1 4t
y (t )  L {Y ( s)}   e  e  e
5
6
30
1
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Ch4_32
4.3 Translation Theorems
Theorem 4.3.1 First Translation Theorem
If L {f} = F(s) and a is any real number, then
L {eatf(t)} = F(s – a)
Proof:
L {eatf(t)} =  e-steatf(t)dt
=  e-(s-a)tf(t)dt = F(s – a)
L {e at f (t )}  L { f (t )}ssa
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Ch4_33
Fig 4.3.1 Shift on s-axis
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Ch4_34
Example 1 Using the First Translation
Theorem
Evaluate (a) L {e5t t 3}
Solution:
(a) L {e t }  L {t }ss5
5t 3
3
(b) L {e2t cos 4t}
3!
6
 4

s ss5 ( s  5) 4
(b) L {e 2t cos 4t}  L {cos 4t}ss( 2)

s
s2

s 2  16 ss2 ( s  2) 2  16
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Ch4_35
Inverse Form of Theorem 4.3.1
 L 1{F (s  a)}  L 1{F (s) ssa }  eat f (t )
(1)
where f (t )  L 1{F (s)}.
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Ch4_36
Example 2 Partial Fractions and Completing
the square
Evaluate (a) L
1
 2s  5 

2
( s  3) 
(b) L
1
 s/2  5/3 
 2

 s  4s  6 
Solution:
(a)
2s  5
A
B


2
( s  3)
s  3 ( s  3) 2
2s  5  A( s  3)  B
we have A = 2, B = 11
2s  5
2
11


2
( s  3)
s  3 ( s  3) 2
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(2)
Ch4_37
Example 2 (2)
And
L
1
 2s  5 
 2L

2
( s  3) 
1
 1 

  11L
 s  3
1
 1 

2
( s  3) 
(3)
From (3), we have
L
1
 1 
L

2
( s  3) 
L
1
1
1
 2
s
 3t
e t
s s 3 
 2s  5 
3t
3t

2
e

11
e
t

2
( s  3) 
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(4)
Ch4_38
Example 2 (3)
(b) s / 2  5 / 3
s / 2  5/3

2
s  4s  6 ( s  2) 2  2
 s / 2  5/3 
L  2

s

4
s

6


1
 s2  2
 L 1
 L
2
2
( s  2)  2  3
(5)
1
1
 L
2
1
1
1




2
 ( s  2)  2 
2
 s

L
 2

 s  2 ss  2  3 2
1
2 2 t
 e 2t cos 2t 
e sin 2t
2
3
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1
 2

 2

s

2

ss  2 
(6)
(7)
Ch4_39
Example 3 IVP
2 3t


y

6
y
'

9
y

t
e ,
Solve
y(0)  2 ,
y' (0)  17
Solution:
2
s Y ( s)  sy (0)  y(0)  6[sY (s)  y(0)]  9Y ( s) 
( s  3)3
2
( s 2  6s  9)Y ( s)  2s  5 
2
( s  3)3
( s  3) 2 Y ( s)  2s  5 
2
( s  3)3
Y (s) 
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2s  5
2

2
( s  3) ( s  3)5
Ch4_40
Example 3 (2)
 Y ( s)  2  11 2 
s  3 ( s  3)
2
( s  3)5
y (t )
 2L
L
1
1
 1   11L


s

3


1
 2
s
1
 1  2
 L

2
 ( s  3)  4!

3t

te
,

s s 3 
L
1
1
 4! 

5
(
s

3
)


(8)
 4!
 4 3t
 5
t e
 s ss3 
1 4 3t
y (t )  2e  11te  t e
12
3t
3t
Copyright © Jones and Bartlett；滄海書局
Ch4_41
Example 4 An IVP
t



y

4
y

6
y

1

e
,
Solve
y(0)  0 ,
y' (0)  0
Solution:
1 1
s Y ( s)  sy (0)  y(0)  4[ sY ( s)  y(0)]  6Y ( s)  
s s 1
2s  1
2
( s  4s  6)Y ( s) 
s ( s  1)
2
2s  1
Y (s) 
s( s  1)(s 2  4s  6)
Y ( s) 
1/ 6 1/ 3 s / 2  5 / 3

 2
s s  1 s  4s  6
Copyright © Jones and Bartlett；滄海書局
Ch4_42
Example 4 (2)
1  1 1 1 
  L 

s
3
s

1
 


1
 s2  2
 L 1
L

2
2
( s  2)  2  3 2
1
Y (s)  L
6
1
1
2




2
( s  2)  2 
1 1 t 1 2t
2 2 t
  e  e cos 2t  e sin 2t
6 3
2
3
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Ch4_43
Definition 4.3.1 Unit Step Function
The Unit Step Function U (t – a) is
0 , 0  t  a
U (t  a)  
ta
1 ,
See Fig 4.3.2.
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Ch4_44
Copyright © Jones and Bartlett；滄海書局
Ch4_45
Copyright © Jones and Bartlett；滄海書局
Ch4_46
Also a function of the type
 g (t ), 0  t  a
f (t )  
ta
 h(t ),
(9)
is the same as
f (t )  g (t )  g (t )U (t  a)  h(t )U (t  a)
(10)
Similarly, a function of the type
0t a
0,

f (t )   g (t ), a  t  b
0,
t b

(11)
can be written as
f (t )  g (t )[U (t  a) U (t  b)]
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(12)
Ch4_47
Example 5 A Piecewise-Defined Function
20t , 0  t  5
Express f (t )  
0 , t  5
in terms of U (t).
Solution:
From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0
f(t) = 20t – 20tU (t – 5)
Consider the function
0t a
0,
f (t  a)U (t  a)  
ta
 f (t  a),
(13)
See Fig 4.3.5.
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Ch4_48
Copyright © Jones and Bartlett；滄海書局
Ch4_49
Fig 4.3.6 Shift on t-axis
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Ch4_50
Theorem 4.3.2 Second Translation Theorem
If F(s) = L {f}, and a > 0, then
L {f(t – a)U (t – a)} = e-asF(s)
Proof:
L { f (t  a)U (t  a)}
a
 e
0

 st

f (t  a)U (t  a)dt   e st f (t  a)U (t  a)dt
a
  e st f (t  a)dt
0
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Ch4_51
Theorem 4.3.2 proof
Let v = t – a, dv = dt, then
L { f (t  a)U (t  a)}

 e
0
s ( va )
f (v)dv  e
as


0
e sv f (v)dv  e asL { f (t )}
If f(t) = 1, then f(t – a) = 1, F(s) = L {1} = 1/s,
e as
L {U (t  a)} 
s
(14)
eg: The L.T. of Fig 4.3.4 is
L { f (t )}  2L {1}  3L {U (t  2)}  L {U (t  3)}
1 e  2 s e 3 s
 2 3

s
s
s
Copyright © Jones and Bartlett；滄海書局
Ch4_52
Inverse Form of Theorem 4.7

L 1{e as F ( s)}  f (t  a)U (t  a)
Copyright © Jones and Bartlett；滄海書局
(15)
Ch4_53
Example 6 Using Formula (15)
Evaluate (a) L
1
 1 e 2 s 


s

4


(b) L
 s e s / 2 
 2

s

9


1
Solution:
(a) a  2, F (s)  1/(s  4), L 1{F (s)}  e4t
then
L
 1 e 2 s   e 4 ( t 2U
)
(t  2)


s  4

1
2
1
a


/
2
,
F
(
s
)

s
/(
s

9
),
L
{F ( s)}  cos 3t
(b)
then
L
1
 s e s / 2   cos 3 t   U


 2

s

9
2




Copyright © Jones and Bartlett；滄海書局
t   


2


Ch4_54
Alternative Form of Theorem 4.3.2
Since t 2  (t  2)2  4(t  2)  4 , then
L {t 2U (t  2)}
 L {(t  2) 2U (t  2)  4(t  2)U (t  2)  4U(t  2)}
The above can be solved. However, we try another
approach.
Let u = t – a,


L {g (t )U (t  a)}   e g (t )dt   e s (ua ) g (u  a)du
a
That is,
 st
0
L {g (t )U (t  a)}  e asL {g (t  a)}
Copyright © Jones and Bartlett；滄海書局
(16)
Ch4_55
Example 7 Second Translation Theorem –
Alternative Form
Find L {cos tU (t   )}
Solution:
With g(t) = cos t, a = , then
g(t + ) = cos(t + )= −cos t
By (16),
s s
L {cos tU (t   )}  e L {cos t}   2 e
s 1
s
Copyright © Jones and Bartlett；滄海書局
Ch4_56
Example 8 An IVP
Solve y' y  f (t ) , y(0)  5
0t 
0 ,
f (t )  
3 sin t , t  
Solution:
We find f(t) = 3 cos tU (t −), then
s s
sY ( s)  y (0)  Y ( s)   3 2 e
s 1
3s s
( s  1)Y ( s)  5  2 e
s 1
5
3  1 s
1 s
s s 
Y ( s) 
 
e  2 e  2 e 
s 1 2  s 1
s 1
s 1

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(17)
Ch4_57
Example 8 (2)
It follows from (15) with a = , then
L
 1 e s   e (t  U
)
(t   ) , L


s

1


1
L
1
 1 e s   sin( t   )U (t   )
 2

s

1


 s e s   cos( t   )U (t   )
 2

s 1

1
Thus
3
3
3
)
y (t )  5e t  e (t  U
(t   )  sin( t   )U (t   )  cos( t   )U (t   )
2
2
2
3
 5e t  [e (t  )  sin t  cos t ]U (t   )
2
5e t ,
0t 
(18)
  t
( t  )
 3 / 2 sin t  3 / 2 cos t ,
t 
5e  3 / 2e
See Fig 4.3.7.
Copyright © Jones and Bartlett；滄海書局
Ch4_58
Fig 4.3.7 Graph of function (18) in Ex 8
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Ch4_59
Beams
Remember that the DE of a beam is
d4y
EI 4  w( x)
dx
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(19)
Ch4_60
Example 9
A beam of length L is embedded at both ends as Fig
4.3.8. Find the deflection of the beam when the load is
given by
  2 
w 1  x , 0  x  L / 2
w( x)   0  L 
0,
L/2  x  L
Solution:
We have the boundary conditions: y(0) = y(L) = 0, y’(0)
= y’(L) = 0. By (10),
2
2
L
w( x)  w0 1  x   w0 1  x U  x  
2
 L 
 L  
2 w0  L
L 
L 


 x   x  U  x  

L 2
2 
2 

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Ch4_61
Fig 4.3.8 Embedded beam with a variable
load in Ex 9
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Ch4_62
Example 9 (2)
Transforming (19) into
EI s 4Y ( s)  s 3 y(0)  s 2 y(0)  sy(0)  y(0) 
2w0  L 2 1 1  Ls 2 

 2 2e


L  s
s s
2 w0  L 2 1 1  Ls 2 
 2 2e


EIL  s
s s
c1 c3 2 w0  L 2 1 1  Ls 2 
Y (s)  3  4 
 6 6e
5


s s EIL  s
s s
s 4Y ( s )  sy" (0)  y ( 3) (0) 
where c1 = y”(0), c3 = y(3)(0)
Copyright © Jones and Bartlett；滄海書局
Ch4_63
Example 9 (3)
Thus
y ( x)
c1
 L
2!
1
 2!   c2 L
 3
 s  3!
2w0  L / 2

L

EIL  4!
1
1
 3! 
 4
s 
 4!  1
 5 L
 s  5!
1
 5!  1
 6 L
 s  5!
1
 5!  Ls / 2 
 6e

s


5
w0  5L 4
c1 2 c2 3
L
x U
5
 x  x 
x

x




2
6
60 EIL  2
2

Copyright © Jones and Bartlett；滄海書局
 x  L 


2 

Ch4_64
Example 9 (4)
Applying y(L) = y’(L) = 0, then
L2
L3 49w0 L4
c1  c2 
0
2
6 1920 EI
L2 85w0 L3
c1 L  c2 
0
2 960 EI
Thus c1  23w0 L2 / 960 EI , c2  9w0 L / 40 EI
23w0 L2 2 3w0 L 3
y ( x) 
x 
x
1920 EI
80 EI
5

w0 5L 4
x LU
5

x

x




60 EIL  2
2

Copyright © Jones and Bartlett；滄海書局
 x  L 


2 

Ch4_65
4.4 Additional Operational Properties
Multiplying a Function by tn
dF d  st
  e f (t )dt
ds ds 0
 

 st

[e f (t )]dt    e st tf (t )dt  L {tf (t )}
0 s
0
that is, L {tf (t )}  
Similarly,
d
L { f (t )}
ds
d
L {tf (t )}
ds
d d
d2

    L { f (t )}  2 L { f (t )}
ds  ds
 ds
L {t 2 f (t )}  L {t  tf (t )}  
Copyright © Jones and Bartlett；滄海書局
Ch4_66
Theorem 4.4.1 Derivatives of Transform
If F(s) =L {f(t)} and n = 1, 2, 3, …, then
n
d
L {t n f (t )}  (1) n n F ( s)
ds
Copyright © Jones and Bartlett；滄海書局
Ch4_67
Example 1 Using Theorem 4.4.1
Find L {t sin kt}
Solution:
With f(t) = sin kt, F(s) = k/(s2 + k2), then
d
L {sin kt}
ds
d k 
2ks
  2
 2
2
ds  s  k  ( s  k 2 ) 2
L {t sin kt}  
Copyright © Jones and Bartlett；滄海書局
Ch4_68
Different approaches
Theorem 4.3.1:
L {te }  L {t}ss3
3t
1
 2
s
s s 3
1

( s  3) 2
Theorem 4.4.1:
d
d 1
1
3t
2
L {te }   L {e }  
 ( s  3) 
ds
ds s  3
( s  3) 2
3t
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Ch4_69
Example 2 An IVP
Solve x  16 x  cos 4t , x(0)  0 , x(0)  1
Solution:
s
( s 2  16) X ( s )  1 
or
s 2  16
1
s
X (s)  2
 2
s  16 ( s  16) 2
From example 1,
Thus
L
1
 2ks 
 t sin kt
 2
2 2
( s  k ) 
 4  1
 2
 L
 s  16  8
1
1
 sin st  t sin 4t
4
8
1
x(t )  L
4
1
Copyright © Jones and Bartlett；滄海書局
1
 8s

 2
2
 ( s  16) 
Ch4_70
Convolution
A special product of f * g is defined by
t
f * g   f ( ) g (t   )d
0
(2)
and is called the convolution of f and g.
The convolution is a function of t, eg:
1
e  sin t   e sin( t   )d  (sin t  cost  et )
0
2
t
t
(3)
Note: f * g = g * f
Copyright © Jones and Bartlett；滄海書局
Ch4_71
Theorem 4.4.2 Convolution Theorem
If f(t) and g(t) are piecewise continuous on [0, ) and
of exponential order, then
L { f  g}  L { f (t )}L {g (t )}  F ( s)G( s)
Proof:
F ( s )G ( s ) 


0


0
 e

 s
0
f ( )d
 e

0
 s
g (  )d

e s (   ) f ( )g (  )dd }


0
0
  f ( )d  e (   ) g (  )d
Copyright © Jones and Bartlett；滄海書局
Ch4_72
Theorem 4.4.2 proof
Holding  fixed, let t =  + , dt = d


F ( s)G( s)   f ( )d  e st g (t   )dt

0
The integrating area is the shaded region in Fig 4.4.1.
Changing the order of integration:

t
F ( s )G ( s )   e dt  f ( ) g (t   )d
 st
0

 e
 f ( ) g (t  )d dt
0
 st
0
t
0
 L { f  g}
Copyright © Jones and Bartlett；滄海書局
Ch4_73
Fig 4.4.1 Changing order of integration from
t first to  first
Copyright © Jones and Bartlett；滄海書局
Ch4_74
Example 3 Transform of a Convolution
Find L
 e sin( t  ) d 
t

0
Solution:
Original statement
= L {et * sin t}
1
1
1

 2

s  1 s  1 ( s  1)(s 2  1)
Copyright © Jones and Bartlett；滄海書局
Ch4_75
Inverse Transform of Theorem 4.9

L -1{F(s)G(s)} = f * g
(4)
Look at the table in Appendix III,
2k 3
L {sin kt  kt cos kt}  2
(s  k 2 )2
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(5)
Ch4_76
Example 4 Inverse Transform as a
Convolution
1


1
 2
2
( s  k2 ) 
Find L
Solution:
Let F ( s)  G ( s) 
1
s2  k 2
1 1  k  1
f (t )  g (t )  L  2
 sin kt
2
k
s  k  k
then
L
1
1

 1 t
 2  sin k sin k (t   )d
 2
2 2
( s  k )  k 0
Copyright © Jones and Bartlett；滄海書局
(6)
Ch4_77
Example 4 (2)
Now recall that
sin A sin B = (1/2) [cos (A – B) – cos (A+B)]
If we set A = k, B = k(t − ), then
L
1
1
1 t


 2  [cos k (2  t )  cos kt ]d
 2
2 2
 ( s  k )  2k 0
t
1 1
 2  sin k (2  t )   cos kt 
2k  2k
0
sin kt  kt cos kt

2k 3
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Ch4_78
Transform of an Integral
When g(t) = 1, G(s) = 1/s, then
L

t
0

t
0

F ( s)
f ( )d 
s
 F ( s) 
f ( )d  L 

 s 
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1
(7)
(8)
Ch4_79
Examples:
 1  t
L  2
  0 sin d  1  cos t
 s ( s  1) 
1
 t
1 
L  2 2
  0 (1  cos )d  t  sin t
 s ( s  1) 
1
L
1
1
1 2

 t
 3 2
  0 (  sin  )d  t  1  cos t
2
 s ( s  1) 
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Ch4_80
Volterra Integral Equation
t
f (t )  g (t )   f ( )h(t   )d
0
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(9)
Ch4_81
Example 5 An Integral Equation
t
Solve f (t )  3t  e  0 f ( )et  d
Solution:
First, h(t-) = e(t-), h(t) = et.
From (9)
2
1
t
2
for f (t )
1
F ( s)  3  3 
 F ( s) 
s s 1
s 1
Solving for F(s) and using partial fractions
F ( s) 
6 6 1
2



s3 s 4 s s  1
2! 

1  3! 
f (t )  3L  3   L  4   L
s 
s 
 3t 2  t 3  1  2e t
1
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1
1   2L
 
s 
1
 1 


 s  1
Ch4_82
Series Circuits
From Fig 4.4.2, we have
di
1 t
L  Ri(t )   i ( )d  E (t )
dt
C 0
(10)
which is called the integrodifferential equation.
Copyright © Jones and Bartlett；滄海書局
Ch4_83
Example 6 An Integrodifferential Equation
Determine i(t) in Fig 4.4.2, when L = 0.1 h, R = 2 , C
= 0.1 f, i(0) = 0, and
E(t) = 120t – 120tU (t – 1)
Solution:
Using the data, (10) becomes
t
di
0.1  2i (t )  10 i ( )d  120t  120U (t  1)
0
dt
And then
I (s)
1 1 s 1 s 

0.1sI ( s )  2 I ( s )  10
 120 2  2 e  e 
s
s 
s s
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Ch4_84
Example 6 (2)
1
1
1

s
s 
I ( s )  1200 

e 
e 
2
2
2
( s  10)
 s ( s  10) s ( s  10)

1 / 10
1 / 100 s
1/100 1 / 100
 1200 



e
2
s  10 ( s  10)
s
 s
1 / 100 s
1 / 10 s
1
s 

e 
e 
e 
2
2
s  10
( s  10)
( s  10)

)
i (t )  12[1 U (t  1)]  12[e 10 t  e 10 ( t 1U
(t  1)]
)
 120te 10 t  1080(t  1)e 10 ( t 1U
(t  1)
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Ch4_85
Example 6 (3)
Written as a piecewise- defined function
12  12e 10 t  120te 10 t ,
i (t )  
10 t
10 ( t 1)
10 t
10 ( t 1)

12
e

12
e

120
te

1080
(
t

1
)
e
,

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0  t 1
t 1
(11)
Ch4_86
Fig 4.4.3
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Ch4_87
Post Script – Green’s Function Redux
By applying the Laplace transform to the initial-value
problem
y  ay  by  f (t ), y (0)  0, y(0)  0,
where a and b are constants, we find that the
transform of y(t) is
F ( s)
Y ( s)  2
s  as  b
where F(s) = L {f(t)}. By rewriting the foregoing
transform as the product
Y ( s) 
1
F ( s)
2
s  as  b
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Ch4_88
We can use the inverse form of the convolution
theorem (4) to write the solution of the IVP as
t
y (t )   g (t   ) f ( )d
0
(12)
1

  g (t ) and 1
where L  2
L {F ( s)}  f (t ).

s

as

b


1
On the other hand, we know from (9) of Section 3.10
that the solution of the IVP is also given by
t
y (t )   G(t ,  ) f ( )d
0
where G(t, ) is the Green’s function for the
differential equation.
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(13)
Ch4_89
By comparing (12) and (13) we see that the Green’s
function for the differential equation is related to
1
  g (t ) by
L 1 


2
 s  as  b 
G(t ,  )  g (t   )
(14)
For example, for the initial-value problem y  4 y  f (t ),
y(0) = 0, y’(0) = 0 we find
1  1
L 1 
 2
  sin 2t  g (t )
s  4 2
Thus from (14) we see that the Green’s function for
the DE y  4 y  f (t ) is G(t, ) = g(t - ) = 1/2 sin 2(t ). See Example 4 in Section 3.10.
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Ch4_90
Periodic Function

f(t + T) = f(t)
Theorem 4.4.3 Transform of a Periodic
Function
If f(t) is piecewise continuous on [0, ), of exponential
order, and periodic with period T, then
1
L { f (t )} 
1  e sT
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T

0
e st f (t )dt
Ch4_91
Theorem 4.4.3 proof
T
L { f (t )}   e
 st
0

f (t )dt   e st f (t )dt
T
Use the same transform method


T
e st f (t )dt  e sT L { f (t )}
T
L { f (t )}   e st f (t )dt  e sT L { f (t )}
0
1
L { f (t )} 
1  e sT
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T

0
e st f (t )dt
Ch4_92
Example 7
Find the L. T. of the function in Fig 4.4.4.
Solution:
We find T = 2 and
1, 0  t  1
E (T )  
0, 1  t  2
From Theorem 4.4.3,
1
L {E (t )} 
1  e 2 s


s
1
1

e
1
 st
e

1
dt

0


0
1  e 2 s s
s(1  e s )
1
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(15)
Ch4_93
Fig 4.4.4
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Ch4_94
Example 8 A Periodic Impressed Voltage
The DE
di
L  Ri  E (t )
dt
(16)
Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.4.4.
Solution:
1
s (1  e s )
1/ L
1
I (s) 

s ( s  R / L) 1  e s
LsI ( s )  RI ( s ) 
or
(17)
1
s
2 s
3 s

1

e

e

e

Because 1  e-s
1
L/ R
L/ R
and


s ( s  R / L)
s
s R/L
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Ch4_95
Example 8 (2)
I (s) 
1 1
1 


1  e s  e 2 s  ...
Rs s  R L
Then i(t) is described as follows and see Fig 4.4.5:
1  e t ,
 t
( t 1)

e

e
,

i (t )  
t
( t 1)
( t  2 )
1

e

e

e
,

 e t  e ( t 1)  e (t 2 )  e (t 3) ,

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0  t 1
1 t  2
2t 3
(15)
3t  4
Ch4_96
Fig 4.4.5
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Ch4_97
4.5 The Dirac Delta Function
Unit Impulse
See Fig 4.5.1(a). Its function is defined by
0  t  t0  a
0,
1
 a (t  t0 )   , t0  a  t  t0  a
 2a
t  t0  a
0,
(1)
where a > 0, t0 > 0.
For a small value of a, a(t – t0) is a constant function
of large magnitude. The behavior of a(t – t0) as a 
0, is called unit impulse, since it has the

property 0  (t  t0 )dt  1 . See Fig 4.5.1(b).
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Ch4_98
Fig 4.5.1 Unit impulse
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Ch4_99
The Dirac Delta Function
This function is defined by
(t – t0) = lima0 a(t – t0)
The two important properties:
(2)
, t  t0
(i)  (t  t0 )  
 0, t  t0
(ii)
x
  (t  t )dt  1 , x > t0
0
0
The unit impulse (t – t0) is called the Dirac delta
function.
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Ch4_100
Theorem 4.5.1 Transform of the Dirac Delta
Function
For t0 > 0,
L { (t  t0 )}  e st0
(3)
Proof:
1
 a (t  t0 )  [U (t  (t0  a) U (t  (t0  a))]
2a
The Laplace Transform is
1  e  s ( t0  a ) e  s ( t0  a ) 
L { a (t  t0 )} 



2a  s
s

e
 st0
(4)
 e sa  e  sa 


 2 sa 
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Ch4_101
When a  0, (4) is 0/0. Use the L’Hopital’s rule, then
(4) becomes 1 as a  0.
Thus ,
L  (t  t0 )  lim L  a (t  t0 )  e
a0
 st0
 e sa  e sa  st0
lim
e
a0
 2sa 
Now when t0 = 0, we have
L  (t )  1
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Ch4_102
Example 1 Two Initial-Value Problems
Solve y" y  4 (t  2 ), subject to
(a) y(0) = 1, y’(0) = 0
(b) y(0) = 0, y’(0) = 0
Solution:
(a)
s2Y – s + Y = 4e-2s
s
4e 2s
Y (s)  2
 2
s 1 s 1
Thus
y(t) = cos t + 4 sin(t – 2)U (t – 2)
Since sin(t – 2) = sin t, then
0  t  2
cos t ,
y (t )  
t  2
cos t  4 sin t ,
(5)
See Fig 4.5.2.
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Ch4_103
Fig 4.5.2
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Ch4_104
Example 1 (2)
(b)
4e 2s
Y ( s)  2
s 1
Thus y(t) = 4 sin(t – 2)U (t – 2)
and
y (t )  4 sin( t  2 )U (t  2 )
0  t  2
0,

t  2
4 sin t ,
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(6)
Ch4_105
Fig 4.5.3
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Ch4_106
4.6 Systems of Linear DEs
Coupled Strings
In example 1, we will deal with
m1 x1  k1 x1  k 2 ( x2  x1 )
m2 x2  k 2 ( x2  x1 )
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(1)
Ch4_107
Example 1 Example 4 of Section 3.12
Revised
Use L.T. to solve
x1  10 x1
 4 x2  0
 4 x1  x2  4 x2  0
(2)
where x1(0) = 0, x1’(0) = 1, x2(0) = 0, x2’(0) = −1.
Solution:
s2X1(s)– sx1(0) – x1’(0) + 10X1(s) – 4X2(s) = 0
−4X1(s) + s2X2(s) – sx2(0) – x2’(0) + 4X2(s) = 0
Rearrange:
(s2 + 10)X1(s)
– 4 X2(s) = 1
−4X1(s) + (s2 + 4)X2(s) = −1
(3)
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Ch4_108
Example 1 (2)
Solving (3) for X1:
s2
1/ 5
6/5
X 1 ( s)  2
 2
 2
2
( s  2)(s  12)
s  2 s  12
x1 (t )  
2
3
sin 2t 
sin 2 3t
10
5
Use X1(s) to get X2(s)
s2  6
2/5
3/ 5
X 2 (s)   2
 2
 2
2
( s  2)(s  12)
s  2 s  12
x2 (t )  
2
3
sin 2t  sin 2 3t
5
10
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Ch4_109
Example 1 (3)
Then
2
3
x1 (t )  
sin 2t 
sin 2 3t
10
5
2
3
x2 (t )  
sin 2t  sin 2 3t
5
10
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(4)
Ch4_110
Networks
From Fig 4.6.1, we have
di
 Ri2  E (t )
dt
di
RC 2  i2  u1  0
dt
L
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(5)
Ch4_111
Example 2 An Electric Network
Solve (5) where E(t) = 60 V, L = 1 h, R = 50 ohm, C =
10-4 f, i1(0) = i2(0) = 0.
Solution:
We have
di
1
 50i2  60
dt
 4 di2
50(10 )
 i2  i1  0
dt
Then
sI1(s) +
50I2(s) = 60/s
−200I1(s) + (s + 200)I2(s) = 0
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Ch4_112
Example 2 (2)
Solving the above
60 s  12000 6 / 5
6/5
60



s ( s  100) 2
s
s  100 ( s  100) 2
12000
6/5
6/5
120
I 2 (s) 



2
s ( s  100)
s
s  100 ( s  100) 2
I1 ( s ) 
Thus
6 6
i1 (t )   e 100 t  60te 100 t
5 5
6 6
i2 (t )   e 100 t  120te 100 t
5 5
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Ch4_113
Double Pendulum
From Fig 4.6.2, we have
(m1  m2 )l121 m2l1l2 2  (m1  m2 )l1 g1  0
m l    m2l1l21 m2l2 g 2  0
2
2 2 2
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(6)
Ch4_114
Example 3 Double Pendulum
Please verify that when
m1  3, m2  1, l1  l2  16, 1 (0)  1,  2 (0)  1,
1 ' (0)  0, 2 ' (0)  0
the solution of (6) is
1
4
1
 2 (t )  cos
2
1 (t )  cos
2
3
t  cos 2t
3 4
2
3
t  cos 2t
3 2
(7)
See Fig 4.6.3.
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Ch4_115
Fig 4.6.3
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Ch4_116