CHAPTER 4
The Laplace Transform
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
Ch4_2
4.1 Definition of Laplace Transform
Basic Definition
If f(t) is defined for t  0, then the improper integral


0
f (t ) dt  lim
b 

b
0
f (t ) dt
(1)
DEFINITION 4.1
Laplace Transform
If f(t) is defined for t  0, then
  st
L { f (t )}   e
0
f (t )dt
(2)
is said to be the Laplace Transform of f.
Ch4_3
Example 1
Evaluate L{1}
Solution:
Here we keep that the bounds of integral are 0 and
 in mind.
From the definition
  st
b  st
L (1)   e (1)dt  lim  e dt
b  0
0
 st b
e
 lim
b  s
0
 e  sb  1 1
 lim
 , s>0
b 
s
s
Since e-st  0 as t , for s > 0.
Ch4_4
Example 2
Evaluate L{t}
Solution
 st 
 te
L {t} 
s
0
1  st
  e dt
s 0
1
11 1
 L{1}     2 , s>0
s
ss s
Ch4_5
Example 3
Evaluate L{e-3t}
Solution
L{e
 3t

}  e
0
 st
e
 3t
 ( s  3) t
e

s3

d t   e ( s 3) t dt
0

0
1

, s  3
s3
Ch4_6
Example 4
Evaluate L{sin 2t}
Solution
  st
L {sin2t}   e
0

e
 st
sin 2t dt

sin 2t
2   st
  e cos 2t dt
0
s
s
0
2   st
  e cos 2t dt , s  0
s 0
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 
0
s
s
s

0

2 4
 2  2 L{sin 2t}
s s
2
L {sin 2t}  2
,s0
s 4
Ch4_8
L.T. is Linear
We can easily verify that
L { f (t )   g (t )}
 L { f (t )}  L {g (t )}
(3)
  F ( s)  G ( s)
Ch4_9
THEOREM 4.1
Transform of Some Basic Functions
1
(a) L {1} 
s
1
(b) L {t }  n1 , n  1, 2, 3,  (c) L {e } 
sa
s
k
s
(d) L {sin kt}  2
(e) L {cos kt}  2
2
s k
s  k2
k
s
L{sinh k t ) 
(f)
(g)
s k
L {cosh kt}  2
2
s k
n
n!
at
2
2
Ch4_10
DEFINITION 4.2
Exponential Order
A function f(t) is said to be of exponential order,
if there exists constants c>0, M > 0, and T > 0, such
That |f(t)|  Mect for all t > T. See Fig 4.2, 4.3.
Ch4_11
Fig 4.2
Ch4_12
Examples See Fig 4.3
| t |  et
| 2 cos t |  2et
tn
 M, t T
ct
e
Ch4_13
Fig 4.4
t2
A function such as e
see Fig 4.4
is not of exponential order,
Ch4_14
THEOREM 4.2
Sufficient Conditions for Existence
If f(t) is piecewise continuous on [0, ) and of
exponential order, then L{f(t)} exists for s > c.
Ch4_15
Fig 4.1
Ch4_16
Example 5
Find L{f(t)} for
0 , 0  t  3
f (t )  
2 , t  3
Solution
3  st
  st
0
3
L { f (t )}   e 0dt   e 3dt
2e  st

s

3
2e 3 s

,s  0
s
Ch4_17
4.2 If F(s)=L(f(t)), then f(t) is
the
inverse
Laplace
1
transform of F(s) and f(t)=L(F(s))
THEOREM 4.3
Some Inverse Transform
11 
(a)1  L  
s 
n! 
(b) t  L  n1  , n  1, 2, 3,  (c) eat  L
s 
n
1
1
k 
(d) sin kt  L  2
2
s  k 
k 
(f)sinh kt  L 1
 2
2
s  k 
1
1 


s  a 
1
s 
(e) cos kt  L  2 2 
s  k 
(g) cosh kt  L 1
s 
2
2
s

k


Ch4_18
Example 1
Find the inverse transform of
1 1 
 5
1
1 
(a) L
(b) L  2

s 
s  7
Solution
(a)
1
1 1 
L  5  L
 s  4!
(b)

1 1
L  2

s  7
1 4! 
 5
1 4
 t
 s  24
1 1 7  1
L  2
sin 7t

7
7
s  7
Ch4_19
L -1 is also linear
We can easily verify that
1
L {F ( s )  G ( s )}
1
1
 L {F ( s )}  L {G ( s )}
(1)
Ch4_20
Example 2
1  2 s
 2
 6
Find L

 s 4 
Solution
6 
1  2 s  6 
1  2 s
L  2
 2
L  2

s  4 s  4
 s 4 
s  6 1 2 
1
 2 L  2
 L  2

s  4 2
s  4
 2 cos 2t  3sin 2t
(2)
Ch4_21
Example 3: Partial Fraction
2


s
 6s  9
1
Find L 

( 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
Ch4_22
Example 3 (2)
A  16/5, B  25/6, c  1/30
(4)
Thus
2


s
 6s  9
1
L 

( s  1)( s  2)( s  4) 
16 1 1  25 1 1  1 1 1 
 L 
 L 
 L 

5
 s  1 6
 s  2  30
s  4
16 t 25 2t 1 4t
 e  e  e
(5)
5
6
30
Ch4_23
Uniqueness of L -1
Suppose that the functions f(t) and g(t) satisfy the
hypotheses of Theorem 4.2, so that their Laplace
transform F(s) and G(s) both exist. If F(s)=G(s) for
all s>c (for some c), then f(t)=g(t) whenever on [0,
+  ) both f and g are continuous.
Ch4_24
Transform of Derivatives
 L { f (t )}
  st
 e
0
f (t )dt e
 st
L { f (t )}  0
e
 st
  st
0
0
 st

f (t )  s  e
  f (0)  sL { f (t )}
, s>0
L { f (t )}  sF ( s)  f (0)


f (t )dt  e
  f (0)  sL { f (t )}
 s[ sF (s)  f (0)]  f (0)
f (t )
0
f (t )dt

 s  est f (t )dt
0
(6)
, s>0
2


L { f (t )}  s F ( s)  sf (0)  f (0)
3
2



L { f (t )}  s F ( s)  s f (0)  sf (0)  f (0)
(7)
(8)
Ch4_25
THEOREM 4.4
Transform of a Derivative
If f , f ' ,  , f ( n1) 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 )}.
Ch4_26
Solving Linear ODEs
n
n1
 an d y  an1 d y    a0 y  g (t )
dt n
dt n1
y (0)  y0 , y(0)  y1 , y ( n1) (0)  yn1
Then
d n y 
d n1 y 
an L  n   an1L  n1     a0 L { y}  L {g (t )} (9)
 dt 
 dt 
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)
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
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
Ch4_29
Example 4: Solving IVP
dy
 3 y  13 sin 2t , y (0)  6
Solve
dt
Solution
dy 

L    3L { y}  13L {sin 2t}
 dt 
26
sY ( s )  6  3Y ( s )  2
s 4
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)
(12)
(13)
Ch4_30
Example 4 (2)
6s 2  50
A
Bs  C

 2
2
( s  3)( s  4) s  3 s  4
6s  50  A( s  4)  ( Bs  C )
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
s 
1 1 
1
1 2 
y (t )  8L 
  2L  2
  3L  2

 s  3
s  4
s  4
3t
y(t )  8e  2 cos 2t  3sin 2t
2
2
Ch4_31
Example 5
4t
y
"

3
y
'

2
y

e
, y(0)  1, y' (0)  5
Solve
Solution
d 2 y 
dy 

L  2   3L    2L { y}  L {e4t }
 dt 
 dt 
1
2
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)
Thus
( s  1)( s  2)( s  4)
16 t 25 2t 1 4t
y (t )  L {Y ( s)}   e  e  e
5
6
30
Ch4_32
1
4.3 Translation Theorems
THEOREM 4.5
Behavior of F(s) as s → 
If f is piecewise continuous on [0, ) and of
exponential order, then lims L{f} = 0.
Proof
  st
L {f}  e
0
 M
| f (t ) | dt
e ( s c ) t
e e dt   M
sc
  st ct
0
M

sc
s 

0
0
Ch4_33
THEOREM 4.6
Translation on the s-axis
If L{f} = F(s) and a is any real number, then
L{eatf(t)} = F(s – a), See Fig 4.10.
Proof
L{eatf(t)} =  e-steatf(t)dt
=  e-(s-a)tf(t)dt = F(s – a): replacing all s in
F(s) by s-a
Ch4_34
Fig 4.10
Ch4_35
Example 1
Find the L.T. of
(a) L {e5t t 3}
Solution
(b) L {e2t cos 4t}
(a) L {e t }  L {t }ss 5
5t 3
3
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
 2

s  16 ss  2 ( s  2) 2  16
Ch4_36
Inverse Form of Theorem 4.6
1
1
at
L
{
F
(
s

a
)}

L
{
F
(
s
)
}

e
f (t )

s s  a
(1)
1
f
(
t
)

L
{F ( s)}.
where
Ch4_37
Parttial Fraction
To perform the inverse transform of R(s)=P(s)/Q(s):
Rule 1: Linear Factor Partial Fractions
Rule 2: Quadratic Factor Partial Fractions
Ch4_38
Example 2
Find the inverse L.T. of
2s  5 
1 s/2  5/3 
(a) L 1
(b)
L  2


2
 s  4s  6 
( s  3) 
Solution
A
B
(a) 2s  5


2
s  3 ( s  3) 2
( s  3)
2 s  5  A( s  3)  B
we have A = 2, B = 11
2s  5
2
11


2
s  3 ( s  3) 2
( s  3)
(2)
Ch4_39
Example 2 (2)
And
1
2s  5 
1 
1 1 
1
L 
 2L 
  11L 
(3)
2
2
s

3


( s  3) 
( s  3) 
From (3), we have
1 2 s  5 
3t
3t
L 
 2e  11e t
2
( s  3) 
(4)
Ch4_40
Example 2 (3)
s / 2  5/3
s / 2  5/3

2
s  4s  6 ( s  2)2  2
1 s / 2  5 / 3 
L  2

 s  4s  6 

1 1 s  2  2 1
1
 L 
 L 

2
2
2
( s  2)  2  3
( s  2)  2 
(b)

1 1 s
2
2

1
 L  2
L  2


2
 s  2 ss  2  3 2
 s  2 ss  2 
1 2t
2 2t
 e cos 2t 
e sin 2t
2
3
(5)
(6)
(7)
Ch4_41
Example 3
2 3t
y
"

6
y
'

9
y

t
e , y(0)  2 , y' (0)  6
Solve
Solution
2
2
s Y ( s)  sy(0)  y(0)  6[ sY ( s)  y(0)]  9Y ( s) 
( s  3)3
2
2
( s  6s  9)Y ( s)  2s  5 
( s  3)3
2
2
( s  3) Y ( s)  2s  5 
( s  3)3
2s  5
2

Y (s ) 
2
( s  3) ( s  3)5
Ch4_42
Example 3 (2)
2
11
2
Y ( s) 


2
s  3 ( s  3) ( s  3)5

y (t )
1 
1  2 1 4!  (8)
1
 2L 
 L 
  11L 
2
5
 s  3
( s  3)  4!
( s  3) 
1
1
1

 4 3t
3t
1 4!
L  2
  te , L  5
t e
 s s  s 3 
 s ss 3 
1 4 3t
y (t )  2e  11te  t e
12
3t
3t
Ch4_43
Example 4
t
Solve y"4 y'6 y  1  e , y(0)  0 , y' (0)  0
Solution
1
1
2
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)
2s  1
Y (s ) 
s ( s  1)( s 2  4s  6)
1/ 6 1/ 3 s / 2  5 / 3
Y (s) 

 2
s
s  1 s  4s  6
Ch4_44
Example 4 (2)
1 11  1 1 1 
Y ( s)  L    L 

6
s  3
 s  1

1 1 s  2 
2
2
1
 L 
L 


2
2
2
( s  2)  2  3 2
( s  2)  2 
1 1 t 1 2t
2 2t
  e  e cos 2t  e sin 2t
6 3
2
3
Ch4_45
DEFINITION 4.3
Unit Step Function
The Unit Step Function U(t – a) defined for t  0 is
0 , 0  t  a
U (t  a )  
ta
1 ,
See Fig 4.11.
Ch4_46
Fig 4.11
Ch4_47
Fig 4.12
Fig 4.13
Fig 4.12 shows the graph of (2t – 3)U(t – 1).
Considering Fig 4.13, it is the same as
f(t) = 2 – 3U(t – 2) + U(t – 3), t  0
Ch4_48
Also a function of the type
 g (t ), 0  t  a
f (t )  
ta
 h(t ),
is the same as
f (t )  g (t )  g (t ) U (t  a)  h(t ) U (t  a)
Similarly, a function of the type
0t a
0,

f (t )   g (t ), a  t  b
0,
t b
can be written as 
f (t )  g (t )[ U (t  a)  U (t  b)]
(9)
(10)
(11)
(12)
Ch4_49
Example 5
20t , 0  t  5
f (t )  
0 , t  5
in terms of U(t). See Fig 4.14.
Express
Solution
From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0
f(t) = 20t – 20tU(t – 5)
Ch4_50
Fig 4.14
Ch4_51
Consider the function
0t a
0,
f (t  a) U (t  a)  
ta
 f (t  a),
(13)
See Fig 4.15.
Ch4_52
Fig 4.15
Ch4_53
THEOREM 4.7
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  st
 e
0


  st
f (t  a) U (t  a)dt   e
a

a
f (t  a) U (t  a)dt
e  st f (t  a)dt
Ch4_54
Let v = t – a, dv = dt, then
L { f (t  a) U (t  a)}
 s (va )
 e
0
f (v)dv  e
 as   sv
0
e
f (v)dv  e as L { f (t )}
If f(t) = 1, then f(t – a) = 1, F(s) = 1/s,
e as
L {U (t  a )} 
(14)
s
eg: The L.T. of Fig 4.13 is
L { f (t )}  2L {1}  3L {U (t  2)}  L {U (t  3)}
1 e2 s e3s
 2 3

s
s
s
Ch4_55
Inverse Form of Theorem 4.7

L 1{eas F ( s)}  f (t  a) U (t  a)
(15)
Ch4_56
Example 6
Find the inverse L.T. of
s
1
s / 2 
1



1

2
s
(a) L 
(b)
L
e
 2

e 
s  9

s  4

Solution
1
4t
a

2
,
F
(
s
)

1
/(
s

4
),
L
{
F
(
s
)}

e
(a)
then
1 1
L 
e2 s   e4(t 2) U (t  2)
s  4

(b) a   / 2, F ( s)  s /( s 2  9), L1{F ( s)}  cos 3t
then
  

1 s
s / 2 
L  2
e
  cos 3 t   U  t  
s  9

 2   2  Ch4_57
Alternative Form of Theorem 4.7
Since t 2  (t  2)2  4(t  2)  4 , then
L {t 2 U (t  2)}
 L {(t  2) U (t  2)  4(t  2) U (t  2)  4U (t  2)}
2
The above can be solved. However, we try another
approach.
Let u = t – a,
  st
 s (u a )
L {g (t ) U (t  a)}   e g (t )dt   e
g (u  a)du
a
0
That is,
(16)
L {g (t ) U (t  a)}  eas L {g (t  a)}
Ch4_58
Example 7
Find L {cos tU (t   )}
Solution
With g(t) = cos t, a = , then
g(t + ) = cos(t + )= −cos t
By (16),
L {cos tU (t   )}  e
s
s s
L {cos t}   2 e
s 1
Ch4_59
Example 8
Solve y ' y  f (t ) , y (0)  5
0t 
0 ,
f (t )  
3 sin t , t  
Solution
We find f(t) = 3 cos t U(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  (17)
s 1 2  s 1
s 1
s 1

Ch4_60
Example 8 (2)
It follows from (15) with a = , then
1
1 s  (t  )
1 1
L 
e e
U (t   ) , L  2 es   sin(t   )U (t   )
s  1

s  1

s
L 1 2 es   cos(t   )U (t   )
s  1

Thus
3 (t  )
3
3
y (t )  5e  e
U (t   )  sin(t   ) U (t   )  cos(t   ) U (t   )
2
2
2
3 (t  )
t
 5e  [e
 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
t
See Fig 4.16
Ch4_61
Fig 4.16
Ch4_62
4.4 Additional Operational Properties
Multiplying a Function by tn
dF d   st
  e f (t )dt
ds ds 0
 
  st
 st

[e f (t )]dt    e tf (t )dt   L {tf (t )}
0 s
0
d
that is, L {tf (t )}   L { f (t )}
ds
Similarly,
L {t 2 f (t )}  L {t  tf (t )}   L {tf (t )}
2
d d
d

    L { f (t )}  2 L { f (t )}
ds  ds
 ds
Ch4_63
THEOREM 4.8
Derivatives of Transform
If F(s) = L{f(t)} and n = 1, 2, 3, …, then
n
d
n
n
L {t f (t )}  (1)
F ( s)
n
ds
Ch4_64
Example 1
Find L{t sin kt}
Solution
With f(t) = sin kt, F(s) = k/(s2 + k2), then
d
L {t sin kt}   L {sin kt}
ds
d k 
2ks
  2
 2
2
ds  s  k  ( s  k 2 ) 2
Ch4_65
Different approaches
Theorem 4.6:
L{te }  L{t}s s 3
3t
1
 2
s
s  s 3
1

( s  3) 2
Theorem 4.8:
d
d 1
1
3t
2
L {te }   L {e }  
 ( s  3) 
2
ds
ds s  3
( s  3)
3t
Ch4_66
Example 2
Solve x"16 x  cos 4t , x(0)  0 , x' (0)  1
Solution
s
2
( s  16) X ( s)  1  2
s  16
or
1
s
X ( s)  2
 2
2
s  16 ( s  16)
From example 1, 1  2ks 
L  2
 t sin kt
2 2
( s  k ) 
Thus

1 1  4  1 1  8s
x(t )  L  2
 L  2
2
4
 s  16  8
( s  16) 
1
1
 sin st  t sin 4t
4
8
Ch4_67
Convolution
A special product of f * g is defined by
f *g 
t
0
f ( ) g (t   )d
(2)
and is called the convolution of f and g.
Note: f * g = g * f
Ch4_68
THEOREM 4.9
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


 s



F ( s)G ( s )    e f ( )d   e  s g (  )d 
 0
 0



0


0
e  s (   ) f ( )g (  )dd }


0
0
  f ( )  e  s (   ) g (  )d d
Ch4_69
Holding  fixed, let t =  + , dt = d


F (s)G(s)   f ( ) est g (t   )dtd

0
The integrating area is the shaded region in Fig 4.32.
Changing the order of integration:

F ( s)G( s)   e
 st
0

 e
0

t
f ( ) g (t   )d dt
 f ( )g(t  )d dt
0
 st
t
0
 L{ f  g}
Ch4_70
Fig 4.32
Ch4_71
Example 3

t 

Find L  0 e sin(t   ) d
Solution
Original statement
= L{et * sin t}
1
1
1

 2

s  1 s  1 ( s  1)( s 2  1)
Ch4_72
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
(5)
Ch4_73
Example 4


1
L 1 2
2 2 
(
s

k
) 

Find
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
1 
 1 t
1
L  2
 2  sin k sin k (t   )d
2 2
( s  k )  k 0
(6)
Ch4_74
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
1 

1
1 t
L  2
 2  [cos k (2  t )  cos kt ]d
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
Ch4_75
Transform of an Integral
When g(t) = 1, G(s) = 1/s, then


F ( s)
L  f ( )d 
0
s
t
t
0 f ( )d  L
1  F ( s ) 


 s 
(7)
(8)
Ch4_76
Examples:
1 
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 
 t
1
1 2
L  3 2
  0 (  sin  )d  t  1  cos t
2
 s ( s  1) 
Ch4_77
Periodic Function

f(t + T) = f(t): periodic with period T
THEOREM 4.10
Transform of a Periodic Function
If f(t) is a periodic function with period T, then
1
L { f (t )} 
1  esT
T
0
est f (t )dt
Ch4_78
Proof
T
L { f (t )}   e
 st
0
  st
f (t )dt   e
T
f (t )dt
Use change of variable
  st
T e
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 )} 
 sT
1 e
T
0 e
 st
f (t )dt
Ch4_79
Example 7
Find the L. T. of the function in Fig 4.35.
Solution
We find T = 2 and
1, 0  t  1
E (T )  
0, 1  t  2
From Theorem 4.10,
1  st
1
L {E (t )} 
e  1dt  0
 2 s 0
1 e
1 1  es
1


2 s
s
1 e
s (1  es )


(12)
Ch4_80
Fig 4.35
Ch4_81
Example 8
di
L  Ri  E (t )
dt
(13)
Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.35.
Solution
1
LsI ( s )  RI ( s ) 
s
s (1  e )
or
1/ L
1
I (s) 

(14)
s
s ( s  R / L) 1  e
1
s
2 s
3 s

1

e

e

e

Because
-s
1 e
and
1
L/ R
L/ R


s ( s  R / L)
s
s  R/ L
Ch4_82
The DE


1 1
1 
s
2 s
I ( s)   
1

e

e
 ...

R s s  R L
Assuming R=1ohm, L=1 henry
Then i(t) is described as follows and see Fig 4.36:
1  e t ,
0  t 1
 t
( t 1)

e

e
,
1 t  2

i(t )  
t
( t 1)
( t  2 )
(15)
1

e

e

e
,
2

t

3

 e t  e (t 1)  e (t 2)  e (t 3) , 3  t  4

Ch4_83
Fig 4.36
Ch4_84
4.5 The Dirac Delta Function
Unit Impulse
See Fig 4.43(a). Its function is defined by
0  t  t0  a
0,
1
 a (t  t0 )   , t0  a  t  t0  a
(1)
2
a

t  t0  a
0,
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   (t  t0 )dt  1 . See Fig 4.43(b).
0
Ch4_85
Fig 4.43
Ch4_86
The Dirac Delta Function
This function is defined by
(t – t0) = lima0 a(t – t0)
The two important properties:
(2)
, t  t0
(1)  (t  t0 )  
 0, t  t0
(2)
x
0  (t  t0 )dt  1
, x > t0
Ch4_87
THEOREM 4.11
Transform of the Dirac Delta Function
For t0 > 0,
L { (t  t0 )}  est0
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

sa
 sa


e

e
 st0

 e 
 2 sa 
(4)
Ch4_88
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 )
a 0
sa
 sa

  st0
e

e
 st0
  e
 e lim 
a 0
 2sa 
Now when t0 = 0, we have
L  (t )  1
Ch4_89
Example 1
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 ,
See Fig 4.44.
(5)
Ch4_90
Fig 4.44
Ch4_91
Example 1 (2)
2s
4
e
(b)
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 t
4 sin t ,
(6)
Ch4_92
Fig 4.45
Ch4_93