Transcript Document
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
11 1
L{1} 2 , s>0
s
ss s
Ch4_5
Example 3
Evaluate L{e-3t}
Solution
L{e
3t
} e
0
st
e
3t
( s 3) t
e
s3
d t e ( s 3) t dt
0
0
1
, s 3
s3
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 est 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
,s0
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 } n1 , n 1, 2, 3, (c) L {e }
sa
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
11
(a)1 L
s
n!
(b) t L n1 , 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 est 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 ( n1) 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 n1 f (0) s n2 f (0) f ( n1) (0)
where F ( s) L { f (t )}.
Ch4_26
Solving Linear ODEs
n
n1
an d y an1 d y a0 y g (t )
dt n
dt n1
y (0) y0 , y(0) y1 , y ( n1) (0) yn1
Then
d n y
d n1 y
an L n an1L n1 a0 L { y} L {g (t )} (9)
dt
dt
an [ s nY ( s ) s n1 y (0) y ( n1) (0)]
an1[ s n1Y ( s ) s n2 y (0) y ( n2 ) (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 an1s n1 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 {e4t }
dt
dt
1
2
s Y ( s ) sy(0) y(0) 3[ sY ( s ) y (0)] 2Y ( s )
s4
1
2
( s 3s 2)Y ( s ) s 2
s4
s2
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
sc
st ct
0
M
sc
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 {e2t cos 4t}
(a) L {e t } L {t }ss 5
5t 3
3
3!
6
4
s ss 5 ( s 5) 4
(b) L {e 2t cos 4t} L {cos 4t}ss ( 2)
s
s2
2
s 16 ss 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 ss 2 3 2
s 2 ss 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 ss 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 11 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 )
ta
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 )
ta
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
0t 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
0t a
0,
f (t a) U (t a)
ta
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 (va )
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 e2 s e3s
2 3
s
s
s
Ch4_55
Inverse Form of Theorem 4.7
L 1{eas 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
e2 s e4(t 2) U (t 2)
s 4
(b) a / 2, F ( s) s /( s 2 9), L1{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)} eas 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
0t
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 es sin(t )U (t )
s 1
s 1
s
L 1 2 es 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 ,
0t
(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 ( )dd }
0
0
f ( ) e s ( ) g ( )d d
Ch4_69
Holding fixed, let t = + , dt = d
F (s)G(s) f ( ) est 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 esT
T
0
est 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 es
1
2 s
s
1 e
s (1 es )
(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) = lima0 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 )} est0
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-2s
s
4e2s
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)
2s
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