The document discusses the Laplace transform, which is used to solve differential equations. It defines the Laplace transform as an integral transform that converts a function f(t) into another function F(s). It provides examples of calculating the Laplace transform of various functions like e^at, sin(at), and discusses properties like linearity. In summary, the document introduces the Laplace transform and provides examples of computing transforms of basic functions through integrals.

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By:
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2. Ch 6.1: Definition of Laplace Transform
Many practical engineering problems involve mechanical or
electrical systems acted upon by discontinuous or impulsive
forcing terms.
For such problems the methods described in Chapter 3 are
difficult to apply.
In this chapter we use the Laplace transform to convert a
problem for an unknown function f into a simpler problem
for F, solve for F, and then recover f from its transform F.
Given a known function K(s,t), an integral transform of a
function f is a relation of the form
= ò ¥ £ a <b £ ¥ b
F(s) K(s,t) f (t)dt,
a
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3. The Laplace Transform
Let f be a function defined for t ³ 0, and satisfies certain
conditions to be named later.
The Laplace Transform of f is defined as
{ } = = ò¥ -
L f (t) F(s) e st f (t)dt
0
Thus the kernel function is K(s,t) = e-st.
Since solutions of linear differential equations with constant
coefficients are based on the exponential function, the
Laplace transform is particularly useful for such equations.
Note that the Laplace Transform is defined by an improper
integral, and thus must be checked for convergence.
On the next few slides, we review examples of improper
integrals admission.and edhole.piecewise com
continuous functions.

4. Example 1
Consider the following improper integral.
ò¥
0
estdt
We can evaluate this integral as follows:
st b
e dt e dt e
lim b st
lim 1 lim( 1)
ò ¥ st ò e
sb
= = = -
b
b
s s
®¥ ®¥ ®¥
0
0 0
b
Note that if s = 0, then est = 1. Thus the following two cases
hold:
1 , if s
0; and
s
diverges, if 0.
ò
0
¥
0
³
= - <
ò
¥
st
e dt
st
e dt s
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5. Example 2
Consider the following improper integral.
ò¥
0
st costdt
We can evaluate this integral using integration by parts:
ò ò
b
st cos tdt =
lim st cos
tdt
¥
0 0
êë é
®¥
b b
st t s tdt
b
= -
ò
lim sin sin
0 0
[ ]
®¥
b b
st t s t
b
= +
lim sin cos
0 0
®¥
b
lim[ sb sin b s (cos b
1)]
= + -
úû ù
®¥
b
Since this limit diverges, so does the original integral.
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6. Piecewise Continuous Functions
A function f is piecewise continuous on an interval [a, b] if
this interval can be partitioned by a finite number of points
a = t0 < t1 < … < tn = b such that
(1) f is continuous on each (tk, tk+1)
f t k n
< ¥ = -
(2) lim ( ) , 0, 
, 1
f t k n
+
k
t t
< ¥ =
®
(3) lim ( ) , 1, ,
-
+
k
t t
1

®
In other words, f is piecewise continuous on [a, b] if it is
continuous there except for a finite number of jump
discontinuities.
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7. Example 3
Consider the following piecewise-defined function f.
ì
t t
ïî
ïí
£ £
, 0 1
2
t t
- < £
3 , 1 2
t t
+ < £
=
1 2 3
f t
( )
From this definition of f, and from the graph of f below, we
see that f is piecewise continuous on [0, 3].
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8. Example 4
Consider the following piecewise-defined function f.
ì
t t
ïî
ïí
+ £ £
1, 0 1
2
( )
= -
4, 2 3
( ) 1
t t
- < £
2 , 1 2
t
< £
f t
From this definition of f, and from the graph of f below, we
see that f is not piecewise continuous on [0, 3].
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9. Theorem 6.1.2
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) | £ Keat when t ³ M, for constants a, K, M, with K, M > 0.
Then the Laplace Transform of f exists for s > a.
0 òL f t = F s = ¥ e-st f t dt
{ ( )} ( ) ( ) finite
Note: A function f that satisfies the conditions specified above
is said to to have exponential order as t ® ¥.
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10. Example 5
Let f (t) = 1 for t ³ 0. Then the Laplace transform F(s) of f
is:
{ }
¥ -
ò
ò
L e dt
lim
®¥
lim
=
=
= -
e dt
-
®¥
-
1 , 0
1
0
0
0
s
= >
s
s
e
st b
b
b st
b
st
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11. Example 6
Let f (t) = eat for t ³ 0. Then the Laplace transform F(s) of f is:
{ }
¥ -
ò
ò
at st at
L e e e dt
e dt
s a t b
s a
lim
lim
s a
e
s a
b
b s a t
b
>
-
=
=
=
-
= -
- -
®¥
- -
®¥
1 ,
0
( )
0
( )
0
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12. Example 7
Let f (t) = sin(at) for t ³ 0. Using integration by parts twice,
the Laplace transform F(s) of f is found as follows:
{ }
ò ò
F ( s ) = L sin( at ) = e sin atdt =
lim e sin
atdt
0 0
e at a s
= é- -
- -
lim ( cos ) / cos
s
1 lim cos
0
0 0
ù
e at a s
s
1 lim ( sin ) / sin
= - é +
2
0 0
F s F s a
a
1 ( ) ( ) , 0
2 2 2
>
+
= - Þ =
ù
úû
êë é
êë
úû ù
= -
úû
êë
ò
ò
ò
- -
®¥
-
®¥
®¥
-
®¥
¥ -
s
s a
s
a
e at
a
a
a
e at
a
a
e at
a
st b b st
b
b st
b
st b b st
b
b st
b
st
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13. Linearity of the Laplace Transform
Suppose f and g 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 c1 f (t) + c2g(t) exists. That is,
{ ( ) ( ) } [ ( ) ( )] is finite
1 2 ò0 1 2 L c f t + c g t = ¥ e-st c f t + c g t dt
{ }
with
+ = ò + ò¥ - ¥ -
L c f ( t ) c g ( t ) c e st f ( t ) dt c e st g ( t )
dt
1 2 1 0 2 0
{ ( )} { ( )}
c L f t c L g t
= +
1 2
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14. Example 8
Let f (t) = 5e-2t - 3sin(4t) for t ³ 0.
Then by linearity of the Laplace transform, and using results
of previous examples, the Laplace transform F(s) of f is:
F s =
L f t
( ) { ( )}
{ -
2
t
}
{ } { }
L e t
= -
5 3sin(4 )
t
L e L t
= -
5 3 sin(4 )
12
, 0
16
2
5
2
2
>
+
-
+
=
-
s
s s
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