Transformada De Laplace

Laplace Transforms: Theory, Problems, and
Solutions
Marcel B. Finan
Arkansas Tech University
c All Rights Reserved

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Contents
41 The Laplace Transform: Basic Definitions and Results
42 Further Studies of Laplace Transform

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15

43 The Laplace Transform and the Method of Partial Fractions 29
44 Laplace Transforms of Periodic Functions

36

46 Convolution Integrals

4647 The Dirac Delta Function and Impulse Response

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48 Solutions to Problems

64

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41

The Laplace Transform: Basic Definitions
and Results

Laplace transform is yet another operational tool for solving constant coefficients linear differential equations. The process of solution consists of three
main steps:
• The given ”hard” problem is transformed into a ”simple” equation.
•This simple equation is solved by purely algebraic manipulations.
• The solution of the simple equation is transformed back to obtain the solution of the given problem.
In this way the Laplace transformation reduces the problem of solving a differential equation to an algebraic problem. The third step is made easier by
tables, whose role is similar to that of integral tables in integration.
Theabove procedure can be summarized by Figure 41.1

Figure 41.1
In this section we introduce the concept of Laplace transform and discuss
some of its properties.
The Laplace transform is defined in the following way. Let f (t) be defined
for t ≥ 0. Then the Laplace transform of f, which is denoted by L[f (t)]
or by F (s), is defined by the following equation
∞

T

f (t)e−st dt =

L[f (t)]= F (s) = lim

T →∞

0

f (t)e−st dt
0

The integral which defined a Laplace transform is an improper integral. An
improper integral may converge or diverge, depending on the integrand.
When the improper integral in convergent then we say that the function f (t)
possesses a Laplace transform. So what types of functions possess Laplace
transforms, that is, what type of functionsguarantees a convergent improper
integral.
Example 41.1
Find the Laplace transform, if it exists, of each of the following functions
(a) f (t) = eat

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(b) f (t) = 1 (c) f (t) = t (d) f (t) = et
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Solution.
(a) Using the definition of Laplace transform we see that
∞

T

e−(s−a)t dt = lim

L[eat ] =

T →∞

0

But

T

e−(s−a)t dt.
0

T

e−(s−a)t dt =

1−e−(s−a)T
s−a0

if s = a
if s = a.

For the improper integral to converge we need s > a. In this case,
L[eat ] = F (s) =

1
, s > a.
s−a

(b) In a similar way to what was done in part (a), we find
∞

T

e−st dt = lim

L[1] =

T →∞

0

0

1
e−st dt = , s > 0.
s

(c) We have
∞

L[t] =

te
0

−st

te−st e−st
−2
dt = −
s
s

∞

=
0

1
, s > 0.
s2

(d) Againusing the definition of Laplace transform we find
∞

2

L[et ] =

2 −st

et

dt.

0
∞

2

2

If s ≤ 0 then t2 − st ≥ 0 so that et −st ≥ 1 and this implies that 0 et −st dt ≥
∞
. Since the integral on the right is divergent, by the comparison theorem
0
of improper integrals (see Theorem 41.1 below) the integral on the left is also
∞
∞
divergent. Now, if s > 0 then 0 et(t−s)dt ≥ s dt. By the same reasoning
2
the integral on the left is divergent. This shows that the function f (t) = et
does not possess a Laplace transform
The above example raises the question of what class or classes of functions
possess a Laplace transform. Looking closely at Example 41.1(a), we notice
∞
that for s > a the integral 0 e−(s−a)t dt is convergent and a critical component forthis convergence is the type of the function f (t). To be more specific,
if f (t) is a continuous function such that
|f (t)| ≤ M eat ,
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t≥C

(1)

where M ≥ 0 and a and C are constants, then this condition yields
∞

∞

C

C

0

0

e−(s−a)t dt.

f (t)e−st dt + M

f (t)e−st dt ≤

Since f (t) is continuous in 0 ≤ t ≤ C, by letting A = max{|f (t)| : 0 ≤ t ≤ C }
we have
C...