Laplace Transform
by www.eletricatotal.com
1 Introduction
In this work we are interested in calculating some Laplace transforms using
the appropriate methods. To do so, let’s define the Laplace transform as:
F(s) = L{f(t)}=Z∞
0
f(t)e−st dt
Example 1 - Calculate the Laplace transform of L{1}.
Solution -
We have that f(t) = 1, so applying the definition we have:
F(s) = L{1}=Z∞
0
1. e−st dt =−1
se−st
∞
0
=1
s
For the simplicity of the Laplace transform of a constant, we can generalize
to any constant, or:
F(s) = L{cte}=cte
s
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Example 2 - Calculate the Laplace transform of L{t}.
Solution -
We have that f(t) = t, so applying the definition we have:
F(s) = L{t}=Z∞
0
t. e−st dt =1
s2
For the case of polynomials we can generalize with the following equation:
F(s) = L{tn}=n!
sn+1
Example 3 - Calculate the Laplace transform of L{eat}.
Solution -
We have that f(t) = eat, so applying the definition we have:
F(s) = L{eat}=Z∞
0
eat e−st dt =Z∞
0
e(a−s)tdt =1
a−se(a−s)t
∞
0
Therefore, we conclude that for the exponential function the Laplace transform
is given by:
F(s) = L{eat}=1
s−a
Now, remembering the Euler formula:
eiθ =cos θ +i sinθ
Using this equation as a reference, let’s find the Laplace transform for the func-
tions sine and cosine.
Example 4 - Calculate the Laplace transform of L{sin ωt}eL{cos ωt}.
Solu¸c˜ao -
We have that f(t) = eiωt, so applying the value found in example 3, we have:
F(s) = L{eiωt}=1
s−iω
For the known property of complex numbers, let’s multiply the numerator and
denominator of the fraction by the complex conjugate of the denominator. In
this way, we are left with:
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F(s) = L{eiωt}=1
s−iω
s+iω
s+iω =s+iω
s2+ω2
Working algebraically the result, we can write:
F(s) = L{eiωt}=s
s2+ω2+iω
s2+ω2
Now, comparing this result with the Euler formula, seen above, we conclude
that:
F(s) = L{sin ω t}=s
s2+ω2
F(s) = L{cos ω t}=ω
s2+ω2
2 Properties
To continue our studies, we are going to list some properties of the Laplace
transforms that will facilitate the calculation of the transforms.
2.1 Linearity
Linearity guarantees us the following equality:
F(s) = L{c.f(t)}=c. L{f(t)}
or, also
L{c1. f(t) + c2. g(t)}=c1.L{f(t)}+c2.L{g(t)}=c1. F (s) + c2. G(s)
Example 5 - Find L{3sin 2t+ 4 e3t}
Solution -
L{3sin 2t+ 4 e3t}= 3 L{sin 2t}+ 4 L{e3t}
= 3 2
s2+ 4+ 4 1
s−3=4s2+ 6 s−2
(s2+ 4)(s−3)
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2.2 First Translation Property
If L{f(t)}=F(s) then
L{eatf(t)}=F(s−a)
Example 6 - Find L{t2e3t}
Solution - We know that:
L{t2}=2!
s3=2
s3
So, applying the first translation property, we will find:
L{t2e3t}=2
(s−3)3
Example 7 - Find L{e−3t(2 cos 4t−4sin 4t)}
Solution - We know that:
L{(2 cos 4t−4sin 4t)}= 2 s
s2+ 16−44
s2+ 16=2s−16
s2+ 16
So, applying the first translation property, we will find:
L{e−3t(2 cos 4t−4sen 4t)}=2 (s+ 3) −16
(s+ 3)2+ 16
Example 8 - Find L{e3tcosh 4t}
Solution - We can write that:
L{e3tcosh 4t}=Le3te4t+e−4t
2=1
2Le7t+e−t
=1
21
s−7+1
s+ 1=s−3
s2−6s−7
Example 9 - Find L{e3tcos24t}
Solution - Let’s remember that:
cos2t=1 + cos2t
2
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Taking this into account, we have
L{cos24t}=L1 + cos 8t
2=1
2L{1 + cos 8t}
=1
21
s+s
s2+ 64
Now applying the first property, we have to replace all the “s” from the pre-
vious expression by “s - 3”, or
L{e3tcos24t}=1
21
s−3+s−3
(s−3)2+ 64
2.3 Second Translation Property
If L{f(t)}=F(s)and
g(t) = f(t−a),if t > a,
0,if t < a.
So, to find the transform of g(t), we use
L{g(t)}=e−asF(s)
Example 10 - Find L{f(t)}if
f(t) =
cos(t−2π
3),if t > 2π
3
0,if t < 2π
3
Solution - We already know that:
L{cos t}=s
s2+1 and as per property 2.3 we have a=2π
3, then:
Lcos (t−2π
3)=e−2π
3ss
s2+ 1=s e−2π
3s
s2+ 1
Example 11 - Find L{f(t)}if
f(t) =
3t, if 0 <=t <= 4
3,if t > 4
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Solution - As:
L{3t}= 3 L{t}=3
s2
this value of the found transform is only valid for the interval given in the problem.
When t= 4, the value of f(t) will be f(4) = 12. But for t > 4 the f(t) will be
equal to 3 (the constant value). Therefore, we must subtract 9 from the value of
f(4). So, we are left with the following solution:
L{f(t)}=3
s2(1 −e−4s)−9
se−4s
note that 9
sis the value of (12 −3) L{1}. And e−4s tells us that this value
is only valid from t > 4.
Figura 1: Function graph
Work in Progress.
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