community project
mathcentre community project
encouraging academics to share maths support resources
All mccp resources are released under an Attribution Non-commerical Share Alike licence
Integration: Laplace Transforms
mccp-Fletcher-006-v2
Suppose f (t) is a function of time defined for t ≥ 0. Its Laplace Transform, which has a wide range
of applications in engineering, is the function F (s) of a new variable s defined by
Z ∞
L(f ) = F (s) =
f (t)e−st dt
0
This note reviews the techniques of integration needed to find and manipulate Laplace Transforms.
Powers
Example If f (t) = t then
Z
F (s) = L(t) =
∞
te−st dt
0
t=∞
Z ∞
−1 −st
−1 −st
=
t×
e
e dt integrating by parts
−
1×
s
s
0
t=0
|
{z
}
= 0 for t = 0 and t = ∞
t=∞
1 −st
integrating again, noting three minus signs
= − 2e
s
t=0
1
= 2 substituting limits t = ∞ and t = 0
s
Exercise Use integration by parts to show that L (t2 ) = (2/s) × L(t). Generalise this to L (tn ).
Exponential and trigonometric functions
Example
L e
−at
=
Z
∞
e
e
−at −st
0
dt =
Z
0
∞
e
−(s+a)t
t=∞
1
1 −(s+a)t
=
e
dt = −
s+a
s+a
t=0
This example with a = −jω can be used to find L(cos ωt) and L(sin ωt):
s + jω
1
=
to make the denominator real
s − jω
(s + jω)(s − jω)
s + jω
s
ω
= 2
= 2
+j 2
2
2
s +ω
s +ω
s + ω2
L (ejωt ) =
Exercise Compare the imaginary parts of Euler’s formula cos(ωt) + j sin(ωt) = ejωt and the final
expression here to show that L(sin(ωt)) = ω/(s2 + ω 2 ). What is L(cos(ωt))?
www.mathcentre.ac.uk
c
Leslie
Fletcher
Liverpool John Moores University
Martin Randles
Liverpool John Moores University
Figure 1: Graph of f (t)
Other types of function
f (t)
Example The Laplace transform of the function f (t) defined by
t if 0 ≤ t ≤ 1
f (t) =
1 if t ≥ 1
mathcentre community project
encouraging academics to share maths support resources
is (1 − e )/s
−s
2
L(f (t)) =
t
All mccp resources are released under an Attribution Non-commerical Share Alike licence
as this calculation shows:
1
Z
0
1
t × e−st dt
Z
using the definition of f (t) for 0 ≤ t ≤ 1
∞
+
1 × e−st dt using the definition of f (t) for t ≥ 1
1
t=1 Z 1
−1 −st
−1 −st
e
e dt integrating by parts for the first integral
= t×
−
1×
s
s
0
t=0
{z
}
|
|
{z
}
(b)
(a)
t=∞
−1 −st
+
evaluating the second integral
e
s
t=1
{z
}
|
(c)
−e−s
=
substituting limits t = 1 and t = 0 in (a)
s
t=1
1 −st
− 2e
integrating (b) again, noting three minus signs
s
t=0
|
{z
}
(d)
e−s
substituting limits t = ∞ and t = 1 in (c)
+
s
−e−s e−s
=
+
already found and cancelling out
s
s
1 − e−s
+
substituting limits t = 1 and t = 0 in (d), giving L(f (t))
s2
The First Shift Theorem
This theorem says that if L(f (t)) = F (s) then L(f (t)e−at ) = F (s + a). To see this compare
these integrals; the second is similar to
the first, but with s replaced by s + a.
Figure 2: Decaying sinusoidal oscilliations
Z ∞
−st
L(f (t)) =
f (t)e dt
1
exp(-2*t)*sin(20*t)
Z0 ∞
0.5
L(f (t)e−at ) =
f (t) |e−at{ze−st} dt
0
0
−(s+a)t
=e
Example The Laplace transform of
the decaying sinusoidal oscillations
20
e−2t sin(20t) is
(s + 2)2 + 400
-0.5
-1
0
0.5
1
Exercise What is the Laplace Transform of e−at cos(ωt)? Answer
www.mathcentre.ac.uk
c
Leslie
Fletcher
Liverpool John Moores University
1.5
2
2.5
3
s+a
(s + a)2 + ω 2
Martin Randles
Liverpool John Moores University