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The Convolution
Theorem



✏

✒

✑

20
...
The convolution is an important construct because of the Convolution Theorem
which gives the inverse Laplace transform of a product of two transformed functions:
L−1 {F (s)G(s)} = (f ∗ g)(t)

✬

✩

x be able to find Laplace and inverse Laplace
transforms of simple functions

Prerequisites
Before starting this Block you should
...


y be able to integrate by parts
z understand how to use step functions in
integration

Learning Style
To achieve what is expected of you
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The Convolution
Let f (t) and g(t) be two functions of t
...


Key Point
If f (t) and g(t) are causal functions then their convolution is defined by:
t

(f ∗ g)(t) =

f (t − x)g(x)dx

0

This is an odd looking definition but it turns out to have considerable use both in Laplace
transform theory and in the modelling of linear engineering systems
...


Example Find the convolution of f and g if f (t) = tu(t) and g(t) = t2 u(t)
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u(t)
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6: The Laplace Transform

2

Solution
Here f (t − x) = (t − x)u(t − x) and g(x) = sin x
...
We find, remembering again that t is a constant in the
integration process,
t

t

(t − x) sin xdx =

−(t − x) cos x

0

t

−
0

(−1)(− cos x)dx
0

t

= [0 + t] −

cos xdx
0
t

= t − sin t

= t − sin x
0

so that
(f ∗ g)(t) = t − sin t or, equivalently, in this case (t ∗ sin t)(t) = t − sin t

Try each part of this exercise
In the last example we found the convolution of f (t) = tu(t) and g(t) = sin t
...
In this
exercise you are asked to find the convolution (g ∗ f )(t) that is, to reverse the order of f and g
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Try each part of this exercise
Obtain the Laplace transforms of f (t) = tu(t) and g(t) = sin t
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What do
you notice?
Part (a) Begin by finding L{f (t)} and L{g(t)}
Answer
Part (b) Now find L{(f ∗ g)(t)}
Answer
3

Engineering Mathematics: Open Learning Unit Level 1
20
...
The Convolution Theorem
Let f (t) and g(t) be causal functions with Laplace transforms F (s) and G(s) respectively, i
...

L{f (t)} = F (s) adn L{g(t)} = G(s)
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+ 9)

s(s2

Solution
In this case we can, of course, find the inverse transform by using partial fractions and then
using the table of transforms
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u(t)
3
3
=

2 −1
L
3

However, we can alternatively use the Convolution Theorem
...
u(t)

So
L−1 {F (s)G(s)} = (f ∗ g)(t)
t

=

by the Convolution Theorem

2u(t − x) sin 3x
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6: The Laplace Transform

4

Solution
Now the variable t can take any value from −∞ to +∞
...
We conclude that
(f ∗ g)(t) = 0 if

t<0

that is, (f ∗ g)(t) is a causal function
...
Now, in the range of integration 0 ≤ x ≤ t and so
u(t − x) = 1

u(x) = 1

since both t − x and x are non-negative
...

Try each part of this exercise
Use the Convolution Theorem to find the inverse transform of H(s) =

s

...
(Treat the cases t < 0 and t ≥ 0
separately
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Find the convolution of
(i) 2tu(t) and t3 u(t) (ii) et u(t) and tu(t) (iii) e−2t u(t)
−t
and e u(t)
...

2
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6: The Laplace Transform

Computer exercise or activity

There is not a special command in DERIVE for obtaining the convolution of two functions
...
Simply author
the appropriate integrand f (x)g(t − x) and integrate between limits 0 and t
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If you then hit the simplify button
DERIVE responds with
t6
60
Use DERIVE to check your answers to Question 1(i),(ii),(iii)
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6: The Laplace Transform

6

End of Block 20
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6: The Laplace Transform

You should find, according to the definition g(t − x) = sin(t − x)
...
xdx

(g ∗ f )(t) =

0

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1
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...
xdx

0
t

=

x cos(t − x)

−
0

= [t − 0] + [sin(t −

t

0
x)]t
0

cos(t − x)dx
= t − sin t

Back to the theory

9

Engineering Mathematics: Open Learning Unit Level 1
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...
6: The Laplace Transform

10

From the example above (f ∗ g)(t) = t − sin t and so
L{(f ∗ g)(t)} = L{t − sin t} =

1
1
− 2
2
s
s +1

Back to the theory

11

Engineering Mathematics: Open Learning Unit Level 1
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...
This, in fact, is a general result which is expressed in the
statement of the convolution theorem which we discuss in the next section
Back to the theory

Engineering Mathematics: Open Learning Unit Level 1
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u(t)

Back to the theory

13

Engineering Mathematics: Open Learning Unit Level 1
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...
u(x)dx

0

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1
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6: The Laplace Transform

1
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(i) (t − 1 + e−t )u(t) (ii) (−et + e2t )u(t) (iii) 12(sin t − t cos t)u(t)
Back to the theory

Engineering Mathematics: Open Learning Unit Level 1
20

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