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© 2007 Paul Dawkins 58 http://tutorial.math.lamar.edu/terms.aspx
Convolution Integrals
On occasion we will run across transforms of the form,
H ( s ) = F ( s G s ) ( )
that can’t be dealt with easily using partial fractions. We would like a way to take the inverse
transform of such a transform. We can use a convolution integral to do this.
Convolution Integral
If f(t) and g(t) are piecewise continuous function on[ 0, • )then the convolution integral of f(t)
and g(t) is,
0
t f * g t = f t - t g t dt
Ú
A nice property of convolution integrals is.
( f^ *^ g^ )( t^ ) =^ ( g^ * f^ )( t )
Or,
0 0
t t f t - t g t d t = f t g t - t dt
Ú Ú
The following fact will allow us to take the inverse transforms of a product of transforms.
Fact
1 f g F s G s F s G s f g t
L * = L = *
Let’s work a quick example to see how this can be used.
Example 1 Use a convolution integral to find the inverse transform of the following transform.
2 2 2
H s
s a
Solution
First note that we could use #11 from out table to do this one so that will be a nice check against
our work here.
Now, since we are going to use a convolution integral here we will need to write it as a product
whose terms are easy to find the inverse transforms of. This is easy to do in this case.
2 2 2 2
H s s a s a
Ê ˆ Ê ˆ
= Á ˜ Á ˜
Ë +^ ¯ Ë + ¯
So, in this case we have,
F s G s f t g t sin at s a a
= = fi = =
Using a convolution integral h(t) is,
© 2007 Paul Dawkins 59 http://tutorial.math.lamar.edu/terms.aspx
( (^ )^ (^ ))
(^2 )
3
sin sin
sin cos 2
t
h t f g t
at a a d a
at at at a
t t t
Ú
This is exactly what we would have gotten by using #11 from the table.
Convolution integrals are very useful in the following kinds of problems.
Example 2 Solve the following IVP
4 y ¢¢ + y = g t ( ) , y ( 0 ) = 3 y ¢( 0 )= - 7
Solution
First, notice that the forcing function in this case has not been specified. Prior to this section we
would not have been able to get a solution to this IVP. With convolution integrals we will be able
to get a solution to this kind of IVP. The solution will be in terms of g(t) but it will be a solution.
Take the Laplace transform of all the terms and plug in the initial conditions.
( (^ )^ (^ )^ (^ )) (^ )^ (^ )
2
2
s Y s sy y Y s G s
s Y s s G s
Notice here that all we could do for the forcing function was to write down G(s) for its transform.
Now, solve for Y(s).
2
(^2 1 ) 4 4
s Y s G s s
s^ G s Y s s s
We factored out a 4 from the denominator in preparation for the inverse transform process. To
take inverse transforms we’ll need to split up the first term and we’ll also rewrite the second term
a little.
(^2 1 ) 4 4
2 2 2 2 (^2 1 2 1 ) 4 4 4
s^ G s Y s
s s
s G s s s s
Now, the first two terms are easy to inverse transform. We’ll need to use a convolution integral
on the last term. The two functions that we will be using are,
( ) ( ) 2sin^
t g t f t
Ê ˆ
= Á ˜
Ë ¯
We can shift either of the two functions in the convolution integral. We’ll shift g(t) in our
solution. Taking the inverse transform gives us,
0
3cos 14sin sin 2 2 2 2
t t t y t g t d
t t t
Ê ˆ Ê ˆ Ê ˆ
= Á ˜ - Á ˜ + Á ˜ -
Ë ¯ Ë ¯ Ë ¯
Û
Ù
ı
© 2007 Paul Dawkins 61 http://tutorial.math.lamar.edu/terms.aspx
Table of Laplace Transforms
1 f t F s
= L F ( s ) = L { f ( t )} ( ) { ( )}
1 f t F s
= L F ( s ) = L{ f ( t )}
- 1
s
at e
s - a
n t n = K 1
n
n
s
p t , p > -
1
p
p
s
G +
- (^) t (^) 3 2 2 s
p
1 t n^ -^2 , n = 1, 2,3,K
1 2
n n
n
s
p
◊ ◊ L -
7. sin^ ( at^ )
2 2
a
s + a
8. cos^ ( at^ )
2 2
s
s + a
9. t^ sin( at^ )
2 2 2
2 as
s + a
10. t^ cos( at^ )
2 2
2 2 2
s a
s a
11. sin^ ( at^ ) -^ at^ cos( at )
3
2 2 2
2 a
s + a
12. sin^ ( at^ ) +^ at^ cos( at )
2
2 2 2
2 as
s + a
13. cos ( at ) - at sin( at )
2 2
2 2 2
s s a
s a
14. cos ( at ) + at sin( at )
2 2
2 2 2
s s 3 a
s a
15. sin ( at + b )
2 2
s sin b a cos b
s a
16. cos ( at + b )
2 2
s cos b a sin b
s a
17. sinh ( at )
2 2
a
s - a
18. cosh ( at )
2 2
s
s - a
19. sin( )
at e bt
(^2 )
b
s - a + b
20. cos( )
at e bt
(^2 )
s a
s a b
21. sinh( )
at e bt
(^2 )
b
s - a - b
22. cosh( )
at e bt
(^2 )
s a
s a b
n at t e n = K
1
n
n
s a
24. f ( ct )
1 s F c c
Ê ˆ
Á ˜
Ë ¯
uc ( t ) = u t ( - c )
Heaviside Function
cs
s
e
d ( t - c )
Dirac Delta Function
27. uc^ ( t^ ) f^ ( t^ -^ c ) ( )
cs F s
e 28. uc ( t ) g t ( ) { ( )}
cs g t c
e L +
ct
e f t F ( s - c ) 30. ( ) , 1, 2,3,
n
t f t n = K ( )
( )
n (^) n
f t t
s
F u du
Ú
0
t f v dv
Ú
F ( s )
s
0
t f t - t g t dt
Ú (^ )^ (^ )
F s G s 34. f ( t + T ) = f ( t )
0
T st
sT
f t dt
Ú
e
e
35. f^ ¢^ ( t ) sF^ ( s^ ) -^ f ( 0 ) 36. f^ ¢¢^ ( t ) ( ) ( ) ( )
2 s F s - sf 0 - f ¢ 0
( )
n
f t ( ) ( ) ( )
( )
( )
1 2 2 1 0 0 0 0
n n n n^ n s F s s f s f sf f
© 2007 Paul Dawkins 62 http://tutorial.math.lamar.edu/terms.aspx
Table Notes
- This list is not inclusive and only contains some of the more commonly used Laplace
transforms and formulas.
- Recall the definition of hyperbolic functions.
cosh ( ) sinh( )
t t t t
t t
= =
e e e e
- Be careful when using “normal” trig function vs. hyperbolic functions. The only
difference in the formulas is the “+ a
2 ” for the “normal” trig functions becomes a “- a
2 ”
for the hyperbolic functions!
- Formula #4 uses the Gamma function which is defined as
1
0
x t t x dx
G = Ú
e
If n is a positive integer then,
G ( n + 1 ) = n!
The Gamma function is an extension of the normal factorial function. Here are a couple
of quick facts for the Gamma function
p p p
p n p p p p n p
p
G + = G
G +
G
Ê ˆ
G Á ˜=
Ë ¯
L
