1. Ch 6.6: The Convolution Integral
Sometimes it is possible to write a Laplace transform H(s) as
H(s) = F(s)G(s), where F(s) and G(s) are the transforms of
known functions f and g, respectively.
In this case we might expect H(s) to be the transform of the
product of f and g. That is, does
H(s) = F(s)G(s) = L{f }L{g} = L{f g}?
On the next slide we give an example that shows that this
equality does not hold, and hence the Laplace transform
cannot in general be commuted with ordinary multiplication.
In this section we examine the convolution of f and g, which
can be viewed as a generalized product, and one for which the
Laplace transform does commute.
2. Example 1
Let f (t) = 1 and g(t) = sin(t). Recall that the Laplace
Transforms of f and g are
Thus
and
Therefore for these functions it follows that
{ } { }
1
1
sin)()( 2
+
==
s
tLtgtfL
{ } { } { } { }
1
1
sin)(,
1
1)( 2
+
====
s
tLtgL
s
LtfL
{ } { } { })()()()( tgLtfLtgtfL ≠
{ } { }
( )1
1
)()( 2
+
=
ss
tgLtfL
3. Theorem 6.6.1
Suppose F(s) = L{f (t)} and G(s) = L{g(t)} both exist for
s > a ≥ 0. Then H(s) = F(s)G(s) = L{h(t)} for s > a, where
The function h(t) is known as the convolution of f and g and
the integrals above are known as convolution integrals.
Note that the equality of the two convolution integrals can be
seen by making the substitution u = t - τ.
The convolution integral defines a “generalized product” and
can be written as h(t) = ( f *g)(t). See text for more details.
∫∫ −=−=
tt
dtgtfdgtfth
00
)()()()()( τττττ
5. Example 2
Find the Laplace Transform of the function h given below.
Solution: Note that f (t) = t and g(t) = sin2t, with
Thus by Theorem 6.6.1,
4
2
}2{sin)}({)(
1
}{)}({)(
2
2
+
===
===
s
tLtgLsG
s
tLtfLsF
∫ −=
t
tdtth
0
2sin)()( ττ
{ }
( )4
2
)()()()( 22
+
===
ss
sGsFsHthL
6. Example 3: Find Inverse Transform (1 of 2)
Find the inverse Laplace Transform of H(s), given below.
Solution: Let F(s) = 2/s2
and G(s) = 1/(s - 2), with
Thus by Theorem 6.6.1,
{ }
{ } t
esGLtg
tsFLtf
21
1
)()(
2)()(
==
==
−
−
)2(
2
)( 2
−
=
ss
sH
{ } ∫ −==−
t
detthsHL
0
21
)(2)()( ττ τ
7. Example 3: Solution h(t) (2 of 2)
We can integrate to simplify h(t), as follows.
[ ] [ ]
2
1
2
1
2
1
2
1
1
2
1
1
22)(2)(
2
222
222
0
2
0
2
0
2
0
2
0
2
0
2
−−=
−+−−=
−−−−=
−−=
−=−=
∫
∫∫∫
te
eettte
eetet
deete
dedetdetth
t
ttt
ttt
ttt
ttt
ττ
τττττ
τττ
τττ
8. Find the solution to the initial value problem
Solution:
or
Letting Y(s) = L{y}, and substituting in initial conditions,
Thus
)}({}{4}{ tgLyLyL =+′′
[ ] )(}{4)0()0(}{2
sGyLysyyLs =+′−−
Example 4: Initial Value Problem (1 of 4)
( ) )(13)(42
sGssYs +−=+
4
)(
4
13
)( 22
+
+
+
−
=
s
sG
s
s
sY
1)0(,3)0(),(4 −=′==+′′ yytgyy
9. We have
Thus
Note that if g(t) is given, then the convolution integral can
be evaluated.
τττ dgtttty
t
)()(2sin
2
1
2sin
2
1
2cos3)(
0∫ −+−=
Example 4: Solution (2 of 4)
)(
4
2
2
1
4
2
2
1
4
3
4
)(
4
13
)(
222
22
sG
sss
s
s
sG
s
s
sY
+
+
+
−
+
=
+
+
+
−
=
10. Recall that the Laplace Transform of the solution y is
Note Φ(s) depends only on system coefficients and initial
conditions, while Ψ(s) depends only on system coefficients
and forcing function g(t).
Further, φ(t) = L-1
{Φ(s)} solves the homogeneous IVP
while ψ(t) = L-1
{Ψ(s)} solves the nonhomogeneous IVP
Example 4:
Laplace Transform of Solution (3 of 4)
)()(
4
)(
4
13
)( 22
sΨsΦ
s
sG
s
s
sY +=
+
+
+
−
=
1)0(,3)0(),(4 −=′==+′′ yytgyy
1)0(,3)0(,04 −=′==+′′ yyyy
0)0(,0)0(),(4 =′==+′′ yytgyy
11. Examining Ψ(s) more closely,
The function H(s) is known as the transfer function, and
depends only on system coefficients.
The function G(s) depends only on external excitation g(t)
applied to system.
If G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1
{H(s)} solves
the nonhomogeneous initial value problem
Thus h(t) is response of system to unit impulse applied at t = 0,
and hence h(t) is called the impulse response of system.
Example 4: Transfer Function (4 of 4)
4
1
)(where),()(
4
)(
)( 22
+
==
+
=
s
sHsGsH
s
sG
sΨ
0)0(,0)0(),(4 =′==+′′ yytyy δ
12. Consider the general initial value problem
This IVP is often called an input-output problem. The
coefficients a, b, c describe properties of physical system,
and g(t) is the input to system. The values y0 and y0' describe
initial state, and solution y is the output at time t.
Using the Laplace transform, we obtain
or
Input-Output Problem (1 of 3)
00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′
[ ] [ ] )()()0()()0()0()(2
sGscYyssYbysysYsa =+−+′−−
)()(
)()(
)( 22
00
sΨsΦ
cbsas
sG
cbsas
yaybas
sY +=
++
+
++
′++
=
13. We have
As before, Φ(s) depends only on system coefficients and
initial conditions, while Ψ(s) depends only on system
coefficients and forcing function g(t).
Further, φ(t) = L-1
{Φ(s)} solves the homogeneous IVP
while ψ(t) = L-1
{Ψ(s)} solves the nonhomogeneous IVP
Laplace Transform of Solution (2 of 3)
00 )0(,)0(,0 yyyycyybya ′=′==+′+′′
0)0(,0)0(),( =′==+′+′′ yytgcyybya
00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′
)()(
)()(
)( 22
00
sΨsΦ
cbsas
sG
cbsas
yaybas
sY +=
++
+
++
′++
=
14. Examining Ψ(s) more closely,
As before, H(s) is the transfer function, and depends only on
system coefficients, while G(s) depends only on external
excitation g(t) applied to system.
Thus if G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1
{H(s)}
solves the nonhomogeneous IVP
Thus h(t) is response of system to unit impulse applied at t = 0,
and hence h(t) is called the impulse response of system, with
Transfer Function (3 of 3)
0)0(,0)0(),( =′==+′+′′ yytcyybya δ
cbsas
sHsGsH
cbsas
sG
sΨ
++
==
++
= 22
1
)(where),()(
)(
)(
{ } τττψ dgthsGsHLt
t
)()()()()(
0
1
∫ −== −
