LECTURE 21: CONVOLUTION
MINGFENG ZHAO
October 24, 2014
Definition 1. Let f (t) be a function on [0, ∞), then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
Z
L[f ](s) =
∞
f (t)e−st dt,
for all s > 0.
0
II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as:
L−1 [F ](t) = f (t),
∀for all t > 0.
Proposition 1. There holds that
I. Linearity:
L[af (t) + bg(t)](s) = aL[f (t)](s) + bL[g(t)](s).
That is,
L−1 [aF (s) + bG(s)](t) = aL−1 [F (s)](t) + bL−1 [G(s)](t).
II. First Shifting Property:
L e−at f (t) (s) = L[f (t)](s + a).
That is,
L−1 [F (s + a)] (t) = e−at L−1 [F (s)](t).
III. Transforms of derivatives:
L[f 0 ](s)
= sL[f ](s) − f (0)
L[f 00 ](s)
= s2 L[f ](s) − sf (0) − f 0 (0).
1
2
MINGFENG ZHAO
IV. Second Shifting Property:
L[ua (t)f (t − a)] = e−as L[f (t)](s),
for all s > 0.
That is,
L−1 e−as G(s) (t) = ua (t)L−1 [G(s)](t − a).
V. Transform of Integrals:
t
Z
L
f (τ ) dτ (s) =
0
L[f (t)](s)
.
s
That is,
L−1
Z t
F (s)
=
L−1 [G(s)](τ ) dτ.
s
0
VI. Transform of Convolution:
L[(f ∗ g)(t)] = L[f (t)] · L[g(t)].
That is,
L−1 [L[f (t)] · L[g(t)]] = f ∗ g(t)
Definition 2. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z t
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ.
0
Remark 1. The Property VI implies Property V: let g(t) = 1, then
Z t
(f ∗ g)(t) =
f (τ ) du.
0
By the transform of convolution, we have
Z t
[{(t)](∫ )
L
.
f (τ ) (s) = L[f (t)](s) · L[1](s) =
s
0
Example 1. Let f (t) = et and g(t) = t for t ≥ 0, find f ∗ g and L[f ∗ g].
By the definition of f ∗ g, we have
Z
(f ∗ g)(t)
t
f (τ )g(t − τ ) dτ
=
0
Z
t
eτ (t − τ ) dτ
=
0
Z
=
t
τ
Z
e dτ −
t
0
0
t
τ eτ dτ
LECTURE 21: CONVOLUTION
=
t
t(et − 1) − τ eτ |0 +
Z
3
t
eτ dτ
Use the integration by parts
0
=
t(et − 1) − tet + et − 1
=
et − t − 1.
So we know that the convolution of f (t) = et and g(t) = t is:
(f ∗ g)(t) = et − t − 1 .
By looking up the table, we have
L[f (t)] = L[et ] =
1
,
s−1
and L[g(t)] = L[t] =
1
.
s2
By the transform of convolution, we have
L[f ∗ g] = L[f (t)] · L[g(t)] =
1
.
s2 (s − 1)
1
.
Example 2. Find L
s2 (s + 1)
By looking up the table, we have
−1
−1
L
1
= t,
s2
By the transform of convolution, then
1
−1
L
=
s2 (s + 1)
=
=
−1
and L
1
= e−t .
s+1
1
1
L
·
s2 s + 1
1
1
L−1 2 ∗ L−1
s
s+1
Z t
τ e−(t−τ ) dτ = e−t +t − 1.
−1
0
Therefore, we have
L−1
1
= e−t +t − 1 .
s2 (s + 1)
Volterra integral equation (NOT in Exams)
The Volterra integral equation is:
Z
x(t) = f (t) + (x ∗ g)(t) = f (t) +
t
g(t − τ )x(τ ) dτ
0
4
MINGFENG ZHAO
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of the above equation, we get
X(s) = L[f (t)] + L[x ∗ g].
By the transform of convolution, we get
X(s) = L[f (t)] + X(s) · L[g(t)].
Then
X(s) =
L[f (t)]
.
1 − L[g(t)]
So
x(t) = L−1 [X(s)] = L−1
L[f (t)]
.
1 − L[g(t)]
Dirac delta and impulse response
Definition 3. For any continuous function f (t) on (−∞, ∞), we have
Z ∞
δ(t)f (t) dt = f (0).
−∞
Proposition 2. For any constant a > 0, we have
L[δ(t − a)] = e−as .
That is,
L−1 [e−as ] = δ(t − a).
Proof. By the definition of the Laplace transform, we have
Z ∞
L[δ(t − a)] =
e−st δ(t − a) dt
0
Z
∞
=
0
t−s(t +a) dt0
Let t0 = t − a
−a
= e−as ,
Since a > 0, by the definition of δ.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca