LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC
DELTA
MINGFENG ZHAO
October 28, 2015
The Laplace transform
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 (t)](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 (s)](t) = f (t),
for all t > 0.
Proposition 1. The followings hold:
I. 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).
II. First Shifting Property:
L e−at f (t) (s)
=
L[f ](s + a)
L−1 [F (s + a)](t)
=
e−at L−1 [F (s)](t).
III. Second Shifting Property:
L[u(t − a)f (t − a)](s)
= e−as L[f (t)](s)
L−1 e−as F (s) =
u(t − a)L−1 [F (s)](t − a).
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2
MINGFENG ZHAO
Remark 1. In practice, we can use the second shifting property in the following way:
L[u(t − a)f (t)](s) = e−as L[f (t + a)](s).
Remark 2. In general, let a0 = 0 < a1 < a2 < a3 < · · · < an < ∞, if


 f1 (t),
if 0 ≤ t < a1 ,







f2 (t),
if a1 ≤ t < a2 ,





 f3 (t),
if a2 ≤ t < a3 ,
f (t) =
.
..
 .

.
.






 fn (t),
if an−1 ≤ t < an ,





 fn+1 (t), if t ≥ an ,
then,
f (t)
=
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 [u(t − an ) − u(t − ∞)]
k=1
=
f1 (t)[1 − u(t − a1 )] +
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 (t)u(t − an ).
k=2
Notice that when a = 0, we know that u(t − 0) = 1 for all t ≥ 0; when a = ∞, we define u(t − ∞) := 0 for all t ≥ 0.
Transform of integrals
Proposition 2 (Transform of integrals). There holds that
Z t
L[f (t)](s)
.
L
f (τ ) dτ (s) =
s
0
That is,
L−1
Z
Proof. Let g(t) =
Z t
F (s)
(t) =
L−1 [F (s)](τ ) dτ.
s
0
t
f (τ ) dτ , by the Fundamental Theorem of Calculus, we know that
0
g(0) = 0,
and g 0 (t) = f (t).
By the Laplace transform of derivative, we have
L[g 0 (t)] = sL[g(t)] − sg(0).
LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
3
Then we get
t
Z
L[f (t)] = sL
f (τ ) dτ .
0
Hence
t
Z
L
0
L[f (t)](s)
f (τ ) dτ (s) =
.
s
1
Example 1. Find L
.
s(s2 + 1)
By looking up the table, we have
−1
L−1
1
= sin(t).
s2 + 1
By the transform of integrals, we have
L
−1
1
2
s(s + 1)
"
−1
1
s2 +1
t
s
L
=
Z
L−1
=
0
#
1
(τ ) dτ
s2 + 1
t
Z
=
sin(τ ) dτ
0
1 − cos(t).
=
So
L−1
Example 2. Solve t2 =
Z
1
(t) = 1 − cos(t) .
s(s2 + 1)
t
eτ x(τ ) dτ .
0
2
Z
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of t =
t
eτ x(τ ) dτ , we have
0
L[t2 ] = L
Z
t
eτ x(τ ) dτ .
0
By the transform of integrals, then
Z
L
t
τ
e x(τ ) dτ
=
0
=
=
L [et x(t)]
s
L[x(t)](s − 1)
s
X(s − 1)
.
s
By the First Shifting Property
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MINGFENG ZHAO
By looking up the table, we have L[t2 ] =
2
, then
s3
X(s − 1)
2
=
.
3
s
s
Then X(s − 1) =
2
, that is,
s2
X(s) =
2
.
(s + 1)2
By the First Shifting Property, we have
x(t) = L−1 [X(s)] = 2e−t t.
2
Z
Therefore, the solution to t =
t
eτ x(τ ) dτ is:
0
x(t) = 2e−t t .
2
Z
Remark 3. Another way to solve the integral equation t =
t
eτ x(τ ) dτ , taking the derivative on the both sides, we
0
get 2t = et x(t), then we have
x(t) = 2te−t .
Transform of Convolution
Definition 2. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z
(f ∗ g)(t) =
t
f (τ )g(t − τ ) dτ.
0
Remark 4. It’s easy to see that the followings are true:
f ∗ g = g ∗ f,
(cf ) ∗ g = f ∗ (cg) = c(f ∗ g),
and
(f ∗ g) ∗ h = f ∗ (g ∗ h).
Proposition 3. There holds that
L[(f ∗ g)(t)] = L[f (t)](s) · L[g(t)](s).
That is,
L−1 [F (s) · G(s)] = L−1 [F (s)] ∗ L−1 [G(s)](t).
LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
5
Proof. In fact, we have
∞
Z
L[f ∗ g]
(f ∗ g)(t)e−st dt
=
0
∞
Z
e−st
=
∞
Z
∞
e−st f (τ )g(t − τ ) dtdτ
τ
0
∞
=
∞
Z
g(t − τ )e
f (τ )
∞
∞
Z
dt dτ
0
g(t0 )e−s(t +τ ) dt0
f (τ )
=
dτ
Let t0 = t − τ
0
0
∞
Z
−st
τ
0
Z
dt
0
=
Z
f (τ )g(t − τ ) dτ
0
Z
t
Z
=
∞
Z
−sτ
0
f (τ )e
g(t )e
0
−st0
0
dt
dτ
0
Z
∞
f (τ )e−sτ dτ
= L[g(t)] ·
0
= L[f (t)] · L[g(t)].
Remark 5. The Convolution Property implies Integral Property: let g(t) = 1, then
Z t
(f ∗ g)(t) =
f (τ ) dτ.
0
By the transform of convolution, we have
Z t
L[f (t)](s)
L
f (τ ) (s) = L[f (t)](s) · L[1](s) =
.
s
0
Example 3. 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
t
τ eτ dτ
e dτ −
t
0
=
Z
0
t
t(et − 1) − τ eτ |0 +
Z
t
0
=
t(et − 1) − tet + et − 1
=
et − t − 1.
eτ dτ
Use the integration by parts
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MINGFENG ZHAO
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
1
,
s−1
L[f (t)] = L[et ] =
and L[g(t)] = L[t] =
1
.
s2
By the transform of convolution, we have
L[f ∗ g](s) = L[f (t)] · L[g(t)] =
1
.
s2 (s − 1)
1
Example 4. Find L
.
s2 (s + 1)
By looking up the table, we have
−1
−1
L
1
= t,
s2
−1
and L
1
= e−t .
s+1
By the transform of convolution, then
L−1
1
s2 (s + 1)
1
1
·
s2 s + 1
1
1
−1
−1
= L
∗L
s2
s+1
Z t
=
τ e−(t−τ ) dτ
= L−1
0
−t
= e +t − 1.
Therefore, we have
L−1
1
= e−t +t − 1 .
s2 (s + 1)
Dirac delta and impulse response
Definition 3. For any continuous function f (t) on (−∞, ∞), we have
Z
∞
δ(t)f (t) dt = f (0).
−∞
LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
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Proposition 4. 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
e−s(t +a) δ(t0 ) dt0
Let t0 = t − a
−a
= e
−as
,
Since a > 0, by the definition of δ.
Problems you can do:
Lebl’s Book [2]: All exercises on Page 267, Page 268, Page 274 and Page 275.
Braun’s Book [1]: All exercises on Page 250, Page 251, Page 256 and Page 257. Read all materials in Section
2.12 and Section 2.13.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
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