Notes of the course: Mathematical Analysis II
preliminary version
Paolo Tilli
Dipartimento di Matematica
Politecnico di Torino
email: [email protected]
March 27, 2008
Chapter 5
Laplace Trasform
5.1
Introduction
Let x(t) be a piecewise continuous function, defined (at least) on the positive half
line [0, +∞). Given a real number s, consider the improper integral
Z ∞
(5.1)
x(t)e−st dt,
0
which can be convergent or non convergent, according to the value of s. The set
D of those number s such that the integral is convergent, represents the domain of
definition of a new function X(s), called the Laplace transform of x(t), defined by
the relation
Z ∞
x(t)e−st dt, s ∈ D.
X(s) =
0
Clearly, the domain D of the Laplace transform X(s) will depend on the original
function x(t).
! Example 5.1.1 If x(t) = et , the integral
Z ∞
Z
−st
X(s) =
x(t)e dt =
0
∞
e(1−s)t dt
0
is convergent only s > 1, hence the domain D of the Laplace transform is the open
half line (1, ∞). If, on the other hand, x(t) = e2t /(1 + t2 ), the integral
Z
X(s) =
∞
−st
x(t)e
Z
dt =
0
0
∞
e(2−s)t
dt
1 + t2
is convergent if and only if s ≥ 2, hence D is the closed half line [2, ∞).
2
Again, if x(t) = e−t , the integral
Z ∞
Z ∞
2
−st
X(s) =
x(t)e dt =
e−t −st dt
0
0
105
2
is convergent for every s, hence we have D = R. Finally, if x(t) = et , the integral
Z ∞
Z ∞
2
−st
x(t)e dt =
et −st dt
X(s) =
0
0
does not converge for any value of s, and the domain D of the Laplace transform
reduces to the empty set in this case. In the previous examples the domain D of the Laplace transfrom is always a half
line of the form (s0 , ∞) (i.e. an open half line) or of the form [s0 , ∞) (i.e. a closed
half line), considering the whole real line and the empty set as degenerate half lines.
It turns out that this is a general fact: in other words, if the integral
Z ∞
(5.2)
x(t)e−s0 t dt
0
is convergent for some s, then it can be proved that also the integral
Z ∞
x(t)e−st dt
0
converges for every values of s such that s > s0 , and hence the domain D of the
laplace transform is always a half line. Let us sum up what we have said in the
following definition.
Definition 5.1.2 Given a function x(t), defined at least for t ≥ 0 and piecewise
continuous, we denote by D the (possibly degenerate) half line defined as
Z ∞
x(t)e−st dt is convergent.
s ∈ D ⇐⇒ the integral
0
We call Laplace transform of the function x(t), and we denote it by one of the
symbols L [x(t)] or X, the new function
Z ∞
def
(5.3)
L [x(t)] (s) = X(s) =
x(t)e−st dt, s ∈ D
0
which is defined over D. When D is not empty, the function x(t) is said to be
Laplace-transformable.
The word transform stresses the fact that one constructs, starting from a given
function x(t), a new function X(s) –the transform of x(t)– according to the diagram
transform
t 7→ x(t), t ≥ 0 −−−−−−→ s 7→ X(s), s ∈ D
The Laplace transform has numerous applications, for instance in the solution of
linear differential equations and systems. The crucial point is the possibility of
reconstructing the original function x(t), from its Laplace transform X(s). This sort
of inverse operation works according to the diagram
inverse transform
t 7→ x(t), t ≥ 0 ←−−−−−−−−−− s 7→ X(s), s ∈ D
106
If we interpret x(t) as some time-dependent signal which is analyzed for positive
times, we see from equation (5.3) that the variable s has the dimensions of a frequence, since the product −st in the exponent must of course be a pure number.
Therefore, we can think of the Laplace transform as a tool which enables one to
represent a signal in the space of frequences.
Let us end this introduction wuith the direct computation of some Laplace transforms. These elementary examples, combined with the properties of the transform
which we will see in the next sections, will finally allow us to compute the Laplace
transform of quite complicated functions as well.
! Example 5.1.3 (step function) Consider teh so called step function, defined by
0 if t < 0,
(5.4)
U (t) =
1 if t ≥ 0.
This function plays a central role in signal processing, as it represents a unitary
signal, which suddenly arises at time zero and persists for all times.
Using the definition 5.1.2, we can compute the Laplace transform. We have
Z ∞
Z ∞
−st
e−st dt
U (t)e dt =
0
0
and the integral is convergent if and only if s > 0: in this case, the value of the
integreal is 1/s, as one can easily check. Then the domain of definition of the
transform L [U (t)] is the open half line (0, ∞), and we have
(5.5)
1
L [U (t)] (s) = ,
s
s > 0.
! Example 5.1.4 (exponential function) Let us compute the Laplace transform
of the exponential function x(t) = eat , where a is a real parameter. We have
Z ∞
Z ∞
1
at −st
e e dt =
e(a−s)t dt =
⇐⇒ s > a,
s−a
0
0
otherwise the integral is not convergent. Hence the domain of the transform is the
open half line (a, ∞) and
(5.6)
L eat (s) =
1
,
s−a
s > a.
! Example 5.1.5 (trigonometric functions) Let us compute the Laplace transform of x(t) = sin t. The integral
Z ∞
(sin t)e−st dt
0
107
converges if and only if s > 0, hence the domain of the Laplace transform of sin t is
the open half line (0, ∞). To evaluate the integral, we first compute the primitive
Z
e−st (−s sin t − cos t)
.
(sin t)e−st dt =
s2 + 1
Then, for s > 0 we find
l
Z ∞
Z l
e−st (−s sin t − cos t) −st
−st
(sin t)e dt = lim
(sin t)e dt = lim
l→∞ 0
l→∞
s2 + 1
0
0
−sl
1
e (−s sin l − cos l) e0 (−s sin 0 − cos 0)
−
= 2
= lim
2
2
l→∞
s +1
s +1
s +1
and the transform is given by
(5.7)
L [sin t] (s) =
s2
1
,
+1
s > 0.
s2
s
,
+1
s > 0.
In a similar way, one can check that
(5.8)
L [cos t] (s) =
! Example 5.1.6 Let us compute the Laplace transform of the function x(t) = t.
The integral
Z ∞
te−st dt
0
converges if and only if s > 0. Moreover, integration by parts reveals that the
function e−st (−st − 1)/s2 is a primitive of te−st , hence for s > 0 we have
l
Z ∞
Z l
e−st (−st − 1) −st
−st
te dt = lim
te dt = lim
l→∞ 0
l→∞
s2
0
0
−sl
e (−sl − 1) e0 (0 − 1)
1
= lim
−
= 2.
2
2
l→∞
s
s
s
Then, the Laplace transform is given by
(5.9)
L [t] (s) =
1
,
s2
s > 0.
Exactly determining the domain of a Laplace transform is not always an easy task.
A useful criterion to check that a given function x(t) is transformable consists in
studying its order of growth.
We say that a function x(t) has exponential growth if we can find two constants
M and α such that
|x(t)| ≤ αeM t ∀t ≥ 0.
108
Moreover, the infimum of those numbers M such that we have such an inequality
is called the order of growth of the function x(t). One can easily check that, if a
function has order of growth s0 , then the funcion is Laplace-transformable, at least
in the open half line (s0 , ∞) (this condition is in general only sufficient, and not
necessary, for the existence of the Laplace transform).
Example 5.1.7 If x(t) is a bounded function, i.e. there exists a constant α such
that |x(t)| ≤ α, then x(t) has exponential growth, and its order of growth (as one
can easily check) is zero. Therefore, any bounded and piecewise continuous function
is Laplace-transformable, at least in the positive half line (0, ∞). ! Example 5.1.8 (order of hyperbolic functions) The hyperbolic functions
def
sinh t =
et − e−t
,
2
def
cosh t =
et + e−t
2
both have exponential growth and order of growth equal to 1. Indeed, we have for
t>0
1
1
| sinh t| = et − e−t ≤ et , t ≥ 0
2
2
and
1
| cosh t| = et + e−t ≤ et , t ≥ 0
2
t
1·t
and the number 1 in e = e cannot be replaced by any smaller number.
The hyperbolic functions sinh t and cosh t are therefore Laplace-transformable in
the half line (1, ∞):
Z ∞
Z ∞
−st
(cosh t)e−st dt, s > 1.
(sinh t)e dt, L [cosh t] (s) =
L [sinh t] (s) =
0
0
The actual computation of the corresponding transforms will be done in the next
section, using the property of linearity of the Laplace transform. Exercises
5.1.1 Determine, using the definition, the Laplace transform of the following functions, indicating
for each of them the domain of the corresponding transform:
sin 2t,
t + 1,
2e4t
3 cos 4t,
U (t − 3),
t2 .
5.1.2 Establish if the following functions have exponential growth and, in this case, find their order
of growth:
2
e5t , 2e−10t + 4, t2 + t, (1 + t3 )e3t , et +2t , et sinh 2t, t sin t
109
5.2
Main properties of the Laplace transform
The Laplace transform has several useful properties, which enable one to considerably
reduce the amount of computation needed to compute the Laplace transform of
a given function, if one should work directly with the definition (5.3). The first
property is the lienarity of the Laplace transform.
Linearity. If x1 (t) and x2 (t) are two functions with Laplace transforms X1 (s) and
X2 (s) in the respective domains D1 and D2 , then any linear combination ax1 (t) +
bx2 (t) is Laplace-transformable, in the domain given by the intersection D1 ∩ D2 ,
and one has
(5.10)
L [ax1 (t) + bx2 (t)] (s) = aX1 (s) + bX2 (s),
s ∈ D 1 ∩ D2 .
Example 5.2.1 Combining (5.6) and (5.7), one has
L 2e3t + 5 sin t (s) =
2
5
+ 2
,
s−3 s +1
s>3
(note that the transformability domain is the intersection of the two domains (3, ∞)
and (0, ∞)). ! Example 5.2.2 (hyperbolic functions) Let us determine the Laplace transform
of the syperbolic functions sinh at and cosh at, where a is a real parameter. We have
sinh at =
1
1
eat − e−at
= eat − e−at ,
2
2
2
hence sinh at is a linear combination of two exponential functions, for which we have
already computed the Laplace transform. In particular, using (5.6) twice (first with
a, then with −a) we find
L eat (s) =
1
,
s−a
s > a,
and
1
, s > −a.
s+a
The two transformability domains are the two half lines (a, ∞) and (−a, ∞), whose
intersection is the half line (|a|, ∞), indipendently of the sign of a. Using the property
of linearity (5.10), we finally obtain
1
1
1
a
(5.11)
L [sinh at] (s) =
, s > |a|.
−
= 2
2 s−a s+a
s − a2
L e−at (s) =
In a similar way, one obtains that
1
1
1
s
+
= 2
,
(5.12)
L [cosh at] (s) =
2 s−a s+a
s − a2
110
s > |a|.
Now we examine some further properties of the Laplace transform, with relative
proofs and examples. Even though these are not the optimal assumptions, for simplicity we will always assume that x(t) is a piecewise continuous function of [0, ∞)
with exponential growth of the kind
|x(t)| ≤ M es0 t
(5.13)
∀t ≥ 0
for some s0 . In this way, for s > s0 the Laplace transform
Z ∞
x(t)e−st dt, s > s0
X(s) =
0
is surely well defined.
Scaling property. If a > 0 is a real parameter, then
(5.14)
1
L [x(at)] (s) = X(s/a),
a
s > as0 .
Proof. It suffices to chang variable z = at in the definition of Laplace transform.
Indeed, if s > as0 , then we have
Z ∞
Z
1 ∞
def 1
def
−st
x(at)e dt =
x(z)e−sz/a dz = X(s/a).
L [x(at)] (s) =
a 0
a
0
! Example 5.2.3 (trigonometric functions) We want to compute the Laplace
transform of the function sin at, where a > 0 is a parameter. Letting x(t) = sin t,
from (5.7) we know that
1
, s > 0.
X(s) = 2
s +1
Using the scaling property (5.14), we obtain
(5.15)
L [sin at] =
1
1
a
·
= 2
,
2
a (s/a) + 1
s + a2
s > 0.
1
s/a
s
·
= 2
,
2
a (s/a) + 1
s + a2
s > 0.
Similarly, from (5.8) we find
(5.16)
L [cos at] =
Shift property. If a > 0 is a real parameter, then
(5.17)
L [x(t − a)U (t − a)] (s) = e−as X(s),
s > s0 .
Note that, recalling (5.4), we have
x(t − a)U (t − a) =
x(t − a) if t ≥ a,
0
if t < a.
111
Therefore, the graph of the function x(t−a)U (t−a) on [0, ∞) is obtained by shifting
the graph of x(t)U (t) to the right of a quantity equal to a (and the shifted function
will be zero on the interval [0, a)). In particular, if x(t)U (t) represents a signal
starting at time t = 0, the function x(t − a)U (t − a) represents a signal which is
delayed in time (i.e. a signal starting at time t = a). For this reason, (5.17) is also
known as the delay formula.
Proof. It is enough to notice that U (t − a) = 0 for t < a and U (t − a) = 1 for
t ≥ a, and then make the change of variable z = t − a in the integral:
Z ∞
def
L [x(t − a)U (t − a)] (s) =
x(t − a)U (t − a)e−st dt
Z0∞
Z ∞
−st
=
x(t − a)e dt =
x(z)e−sa−sz dz
a
0
Z ∞
def
= e−as
x(z)e−sz dz = e−as X(s), s > s0 .
0
! Example 5.2.4 (shifted step function) For fixed a ≥ 0, let us compute the
Laplace transform of the shifted step function, i.e.
0 if t < a,
U (t − a) =
1 if t ≥ a.
Note that U (t − a) = U (t − a)U (t − a), hence we can use the shift formula (5.17)
with x(t) = U (t), thus obtaining
L [U (t − a)] (s) = L [U (t − a)U (t − a)] (s) = e−as L [U (t)] (s).
On the other hand, the transform L [U (t)] (s) of the step function has already been
computed in (5.5), hence we obtain
(5.18)
L [U (t − a)] (s) =
e−as
,
s
s > 0.
! Example 5.2.5 Let us compute the Laplace transform of the function
0
if t < 1,
x(t) =
3t
2e if t ≥ 1.
To do this, we observe that
x(t) = 2e3t U (t − 1) = 2e3 e3(t−1) U (t − 1).
Using first the linearity property and then the shift property with a = 1, we obtain
L [x(t)] (s) = 2e3 L e3(t−1) U (t − 1) (s) = 2e3 e−s L e3t (s),
112
and it suffices to compute the transform of e3t . On the other hand, using (5.6) we
find
1
L e3t (s) =
, s>3
s−3
and hence
e−s
, s > 3.
L [x(t)] (s) = 2e3
s−3
Modulation. If a is a real parameter, then
(5.19)
L eat x(t) (s) = X(s − a),
s > s0 + a.
The meaning of (5.19) is the following: the multiplication of a function x(t) by
an exponential reflects (in the Laplace transform) into a shift in the domain of
frequences. For this reason, the modulation property is sometimes called the second
delay formula.
Proof. It is a simple verification, using the definition of Laplace transform:
Z ∞
Z ∞
at
def
def
at
−st
x(t)e−(s−a)t dt = X(s − a), s > s0 + a.
e x(t)e dt =
L e x(t) (s) =
0
0
! Example 5.2.6 Let us transform the function x(t) = e2t sin 3t. Using first the
modulation property (5.19) (with a = 2) and then formula (5.15) (with a = 3), we
obtain
3
3
=
, s > 2.
L e2t sin 3t (s) = L [sin 3t] (s − 2) =
2
2
(s − 2) + 3
(s − 2)2 + 9
Multiplication by t. There holds
(5.20)
L [tx(t)] (s) = −X 0 (s),
s > s0 .
Iterating (5.20), one obtains
dn
X(s), s > s0 , n positive integrer.
dsn
A full proof of property (5.20) is beyound the scope of these notes, since it requires
some advanced mathematics. However, it is easy to give an intuitive explanation of
(5.20), obtained interchanging the symbols of derivative and integral:
Z ∞
Z ∞
d
def d
0
−st
X (s) =
x(t)e dt =
x(t)e−st dt
ds
ds
0
Z ∞0
def
=−
x(t)te−st dt = −L [tx(t)] (s).
(5.21)
L [tn x(t)] (s) = (−1)n
0
To make this proof rigorous, one should properly justify the second equality, that is,
the fact that differentiation with respect to s of the integral in dt (where s apears
as a parameter), is equal to the integral of the derivative.
113
! Example 5.2.7 (powers of t) Given a positive integrer n, let us compute the
Laplace transform of the function tn . It is sufficient to use (5.21) with x(t) = U (t),
i.e. the step function whose transform has already been computed in (5.5):
n
1
n!
n
n
n d
=
L [t ] (s) = L [t U (t)] (s) = (−1)
, s > 0.
dsn s
sn+1
For instance, we have choosing n = 1, 2, 3
0
00
2
1
2
1
1
= 2, L t =
= 3,
L [t] = −
s
s
s
s
L t3 = −
000
6
1
= 4,
s
s
s > 0.
! Example 5.2.8 Let us compute the Laplace transform of t sin 3t. From (5.15) with
a = 3, we have
3
, s>0
L [sin 3t] (s) = 2
s +9
and hence using (5.20) we find
0
3
6s
L [t sin 3t] (s) = − 2
, s > 0.
= 2
s +9
(s + 9)2
In the same way as multiplication by t becomes (in the Laplace transform) a derivative and in a change of sign, division by t reflects into an integration: Division by
t. If the limit from the right
def
x(0+ ) = lim+
t→0
x(t)
t
exists and is finite, then
Z ∞
x(t)
L
(s) =
X(S) dS,
t
s
(5.22)
s > s0 .
! Example 5.2.9 Let us compute the Laplace transform of the function sin(t)/t. The
limit from the right
sin t
lim+
t→0
t
is finite, hence we can apply (5.22):
Z ∞
sin t
L
(s) =
L [sin t] (S) dS.
t
s
Recalling (5.7), we find
Z ∞
∞
sin t
1
L
(s) =
dS
=
arctan(S)
2+1
t
S
s
s
(5.23)
π
= − arctan s = arctan(1/s), s > 0.
2
114
Remark 5.2.10 If in (5.13) we have s0 < 0, then we can set s = 0 in (5.22), thus
obtaining
Z ∞
x(t)
X(s) ds.
L
(0) =
t
0
But recalling the definition of Laplace transform (5.3), we can rewrite this identity
in the following way:
Z
∞
(5.24)
0
x(t)
dt =
t
Z
∞
X(s) ds.
0
Equality (5.24) is still true when s0 = 0, but in this case one must assume that both
integrals are convergent. ! Example 5.2.11 To evaluate the integral
Z ∞
sin t
dt,
t
0
which we know is convergent, we can use (5.24). Indeed, letting x(t) = sin t (5.13)
is satisfied with s0 = 0, and we have X(s) = 1/(s2 + 1) for s > 0 thanks to (5.7).
Applying (5.24), we obtain
Z ∞
Z ∞
∞ π
1
sin t
dt =
ds = arctan s = .
2
t
s +1
2
0
0
0
Trasform of the derivative. If x(t) is continuous on [0, ∞) and of class C 1 in
(0, +∞), then
(5.25)
L [x0 (t)] (s) = sX(s) − x(0),
s > s0 .
Proof. Fix a number l > 0. Integrating by parts we find
l
Z l
Z l
Z l
0
−st
−st −st
−sl
x (t)e dt = x(t)e + s
x(t)e dt = x(l)e − x(0) + s
x(t)e−st dt.
0
0
0
0
Due to (5.13), we have
lim x(l)e−sl = 0,
l→∞
s > s0 ,
and hence by definition of Laplace transform
Z l
Z l
def
0
0
−st
−sl
−st
L [x (t)] (s) = lim
x (t)e dt = lim x(l)e − x(0) + s
x(t)e dt
l→∞ 0
l→∞
0
Z l
def
= −x(0) + s lim
x(t)e−st dt = −x(0) + sX(s), s > s0 ,
l→∞
0
that is (5.25). 115
Remark 5.2.12 Keeping all the other assumptions, if x(t) is continuous in (0, ∞)
and the limit from the right
def
x(0+ ) = lim+ x(t)
t→0
exists and is finite, then (5.25) is still valid in the form
L [x0 (t)] (s) = sX(s) − x(0+ ),
(5.26)
s > s0 .
Moreover, if one makes analogous assumptions on derivatives of higher order, (5.26)
can be iterated, to compute the Laplace transform of derivetives of higher order. For
instance, if also x0 (t) satisfies the same assumptions as x(t), then we have
0
L [x00 (t)] (s) = L (x0 (t)) (s) = sL [x0 (t)] (s) − x0 (0+ ) = s2 X(s) − sx(0+ ) − x0 (0+ )
and hence
(5.27)
L [x00 (t)] = s2 X(s) − sx(0+ ) − x0 (0+ ),
s > s0 .
We will use this relation later, for the explicit solution of some differential equations.
! Example 5.2.13 Let us compute again, in a different way, the Laplace transform
of the function x(t) = sin at, where now a is any real number. We have
x00 (t) = −a2 sin at,
x(0) = 0,
x0 (0) = a.
By linearity, using (5.27) we have
−a2 L [sin at] (s) = L [x00 (t)] (s) = s2 L [x(t)] (s) − a,
and hence solving with respect to L [x(t)] (s) we obtain
L [sin at] (s) = L [x(t)] (s) =
s2
a
,
+ a2
s > 0,
in accordance with (5.15) (which had been obtained only for a > 0). Trasform of the primitive. There holds
Z
(5.28)
t
L
x(r) dr (s) =
0
Proof. Fix l > 0 and let
X(s)
,
s
Z
g(t) =
s > max{s0 , 0}.
t
x(r) dr.
0
Since g 0 (t) = x(t) and g(0) = 0, integrating by parts we have
l
Z l
Z
Z
g(t)e−st 1 l 0
g(l)e−sl 1 l
−st
−st
g(t)e dt = −
g (t)e dt = −
+
x(t)e−st dt.
+s
s
l
s
0
0
0
0
116
Taking the limit as l → ∞ we obtain (5.28) because, thanks to (5.13), one has
(
Z l
Z l
−M/s0 if s0 < 0,
es0 t dt ≤
|g(l)| = x(t) dt ≤ M
M les0 t if s0 ≥ 0,
0
0
and hence if s > max{s0 , 0} one has
g(l)e−sl
= 0.
l→∞
l
lim
Note that, in (5.28), the assumption that s > max{s0 , 0} cannot be replaced by the
weaker condition that s > s0 . Indeed, consider x(t) = e−t , which satisfies (5.13)
with s0 = −1. Then we have g(t) = 1 − e−t , and the Laplace transform of g(t) has
a domain of definition given by (0, ∞), not by (s0 , ∞).
! Example 5.2.14 (integral sine) Let us compute tha Laplace transform of the
integral sine:
Z t
sin r
dr, t ≥ 0.
sI(t) =
r
0
Letting x(t) = (sin t)/t, (5.13) is true with s0 = 0; hence, using (5.28) and recalling
(5.23), we have
arctan(1/s)
, s > 0.
L [sI(t)] (s) =
s
periodic functions. If x(t) is periodic with period T on [0, ∞), then
RT
L [x(t)] (s) =
(5.29)
0
x(t)e−st dt
,
1 − esT
s > 0.
Proof. First note that a function which is periodic and piecewise continuous is
bounded, hence we can transform for s > 0. If n is a positive integer, with the
change of variable t = z + nT we find
Z nT +T
Z T
Z T
−st
−s(z+nT )
−snT
x(t)e dt =
x(z + nT )e
dz = e
x(z)e−sz dz,
nT
0
0
having used periodicity. Summing over N consecutive periods, we find
Z
NT
−st
x(t)e
0
dt =
N
−1 Z nT +T
X
n=0
−st
x(t)e
nT
Z
dt =
T
−sz
x(z)e
0
dz
N
−1
X
e−snT ,
n=0
and (5.29) follows letting N → ∞, and recalling the formula for the sum of a
geometric series of ratio e−sT . 117
! Example 5.2.15 Let us compute the Laplace transform of the square wave of period
T and amplitude A, defined on the interval [0, T ) as
A if 0 ≤ t < T /2,
x(t) =
−A if T /2 ≤ t < T .
To apply (5.29), we first compute
Z
T
−st
x(t)e
0
Z
T
2
−st
e
dt = A
Z
T
−st
dt−A
e
T
2
0
1 − e−sT /2
1 − 2e−sT /2 + e−sT
dt = A
=A
s
s
and then using (5.29)
2
1 − e−sT /2
1 − e−sT /2
=
A
,
L [x(t)] (s) = A
s (1 − e−sT )
s (1 + e−sT /2 )
s > 0.
We end this section with two properties which relate the limit values, at zero and at
infinity, of x(t) and its transform X(s).
Initial value. There holds
def
x(0+ ) = lim+ x(t) = lim sX(s).
(5.30)
s→∞
t→0
Proof. First note that the first limit exists and is finite, since teh function x9t) is
by assumption piecewise continuous on [0, ∞). With the change of variable z = st,
we have
Z ∞
Z ∞
def
−st
x(z/s)e−z dz.
x(t)e dt =
sX(s) = s
0
0
On the other hand, for every z > 0 we have
lim x(z/s) = lim+ x(t) = x(0+ ).
s→∞
t→0
Therefore, exchanging the limit with the integral, one has
Z ∞
Z ∞
−z
lim sX(s) = lim
x(z/s)e dz =
lim x(z/s)e−z dz
s→∞
s→∞
s→∞
Z ∞ 0
Z ∞0
=
x(0+ )e−z dz = x(0+ )
e−z dz = x(0+ )
0
0
and (5.30) follows. We point out that in the second equality one should justify the
possibility of exchanging the limit with the integral: a rigorous justification, however,
is beyound the scope of these notes and is omitted. Final value. If s0 ≤ 0, then
(5.31)
lim x(t) = lim+ sX(s)
t→∞
s→0
118
2
provided that the limits exist.
This property can be proved in a way very similar to the proof of the initial value
property; note, however, that here one must assume that the limits exist.
We sum up the main properties of the Laplace transform in Table 5.1. Keep in
mind, however, that some properties require some additional assumptions in addition
to (5.13): for the correct assumption, we refer to the detailed description of each
property which we have given above.
Linearity
L [ax1 (t) + bx2 (t)] (s) = aX1 (s) + bX2 (s)
Scaling
1
L [x(at)] (s) = X(s/a),
a
Shift
L [x(t − a)U (t − a)] (s) = e−as X(s),
Modulation
L [eat x(t)] (s) = X(s − a),
Multiplication by t
L [tx(t)] (s) = −X 0 (s), s > s0
Z ∞
x(t)
L
(s) =
X(S) dS, s > s0
t
s
Division by t
Trasform of the derivative
Trasform of the primitive
Initial value
Final value
s > as0
s > s0
s > s0 + a
L [x0 (t)] (s) = sX(s) − x(0+ ), s > s0
Z t
X(s)
x(r) dr (s) =
L
, s > max{s0 , 0}
s
0
lim x(t) = lim sX(s)
s→∞
t→0+
lim x(t) = lim+ sX(s)
t→∞
s→0
Table 5.1: Main properties of the Laplace transform.
5.3
Inverse Laplace transform
The Laplace transform would be of little use if one could not reconstruct the original
function (9t), knowing its Laplace transform X(s). The operation of passing from
X(s) to x(t) is called inverse Laplace transform: a detailed definition of the inverse
transform is beyond our scope, since it requires the theory of functions of one complex
variable. Here, we limit ourselves to give some examples and some techniques for
the computation of the inverse transform in some cases.
119
Definition 5.3.1 Let X(s) be a function defined at least of an open half line of the
kind (s0 , ∞). We say that x(t) is an inverse Laplace transform of X(s) and we write
x(t) = L−1 [X(s)] (t),
t≥0
if x(t) is a piecewise continuous function on [0, ∞) which is Laplace-transformable
and such that
X(s) = L [x(t)] (s), s > s0 .
We should first point out that the inverse transform of X(s), when it exists, is
never uniquely determined. Indeed, since the Laplace transform is defined by the
integral (5.3), if x(t) is an inverse trasnform of X(s), then any function obtained by
changing the values of x(t) at finitely many points is also an inverse transform of
X(s). However, this is not a severe limitation, since one can prove that, if s(t) is an
inverse transform, then the regularization of x(t) is uniquely determined. For this
reason, we will often speak of the inverse Laplace transform, instead of an inverse
Laplace transform.
Note that, in the definition of Laplace transform of x(t) (5.3), only the values of
x(t) with positive t are relevant: therefore, knowing the Laplace transform X(s), we
can have no information on the values of (t) for negative t. Therefore, in the sequel,
all the inverse transforms will be considered as defined only on the positive half line.
It is clear that the properties of the inverse Laplace transform can be obtained
from the corresponding properties of the Laplace transform, even though some care
is needed.
For further reference, we include the properties of the inverse transform in Tabella
5.2: note however that these properties should be used with some care, paying some
attention to the assumptions and to the domains of defitinion. On comparing the
two tables 5.1 and 5.2, one should observe that the property of multiplication by t of
the transform becomes the property of the inverse transform of a derivative, whereas
the division by t of the transform becomes the inverse transform of the primitive.
(and vice versa, for multiplication and division by s).
In the computation of Laplace transforms, Table 5.3 can be useful.
! Example 5.3.2 Let us compute the inverse Laplace transform of the function
X(s) =
3
.
s−4
Recalling (5.6) (see also Table 5.3), we have by linearity
1
3
−1
(t) = 3L
(t) = 3e4t .
L
s−4
s−4
! Example 5.3.3 Let us compute the inverse Laplace transform of the function
s2
1
.
−1
120
Linearity
L−1 [aX1 (s) + bX2 (s)] (t) = ax1 (t) + bx2 (t)
Scaling
1
L−1 [X(as)] (t) = x(t/a)
a
Shift
L−1 [e−as X(s)] (t) = x(t − a)U (t − a)
Modulation
L−1 [X(s − a)] (t) = eat x(t)
Inverse transform of the derivate
L−1 [X 0 (s)] (t) = −tx(t)
Z ∞
x(t)
−1
X(S) dS (t) =
L
t
s
Inverse transform of the primitive
Multiplication by s
Division by s
L−1 [sX(s) − x(0+ )] (t) = x0 (t)
Z t
−1 X(s)
x(r) dr
L
(t) =
s
0
Table 5.2: Main properties of the inverse Laplace transform.
Looking at Table (5.3), we immediately find
1
−1
L
(t) = sinh t.
s2 − 1
Alternatively, we can decompose the function into simple fractions as:
1
1
1
1
1
.
=
=
−
s2 − 1
(s + 1)(s − 1)
2 s−1 s+1
Then, by linearity of the inverse transform, we have
1
1
1
−1
−1 1
L
(t) = L
−
(t)
s2 − 1
2 s−1 s+1
1 −1
1
1 −1
1
= L
(t) − L
(t).
2
s−1
2
s+1
Since 1/(s − 1) is the transform of et whereas 1/(s + 1) is the transform of e−t , we
obtain
1
1
L−1 1/(s2 − 1) (t) = et − e−t = sinh t.
2
2
121
Function x(t)
Laplace transform X(s) Domain
U (t)
1
s
eαt
tn
(a ∈ R)
(n ∈ N)
sin(at) (a ∈ R)
cos(at) (a ∈ R)
sinh(at) (a ∈ R)
cosh(at) (a ∈ R)
eat sin(bt) (a, b ∈ R)
eat cos(bt) (a, b ∈ R)
sin t
t
s>0
1
s−a
n!
n+1
s
a
s 2 + a2
s
2
s + a2
a
2
s − a2
a
2
s − a2
b
(s − a)2 + b2
s−a
(s − a)2 + b2
arctan(1/s)
s>a
s>0
s>0
s>0
s > |a|
s > |a|
s>a
s>a
s>0
Table 5.3: Some functions and their Laplace transforms.
We end this dection with a rather long list of examples of computation of the inverse
transform, pointing out the properties of the inverse trasnform which are used. We
will implicitly use table 5.3, for what concerns the Laplace transform of elementary
functions.
! Example 5.3.4 (linearity) Let us compute the inverse Laplace transform of the
function
4
3s
5
− 2
+ 2
.
s − 2 s + 16 s + 4
122
By linearity, we obtain
4
3s
5
−1
−
+
(t)
L
s − 2 s2 + 16 s2 + 4
1
s
1
−1
−1
−1
= 4L
(t) − 3L
(t) + 5L
(t)
s−2
s2 + 16
s2 + 4
s
2
1
5 −1
−1
−1
(t) − 3L
(t) + L
(t)
= 4L
s−2
s2 + 42
2
s2 + 22
5
= 4e2t − 3 cos 4t + sin 2t.
2
! Example 5.3.5 (scaling) Let us compute the inverse Laplace transform of the
function
s
.
2
9s + 16
Since
s
s
−1
−1
L
(t) = L
(t) = cos 4t,
s2 + 16
s2 + 42
by the scaling property of the inverse transform we obtain
s
1 −1
3s
1
4
s
−1
−1
L
(t) = L
(t) = L
(t) = cos
t .
2
2
2
9s + 16
(3s) + 16
3
(3s) + 16
3
3
! Example 5.3.6 (shift) Let us compute the inverse Laplace transform of the function
e−πs/3
.
1 + s2
Since
1
−1
(t) = sin t,
L
s2 + 1
from the scaling property of the inverse transform we find
−πs/3 sin(t − π/3) se t ≥ π/3,
−1 e
L
(t) = U (t − π/3) sin(t − π/3) =
2
0
se t < π/3.
1+s
! Example 5.3.7 (modulation) Let us compute the inverse Laplace transform of
the function
1
.
2
s − 2s + 5
First, let us complete the square in the numerator:
s2
1
1
=
.
− 2s + 5
(s − 1)2 + 4
123
Since
1
1
(t)
=
sin 2t,
L
s2 + 4
2
by the modulation property we have
1
1
1
−1
−1
L
(t) = L
(t) = et sin 2t.
2
2
s − 2s + 5
(s − 1) + 4
2
−1
! Example 5.3.8 (inverse transform of the derivative) Let us compute the inverse Laplace transform of the function
s
.
2
(s + 1)2
Since
−1
L
1
(t) = sin t e
s2 + 1
we have
−1
d
ds
1
2
s +1
−2s
(t) = −t sin t
(s2 + 1)2
1
s
(t) = t sin t.
2
2
(s + 1)
2
L
=
−2s
,
+ 1)2
(s2
and hence by linearity
−1
L
! Example 5.3.9 (inverse transform of the drrivative) Let us compute the inverse Laplace transform of the function
1
,
(s − a)n+1
a ∈ R, ed n intero positivo.
If n = 0, we clearly have
−1
L
1
(t) = eat .
s−a
If n > 0, it is important to recognize in the function 1/(s − a)n+1 the n-th derivative
(up to a multiplicative constant) n of the function 1/(s − a). More precisely, we have
(−1)n dn
1
1
=
.
(s − a)n+1
n! dsn s − a
Hence, using n times the property of the inverse transform of the derivative, we
obtain
1
tn at
−1
L
(t)
=
e .
(s − a)n+1
n!
For instance, for n = 1 and n = 2 we find
1
1
t2 at
−1
at
−1
(5.32)
L
(t)
=
te
,
L
(t)
=
e .
(s − a)2
(s − a)3
2
124
! Example 5.3.10 (inverse transform of the integral) Let us compute the inverse
Laplace transform of the function
s+1
.
log
s
Observe that
log
s+1
s
∞
Z
=
s
1
1
−
S S+1
dS,
and we can compute the inverse transform of the integrand function, since
1
−1 1
L
−
(t) = U (t) − e−t .
s s+1
Therefore, by the property of the inverse transform of the integral, we obtain
s+1
1 − e−t
−1
L
log
.
(t) =
s
t
! Example 5.3.11 (multiplication by s) Let us compute the inverse Laplace transform of the function
s
.
(s − 1)2 + 1
Since, according to table 5.3, one has
1
−1
L
(t) = et sin t
(s − 1)2 + 1
and x(t) = et sin t satisfies x+ (0) = 0, we can use the multiplication by s, thus
obtaining
d t
s
−1
(t) =
e sin t = et (cos t + sin t).
L
2
(s − 1) + 1
dt
! Example 5.3.12 (division by s) Let us compute the inverse Laplace transform of
the function
1
.
2
s(s + 4)
Since
−1
L
1
1
(t) = sin 2t,
2
s +4
2
using the division by s we find
Z t
1
1
1 1
−1
L
(t) =
sin 2r dr = − cos 2t.
2
s(s + 4)
4 4
0 2
125
To compute the inverse transform of
1
,
+ 4)
s3 (s2
it suffices to apply two more times the division property:
Z t
1 1
1
t 1
−1
L
(t) =
− cos 2r dr = − sin 2t,
2
2
s (s + 4)
4 4
4 8
0
and hence
−1
L
Z t
1
r 1
t2
1
1
(t)
=
−
sin
2r
dr
=
+
cos 2t − .
3
2
s (s + 4)
4 8
8
16
16
0
! Example 5.3.13 (reduction to simple fractions) To compute the inverse transform of the rational function
s2
3s + 7
3s + 7
=
− 2s − 3
(s − 3)(s + 1)
we decompose it as a sum of simple fractions, that is, we determine numbers A and
B such that
A
B
3s + 7
=
+
.
(s − 3)(s + 1)
s−3 s+1
(5.33)
To find A and B, we can solve a linear system of two equations. Altgernatively
(which is faster), we can multiply (5.33) by s − 3 and then take the limit as s → 3,
thus obtaining
(s − 3)B
3s + 7
= A + lim
=A
lim
s→3
s→3 s + 1
s+1
and hence A = 4. To find B, we multiply (5.33) by s + 1 and then take the limit as
s → −1, thus finding
lim
s→−1
3s + 7
A(s + 1)
= lim s → 3
+B =B
s−3
s−3
and hence B = −1. Then we have
3s + 7
4
1
−1
−1
L
(t) = L
−
(t)
(s − 3)(2 + 1)
s−3 s+1
1
1
−1
−1
(t) − L
(t) = 4e3t − e−t .
= 4L
s−3
s+1
126
! Example 5.3.14 (simple fractions again) To compute the inverse transform of
the rational function
5s2 − 15s − 11
(s + 1)(s − 2)3
we decompose it into simple fractions, i.e. we find A, B, C and D such that
(5.34)
B
5s2 − 15s − 11
A
C
D
+
.
=
+
+
3
3
2
(s + 1)(s − 2)
s + 1 (s − 2)
(s − 2)
s−2
To find A, we proceed as in the previous example: multiply by s + 1 and take the
limit as s → −1, finding
5s2 − 15s − 11
1
A = lim
=− .
3
s→−1
(s − 2)
3
To find B, multiply by (s − 2)3 and then take the limit as s → 2, finding
5s2 − 15s − 11
= −7.
s→2
(s − 2)3
B = lim
Once we have found A and B, (5.34) becomes
5s2 − 15s − 11
−1/3
−7
C
D
=
+
+
+
.
3
3
2
(s + 1)(s − 2)
s + 1 (s − 2)
(s − 2)
s−2
To find C, we first take to the left the fraction −7/(s−2)3 and then we collect terms,
finding
(5.35)
5s2 − 8s − 4
−1/3
C
D
=
+
+
.
3
2
(s + 1)(s − 2)
s + 1 (s − 2)
s−2
Now we can multiply by (s − 2)2 and take the limit as s → 2, thus finding
5s2 − 8s − 4
10s − 8
= lim
=4
s→2 (s + 1)(s − 2)
s→2 (s − 2) + (s + 1)
C = lim
(the limit has been computed with the rule of De L’Hôpital). Then (5.35) becomes
5s2 − 8s − 4
−1/3
4
D
=
+
+
(s + 1)(s − 2)3
s + 1 (s − 2)2 s − 2
and to find D we take to the left the fraction 4/(s − 2)2 , thus obtaining
5s2 − 8s − 4 − 4(s + 1)(s − 2)
−1/3
D
=
+
.
(s + 1)(s − 2)3
s+1 s−2
Now, multiplying by s − 2 and taking the limit as s → 2, using two times the rule
of De L’Hôpital we find
5s2 − 8s − 4 − 4(s + 1)(s − 2)
s→2
(s + 1)(s − 2)2
10s − 8 − 4(s + 1) − 4(s − 2)
= lim
s→2
(s − 2)2 + 2(s + 1)(s − 2)
10 − 4 − 4
1
= lim
= .
s→2 2(s − 2) + 2(s + 1) + 2(s − 2)
3
D = lim
127
Therefore, recalling (5.32) we have
2
−1 5s − 15s − 11
(t)
L
(s + 1)(s − 2)3
−7
4
1/3
−1 −1/3
−1
−1
−1
=L
(t) + L
(t) + L
(t) + L
(t)
s+1
(s − 2)3
(s − 2)2
s−2
1
1
1
1 −1
1
1 −1
−1
−1
(t) − 7L
(t) + 4L
(t) + L
(t)
=− L
3
s+1
(s − 2)3
(s − 2)2
3
s−2
1
7
1
= − e−t − t2 e2t + 4te2t + e2t .
3
2
3
Exercises
5.3.1 Find the inverse Laplace transform of the following functions (the answer is within brackets):
a)
3
s+4
(3e−4t )
b)
1
2s − 5
( 12 e5t/2 )
c)
8s
s2 + 16
(8 cos 4t)
d)
1
s5
(t4 /24)
5.3.2 Check that the functions in the previous exercise are the Laplace transforms of the corresponding functions within brackets.
5.3.3 Find the inverse Laplace transform of the following functions (the answer is within brackets):
a)
s2
6
+4
(3 sin 2t)
b)
3s − 12
s2 + 8
√
√
√
(3 cos 2 2t − 3 2 sin 2 2t)
c)
2s − 5
s2 − 9
(2 cosh 3t −
5
3
sinh 3t)
5.3.4 Check that the functions in the previous exercise are the Laplace transforms of the corresponding functions within brackets.
5.3.5 Find the inverse Laplace transform of the following functions (the answer is within brackets):
a)
3s − 8
4s2 + 25
( 43 cos 5t/2 −
4
5
sin 5t/2)
b)
5s + 10
9s2 − 16
( 59 cosh 4t/3 +
5
6
sinh 4t/3)
5.3.6 Check that the functions in the previous exercise are the Laplace transforms of the corresponding functions within brackets.
5.3.7 Find the inverse Laplace transform of the following functions (the answer is within brackets):
a)
3s − 8 4s − 24
− 2
s2 + 4
s − 16
(3 cos 2t − 4 sin 2t − 4 cosh 4t + 6 sinh 4t)
b) −
7
3s + 2
(− 37 e−2t/3 )
5.3.8 Check that the functions in the previous exercise are the Laplace transforms of the corresponding functions within brackets.
5.3.9 Find the inverse Laplace transform of the following functions (the answer is within brackets):
a)
3s − 14
s2 − 4s + 8
(e2t (3 cos 2t − 4 sin 2t))
b)
8s + 20
s2 − 12s + 32
(−13e4t + 21e8t )
5.3.10 Check that the functions in the previous exercise are the Laplace transforms of the corresponding functions within brackets.
128