Chapter 16
Basic Laplace Transform
Techniques
16.1
Laplace transform
The variation of the constant approach transforms a linear first-order ODE into
an equivalent but simpler integration problem. In contrast, the Laplace transform
transforms an ODE into an equivalent algebraic equation (without derivative). To
do this the integration process must certainly be part of the way the Laplace transform is defined (integration has to occur somewhere!).
16.1.1
Definition
Definition 16.1. The Laplace transform of f (t) is a function F (s) = L{f (t)}(s)
defined by
Z ∞
F (s) =
f (t)e−st dt.
(16.1)
0
The Laplace transform is an integral transform. Another notation for the
Laplace transform of a function f (t) is L{f (t)}(s), which makes it more clear as to
which function f (t) we are talking about, but is a bit more cumbersome. Sometimes
we simply write L{f (t)}, but keep in mind that the Laplace transform of f (t) is a
function of s only!
We can think of s as a real number which is fixed when evaluating the integral
(16.1). Values of the Laplace transform for complex values of s can however be
useful in practical applications. In particular, the Laplace transform becomes the
Fourier transform when s = iω with ω real (corresponding to a given frequency
ω
2π ).
Note that the Laplace transform only uses values of f (t) for t ≥ 0.
The Laplace transform is used extensively in Electrical Engineering and Control Theory. The Fourier transform and other integral transforms are used in numerous applications (stability of algorithms, medical imaging,...)
61
62
16.1.2
Chapter 16. Basic Laplace Transform Techniques
Laplace transform of standard functions
Let’s compute the Laplace transform of a few simple functions.
Example 16.2 The Laplace transform of f (t) = 1 is
Z
∞
1 · e−st dt
0
· −st ¸∞
e
=
−s 0
¯
¶
µ
e−st ¯¯
e−st
−
= lim
t→∞ −s
−s ¯t=0
1
=0−
−s
1
= .
s
(16.2a)
L{1} =
(assume s 6= 0)
(assume s > 0)
(16.2b)
(16.2c)
(16.2d)
(16.2e)
Line (16.2a) uses the definition (16.1) of the Laplace transform, substituting f (t) =
1. On Line (16.2b) the result of the integration is evaluated between t = 0 and
t = ∞ (understood in the limit sense). This yields (16.2c). Note that (16.2b)
assumes s 6= 0 because of a division by 0 (if s = 0 then (16.2a) diverges (equal ∞)).
Some additional restriction must be placed on s for the limit appearing in (16.2c) to
exist (be finite), namely s > 0, in which case the limit vanishes. The result (16.2e)
is recorded as entry #1 in Table 16.1.
Example 16.3 The Laplace transform of f (t) = t (a given real number) is
Z
∞
t · e−st dt
(16.3a)
0
·
¸∞ Z ∞ −st
e−st
e
= t·
−
dt
(16.3b)
−s 0
−s
0
¯
µ
¶
Z
te−st ¯¯
te−st
1 ∞ −st
= lim
−
e dt (16.3c)
+
t→∞ −s
−s ¯t=0 s 0
1
= 0 − 0 + L{1}
(16.3d)
s
1 1
(16.3e)
=
s s
1
(16.3f)
= 2.
s
L{t} =
(assume s 6= 0)
(assume s > 0)
Line (16.3a) uses the definition (16.1) of the Laplace transform, substituting f (t) =
t. Integration by parts yields (16.3b), which can be written as (16.3c). To obtain
(16.3d) note that the limit vanishes provided s > 0, and the integral was computed
in Example 16.2 as the Laplace transform of f (t) = 1. The result (16.3f) is recorded
as entry #2 in Table 16.1.
16.1. Laplace transform
63
Example 16.4 The Laplace transform of f (t) = eat (a given real number) is
Z
∞
L{eat } =
Z
0
=
·
∞
eat · e−st dt
(16.4a)
e(a−s)t dt
(16.4b)
0
¸∞
e(a−s)t
a−s 0
¯
µ
¶
e(a−s)t
e(a−s)t ¯¯
= lim
−
t→∞ a − s
a − s ¯t=0
1
=0−
a−s
1
=
.
s−a
=
(16.4c)
(16.4d)
(16.4e)
(16.4f)
Line (16.4a) uses the definition (16.1) of the Laplace transform, substituting f (t) =
eat . The property (1.1) of exponential functions yields (16.4b), from which (16.4c)
and (16.4d) follow (assuming s 6= a). The limit exists (and vanishes) provided s > a.
The result (16.4f) is recorded as entry #3 in Table 16.1.
Example 16.5 The Laplace transform of f (t) = sin(bt) is
Z
L{sin(bt)} =
∞
sin(bt) · e−st dt.
(16.5a)
0
From (??) we obtain (with a → −s)
·
¸
¢ ∞
e−st ¡
−
b
cos(bt)
+
(−s)
sin(bt)
s2 + b2
0
µ
¶
¢
−b
e−st ¡
= lim 2
−
b
cos(bt)
+
(−s)
sin(bt)
− 2
t→∞ s + b2
s + b2
−b
=0− 2
s + b2
b
= 2
.
s + b2
=
(16.5b)
(16.5c)
(16.5d)
(16.5e)
Note that the limit in (16.5c) exists as long as the exponential factor eat has a finite
limit, which is the case for s > 0.
An alternate way to obtain (16.5e) utilizes Euler’s formula (1.10) and recog-
64
Chapter 16. Basic Laplace Transform Techniques
nizing sin(bt) as =eibt . From Example 16.4 we obtain, with a = ib,
L{sin(bt)} = L{=eibt }
= =L{eibt }
1
==
s − ib
s + ib
==
(s − ib)(s + ib)
s + ib
== 2
s + b2
b
= 2
.
s + b2
The result (16.5e) is recorded as entry #4 in Table 16.1.
The existence of the Laplace transform for certain values of s hinges upon the
convergence of the integral (16.1), i.e., on the behavior of the integrand f (t)e−st as
t → 0 and t → ∞.
2
Example 16.6 If f (t) = et then
Z ∞
Z ∞
Z
2
2
et · e−st dt ≥
et · e−st dt ≥
0
s
∞
Z
est · e−st dt =
s
∞
dt = ∞
s
for any (fixed) value of s. Thus the integral (16.1) diverges for all values of s, i.e.,
2
f (t) = et does NOT have a Laplace transform.
Example 16.7 If f (t) = 1t then (assume s > 0)
Z ∞
Z ∞
1 −st
1 − e−s
0≤
· e dt ≤
·e−st dt =
t
s
1
1
converges but
Z 1
Z 1
Z 1
1 −s
dt
1 −st
−s
· e dt ≥
· e dt = e
= e−s [ln t]10 = e−s (0 − (−∞)) = ∞
t
t
0
0
0 t
diverges for any s > 0. Thus f (t) =
1
t
does NOT have a Laplace transform.
for L{f (t)} to exist the function f (t) must be well-behaved around t = 0 (for example finite and continuous at t = 0) and not grow faster than regular exponentials
eαt as t → ∞ (such functions are of exponential order). When a Laplace transform exists the limit at ∞ such as the ones appearing in (16.2a), (16.2a) or (16.2a)
typically vanishes provided suitable restrictions are placed on s.
16.2
Basic properties
Two basic properties make the Laplace transform well-suited for solving linear
ODEs.
16.2. Basic properties
16.2.1
65
Linearity
The Laplace transform of a linear combination of two (or more) functions is equal
to the linear combination of the respective Laplace transforms. Mathematically,
L{αf (t) + βg(t)} = αL{f (t))} + βL{g(t)} = αF (s) + βG(s).
(16.6)
This linear property easily follows from the linearity property for integrals.
Example 16.8
L{2t − 3 sin t} = 2L{t} − 3L{sin t} =
2
3
− 2
s2
s +1
using Table 16.1, entries #2 and #4 (with b = 1).
16.2.2
Derivative formula
The derivative formula relates the Laplace transform of the derivative f 0 (t) of a
function f (t) to the Laplace transform F (s) of the function f (t) itself, namely
L{f 0 (t)} = s L{f (t)} − f (0) = sF (s) − f (0).
(16.7)
The derivative formula (16.7) is recorded in Table 16.1 as entry #15.
Example 16.9
L{1} = s L{t} − 0
is easily verified using Table 16.1, entries #1 and #2 (f (t) = t).
Example 16.10 For f (t) = sin(bt) we have f 0 (t) = b cos(bt) and f (0) = 0, so that
L{b cos(bt)} = s L{sin(bt)} − 0 = s
b
s2 + b2
using Table 16.1, entry #4. Using the linear property (16.6) we then obtain
bL{cos(bt)} = s
i.e.,
L{cos(bt)} =
s2
b
,
+ b2
s
,
s2 + b2
which is recorded as entry #5 in Table 16.1.
Example 16.11 For f (t) = eat we have f 0 (t) = aeat and f (0) = 1, so that
L{aeat } = s L{eat } − 1.
Using the linear property (16.6) and Table 16.1, entries #2 we verify that
a
1
1
=s
−1
s−a
s−a
(16.8)
66
Chapter 16. Basic Laplace Transform Techniques
indeed.
Note that we could have used (16.8) to find L{eat }, had we not known its
expression from a previous calculation:
aL{eat } = L{aeat } = s L{eat } − 1
⇒
L{eat } =
1
.
s−a
We include here a proof of the derivative formula (16.7) because it is short
and simple:
Z ∞
(definition)
L{f 0 (t)} =
f 0 (t)e−st dt
0
Z ∞
d ¡ −st ¢
e
(integration by parts)
= [f (t)e−st ]∞
−
f (t)
dt
0
dt
0
Z
∞
³
´
= lim f (t)e−st − f (0) + s
f (t)e−st dt (16.9)
t→∞
0
= −f (0) + s L{f (t)}.
The limit in (16.9) is guaranteed to exist if |f (t)| ≤ Ceαt for all t ≥ 0 (exponential
order). The derivative formula then holds for s > α.
The derivative formula can be applied recursively to yield formulas for the
Laplace transform of higher order derivatives of f (t). In particular,
L{f 00 (t)} = L{(f 0 )0 (t)}
= s L{f 0 (t)} − f 0 (0)
= s (s L{f (t)} − f (0)) − f 0 (0)
= s2 L{f (t)} − sf (0) − f 0 (0)
= s2 F (s) − sf (0) − f 0 (0).
(16.10)
This formula is recorded in Table 16.1, entry #16.
Example 16.12 Take f (t) = sin(bt), so that f (0) = 0, f 0 (t) = b cos(bt), f 0 (0) = b
and f 00 = −b2 sin(bt). Then the formula (16.10) reduces to
L{−b2 sin(bt)} = s2 L{sin(bt)} − s · 0 − b,
(16.11)
i.e., by linearity and Table 16.1, entries #4,
−b2
b
b
= s2 2
− b,
s2 + b2
s + b2
which is easily verified.
Note that (16.11) could be used to determine L{sin(bt)}:
−b2 L{sin(bt)} = s2 L{sin(bt)} − s · 0 − b
⇒
L{sin(bt)} =
b
.
s2 + b 2
16.3. Solution of linear IPVs
16.3
67
Solution of linear IPVs
Examples 16.11 and 16.12 show that the Laplace transform of certain functions can
easily be obtained by using the linearity property (16.6) and the derivative formulas
(16.7), (16.10). Here we formalize this idea by showing that the Laplace transform
of the solution of the second-order constant coefficients linear IVP

 au00 + bu0 + cu = g(t),
u(0) = α,
 0
u (0) = β
(if a 6= 0)
can be obtained this way.
16.3.1
Obtaining the Laplace transform of the solution
Apply the Laplace transform to both sides of the ODE and use linearity (#14) and
the derivative formulas (#15, #16):
au00
+
bu0
+
cu
a L{u00 } +b L{u0 } +c L{u}
?
= s2 L{u} − su(0) − u0 (0)
= s2 U (s) − αs − β
=
↓L
=
?
= U (s)
g(t)
L{g(t)}
?
= G(s)
?
= sL{u} − u(0)
= sU (s) − α
Note how the ICs are used right away in the determination of U (s). Collecting
terms yields
(as2 + bs + c)U (s) − aαs − (aβ + bα) = G(s),
i.e.,
U (s) =
The quantity
G(s)
+ bs + c
↓
U (s) with trivial ICs
α = 0, β = 0
(start at rest)
as2
+
aαs + aβ + bα
.
as2 + bs + c
↓
U (s) with no forcing
g(t) = 0, G(s) = 0
(HODE)
1
is called a transfer function.
as2 + bs + c
Example 16.13 Find U (s) if u(t) is the solution of the IVP
½ 0
u = 2u,
u(0) = 1.
(16.12)
68
Chapter 16. Basic Laplace Transform Techniques
u0
−
2u
=
↓L
=
L{u0 } −2 L{u}
?
= sL{u} − u(0)
= sU (s) − 1
0
L{0}
?
= U (s)
?
=0
Thus
sU (s) − 1 − 2U (s) = 0
⇒
U (s) =
1
.
s−2
This expression is of course the Laplace transform of u(t) = e2t , according to entry
#3 of Table 16.1.
Example 16.14 Find U (s) if u(t) is the solution of the IVP
½ 0
u − 2u = eat ,
u(0) = 1,
(16.13)
where a is a (constant) parameter.
u0
−
2u
L{u0 } −2 L{u}
?
= sL{u} − u(0)
= sU (s) − 1
Thus
sU (s) − 1 − 2U (s) =
1
s−2
=
↓L
L{e2t }
=
?
= U (s)
⇒
e2t
?
1
=
(#3)
s−2
U (s) =
1
1
+
.
2
(s − 2)
s−2
This expression can be recognized as
L{te2t + e2t } = L{te2t } + L{e2t } =
1
1
+
(s − 2)2
s−2
according to entries #8 and #3 of Table 16.1, respectively. A direct verification
shows that u(t) = (t + 1)et is indeed the solution of the IVP (16.13).
Example 16.15 Find U (s) if u(t) is the solution of the IVP

 2u00 + 3u0 + u = 10 cos t,
u(0) = 1,
 0
u (0) = −1.
(16.14)
16.3. Solution of linear IPVs
69
2u00
+
3u0 +
u
=
10 cos t
↓L
=
10 L{cos t}
2 L{u00 } +3 L{u0 } + L{u}
?
= s2 L{u} − su(0) − u0 (0)
= s2 U (s) − s + 1
?
= U (s)
?
1
= 2
s +1
?
= sL{u} − u(0)
= sU (s) − 1
Thus
2(s2 U (s) − s + 1)) + 3(sU (s) − 1) + U (s) =
i.e.,
(2s2 + 3s + 1)U (s) − 2s − 1 =
or
U (s) =
10
,
+1
s2
10
,
+1
s2
10
2s + 1
+
.
(s2 + 1)(2s2 + 3s + 1) 2s2 + 3s + 1
(16.15)
The second term in (16.15) can be simplified if we factorize the denominator:
2s + 1
1
2s + 1
=
=
= L{e−t },
2s2 + 3s + 1
(2s + 1)(s + 1)
s+1
according to entry #3 of Table 16.1. Unfortunately the first term in (16.15) cannot
be (as) easily written as the Laplace transform of a recognizable function in Table
16.1.
16.3.2
Inverse transform
In order to determine the solution u(t) in Example (16.15) we need to find a function
of t whose Laplace transform is equal to the first term on the right-hand side of
(16.15).
The good news is that the Laplace transform is a one-to-one map, i.e., two
distinct functions will have two distinct Laplace transforms. This permits us to
define an inverse transform:
L
u(t) −→
←− U (s),
L−1
L{u(t)} = U (s)
⇔
u(t) = L−1 {U (s)}.
The formula for the inverse transform of a function F (s) looks deceptively similar
to the formula (16.1) for the Laplace transform:
Z
1
−1
L {F (s)} =
F (s)est ds.
(16.16)
2iπ
70
Chapter 16. Basic Laplace Transform Techniques
1
Except for the constant factor 2iπ
, this formula looks very “symmetric” from (16.1):
exchange the roles of t and s, change the sign in the exponential. The bad news is
that the integration must be carried out along a path in the complex plane, rather
than over a range of real values (integration “limits” were left out in (16.16)). As
a result, this formula is not practical in this course, even for functions as simple as
F (s) = 1s = L{1}.
The alternative to formula (16.16) is to use a table of transforms with sufficiently many formulas to provide a way to determine an inverse transform. Fortunately, many Laplace transform functions F (s) are rational functions (see Table
16.1) which can be written as linear combinations of simpler rational functions using the partial fraction decomposition (PFD). It is advisable at this point to review
how a PFD can be determined by consulting Section 3.4 page 23.
¾
½
1
.
Example 16.16 Determine f (t) = L−1
s2 + 3s + 2
1
1
1
=
−
(see Example 3.6)
+ 3s + 2
s+1 s+2
2. Inverse transform each term in the PFD of F (s) using Table 16.1:
1. PFD:
s2
F (s) =
1
s+1
1
s+2
−
↓ L−1
↓ L−1
e−t
(#3)
e−2t
(#3)
⇒ f (t) = L−1 {F (s)} = e−t − e−2t .
½
Example 16.17 Determine f (t) = L−1
¾
s+7
.
s2 + 2s + 10
1. Factor denominator: cannot be done with real coefficients since
s2 + 2s + 10 = (s + 1)2 − 1 + 10 = (s + 1)2 + 9.
Thus
s2
s+7
s+7
=
.
+ 2s + 10
(s + 1)2 + 9
2. PFD setup: already in PFD form.
3. Inverse transform each term in the PFD of F (s) using Table 16.1: express the
numerator in the form s + 7 = (s + 1) + 2(3). Then
F (s) =
s+1
(s + 1)2 + 9
+ 2
3
(s + 1)2 + 9
↓ L−1
↓ L−1
e−t cos(3t)
(#12)
e−t sin(3t)
(#11)
⇒ f (t) = e−t (cos(3t) + 2 sin(3t)).
16.3. Solution of linear IPVs
71
Example 16.18 We revisit Example 16.15 and consider
F (s) =
10
.
(s2 + 1)(2s2 + 3s + 1)
1. Factor denominator (real coefficients):
10
10
= 2
.
(s2 + 1)(2s2 + 3s + 1)
(s + 1)(2s + 1)(s + 1)
2. PFD setup:
(s2
10
as + b
c
d
= 2
+
+
.
+ 1)(2s + 1)(s + 1)
s + 1 2s + 1 s + 1
(16.17)
3. Determine unknown coefficients:
(i) Multiply by common denominator and simplify factors:
10 = (as + b)(2s + 1)(s + 1) + c(s2 + 1)(s + 1) + d(s2 + 1)(2s + 1). (16.18)
(ii) Expand and compare powers of s:
10 = (2as3 +(3a+2b)s2 +(a+3b)s+b)+c(s3 +s2 +s+1)+d(2s3 +s2 +2s+1).
s3 : 0 =
s2 :
0 =
s: 0 =
1 : 10 =
2a + c + 2d,
3a + 2b + c + d,
a + 3b + c + 2d,
b + c + d.
This system is not trivial. We can set it up in matrix form

2
3

1
0
0
2
3
1
1
1
1
1
   
2
a
0
b  0 
1
  =  
2  c   0 
1
d
10
and use the row reduction method of Section ?? page ??:
(16.19)
72
Chapter 16. Basic Laplace Transform Techniques

r2 → r2 − 3r1
r3 → r3 − r1
−→
r3 → r3 − 3r2
r4 → r4 − r2
−→
r4 → r4 − 54 r3
−→
r1 → r1 − r4
r2 → r2 + r4
r3 → r3 − 16
r
5 4
−→
2 0 1
 3 2 1

 1 3 1
 0 1 11
1 0
2
 0 2 −1
2

1
 0 3
2
1
 0 1
1
1 0
2
 0 1 −1
4

5
 0 0
4
5
4
 0 0
1
1 0
2
 0 1 −1
4

 0 0
1
0
0 0
1
1 0
2
 0 1 −1
4

 0 0
1
0 0
0

2 0
1 0 
r1 → 12 r1

2 0 
−→
1 10 
1
0
−2 0 
r2 → 12 r2


1
0
−→
1
10 
0
1
r3 → 45 r3
−1 0 


4
0
−→
2
10 
1
0
1
−1 0 
 r4 → − 2 r4
16

0
−→
5
−2 10 
0
5
r1 → r1 − 12 r3
0 −5 
 r2 → r2 + 1 r3
4
0 16 
−→
1 −5

1 0 21
 3 2 1

 1 3 1
 0 1 11
1 0
2
 0 1 −1
4

1
 0 3
2
1
 0 1
1
1 0
2
 0 1 −1
4

 0 0
1
5
4
 0 0
1
1 0
2
 0 1 −1
4

 0 0
1
0
0 0
1 0 0
 0 1 0

 0 0 1
0 0 0

0
0 

0 
10 
0
0 

0 
10 
0
0 

16
0 
5
2
10 
1
0
−1
0 

16
0 
5
1
−5
0 −3
0 −1 

0 16 
1 −5
1
1
2
1
1
−1
1
1
1
−1
Therefore
a = −3,
b = −1,
c = 16,
d = −5,
(ii’) The alternate approach of picking suitable values of s in (16.18) can right
away deliver some of the coefficients:
s = −1 : 10 = −2d ⇒
s = − 12 : 10 = 58 c
⇒
d = −5,
c = 16.
Obtaining a and b (which correspond to a quadratic denominator s2 +
1) would require substituting s = i (so that s2 + 1 = 0) and lead to
calculations with complex numbers. Alternately we can substitute other
interesting values:
s = 0 : 10 = b + c + d
⇒
s = 1 : 10 = 6(a + b) + 4c + 6d ⇒
b = −1,
a = −3.
4. Inverse transform each term in the PFD of F (s) using Table 16.1:
−3s − 1
s2 + 1
s
1
=−3 2
− 2
s +1
s +1
F (s) =
↓ L−1
cos t
(#5)
16
2s + 1
1
+ 8
s + 12
+
↓ L−1
↓ L−1
sin t
(#4)
e− 2 t
(#3)
1
+
−
−5
s+1
1
5
s+1
↓ L−1
e−t
(#3)
1
⇒ f (t) = L−1 {F (s)} = −3 cos t − sin t + 8e− 2 t − 5e−t .
16.3. Solution of linear IPVs
73
The solution of the IVP (16.14) of Example 16.15 thus becomes
1
u(t) = f (t) + e−t = −3 cos t − sin t + 8e− 2 t − 4e−t .
16.3.3
(16.20)
Application to linear systems
Short table of Laplace transforms
Table 16.1 includes the Laplace transform of standard functions. Conversely, the
table can be used to identify the inverse transform of functions of s listed in the
column labelled “U (s)”.
Typical mistakes and useful tips
• [mistake] t should NOT appear in the Laplace transform U (s) of u(t). U (s)
only depends on s. The confusion sometimes occurR because the dt in the
∞
integral (16.1) is missing. For example, when writing 0 e−st it becomes easy
to forget that the integrand is a function of t and integrate as a function
of s instead. Another opportunity for an error occurs when evaluating an
expression such as (16.2b). Students sometimes substitute 0 and ∞ for s
instead of t.
• [mistake] The Laplace transform of a product f (t)g(t) of two functions is
NOT the product F (s)G(s) of the Laplace transforms. For example
f (t) = g(t) = 1
⇒
F (s) = G(s) =
1
,
s
but
1
6= F (s)G(s).
s
A mistake is often made the other way around. Consider for example
f (t)g(t) = 1
⇒
L{f (t)g(t)}(s) =
U (s) =
2
.
(s + 1)(s2 + 1)
It is tempting, but INCORRECT, to do the following:
U (s) =
2
s+1
1
s2 + 1
↓ L−1
↓ L−1
2e−t
(#3)
sin t
(#4)
⇒ u(t) = 2e−t sin t.
2
2
If this were the case we would have U (s) = (s+1)
2 +1 = s2 +2s+2 according to
formula #11, which is not the case. The correct function u(t) can be obtained
74
Chapter 16. Basic Laplace Transform Techniques
here using a PFD of U (s). From Example ... we have
U (s) =
1
−s + 1
1
+ 2
=
s+1
s +1
s+1
+
↓ L−1
e−t
(#3)
−s
+1
+
s2
↓ L−1
− cos(t)
(#5)
s2
1
+1
↓ L−1
sin(t)
(#4)
⇒ u(t) = e−t − cos t + sin t.
• [mistake] The Laplace transform of f 0 (t) is NOT F 0 (s) (but is sF (s) − f (0)).
• [tip] The solution of an IVP obtained using the Laplace transform can be
checked using the IC(s). Remember that the ICs are used early in the solution
process. Many errors can occur in the calculations until the final answer. Thus
check whether your solution satisfies the ICs! (at least the first one, u(0). For
example the solution of the IVP (16.14) given by (16.17) satisfies
u(0) = −3 − 0 + 8 − 4 = 1,
u0 (0) = 0 − 1 − 4 + 4 = −1.
X
X
• [tip] The Laplace inversion process of a rational function F (s) can be carried
out without explicit knowledge of the coefficients of the PFD, provided it is
set-up correctly. In Example 16.18 the set-up (16.17) yields
as + b
s2 + 1
F (s) =
=a
s2
+
s
1
+b 2
+1
s +1
↓ L−1
cos t
(#5)
+
c
2s + 1
1
c
2
s + 12
↓ L−1
↓ L−1
sin t
(#4)
e− 2 t
(#3)
1
+
+
d
s+1
1
d
s+1
↓ L−1
e−t
(#3)
c 1
⇒ f (t) = L−1 {F (s)} = a cos t + b sin t + e− 2 t + de−t .
2
So you can still proceed if you get stuck in the system solution for a, b, c, d.
• [tip] Large values of s correspond to small values of t. This remark is sometimes useful to check whether a Laplace transform or inverse transform makes
sense. The diagram
1
L
eat −−−−→
s
−

a


y≈ for s large
≈ for t ' 0y
1
s
illustrates this. If s = iω, large s means large ω, i.e., high frequency. A
function which varies with a high frequency must be observed on a small time
scale.
L
1 −−−−→
Exercises
75
Exercises
16.1. Use
a) either the expression of sinh(bt) and cosh(bt) in terms of ebt and e−bt
b) or Problem 1.54
as well as Table 16.1 to show the following formulas:
2bs
− b2 )2
2
s + b2
(ii) L{t cosh(bt)} = 2
(s − b2 )2
(i) L{t sinh(bt)} =
b
(s − a)2 − b2
b
(iv) L{eat cosh(bt)} =
(s − a)2 − b2
(iii) L{eat sinh(bt)} =
(s2
In Problems 16.2-16.3 determine the inverse Laplace transform of the given function.
16.2.
4s
(s2 − 4)2
16.3.
Z
2
(s − 2)2 − 4
t
16.4. Consider the function g(t) =
f (τ )dτ .
0
a) Verify that g 0 (t) = f (t). What is g(0)?
b) Apply the derivative formula (16.7) to g(t) and show that
1
L{g(t)} = F (s) ,
s
i.e.,
F (s)
1
s
L−1
Z
−−−−→
t
f (τ )dτ.
0
(16.21)
76
Chapter 16. Basic Laplace Transform Techniques
Table 16.1. Short table of Laplace transforms
u(t)
U (s)
1
s
1
s2
1
s−a
b
s2 + b2
s
s2 + b2
a
2
s − a2
s
2
s − a2
1
(s − a)2
2bs
(s2 + b2 )2
s2 − b2
(s2 + b2 )2
b
(s − a)2 + b2
s−a
(s − a)2 + b2
2b3
(s2 + b2 )2
#1
1
#2
t
#3
eat
#4
sin(bt)
#5
cos(bt)
#6
sinh(at)
#7
cosh(at)
#8
teat
#9
t sin(bt)
#10
t cos(bt)
#11
eat sin(bt)
#12
eat cos(bt)
#13
sin(bt) − bt cos(bt)
#14
αf (t) + βg(t)
αF (s) + βG(s)
#15
f 0 (t)
sF (s) − f (0)
#16
f 00 (t)
s2 F (s) − sf (0) − f 0 (0)
Reference
Chapter 17
Advanced Laplace
Transform Techniques
17.1
Laplace transform of special functions
17.1.1
Piecewise functions
Let c ≥ 0 fixed and consider the function uc (t):
6
½
uc (t)
1
uc (t) =
0 if t ≤ c
1 if t > c
- t
0
c
The function uc (t) is piecewise constant and is a classical model for physical phenomena which can toggle between “off” (0) and “on” (1) states (e.g., a switch).
The function uc (t) is called a Heaviside function and can be used to express general
piecewise functions.
Example 17.1 The function
26






f (t)
1
3
0
1
2
4
-t
f (t) =
−1
77





0
1
2
−1
0
if
if
if
if
if
t≤1
1<t≤2
2<t≤3
3<t≤4
t>4
78
Chapter 17. Advanced Laplace Transform Techniques
is expressed as
f (t) = 0 + u1 (t)(1 − 0) + u2 (t)(2 − 1) + u3 (t)((−1) − 2) + u4 (t)(0 − (−1))
= u1 (t) + u2 (t) − 3u3 (t) + u4 (t).
For example f (2.5) = 1 + 1 − 3(0) + 0 = 2 since u1 (t) = 1 for t > 1, u2 (t) = 1 for
t > 2, u3 (t) = 0 for t ≤ 3 and u4 (t) = 0 for t ≤ 4.
In general the expression of f (t) starts with the first expression of the function (at
t = 0). Whenever the function f (t) exhibits a discontinuity at some time t = c a
jump term uc (t)(right function − left function) is added.
Example 17.2 The function
26
g(t)
1
3
0
1
2
4
-t

t



8 − 3t
g(t) =
t
 −4


0
if
if
if
if
t≤2
2<t≤3
3<t≤4
t>4
−1
is expressed as
g(t) = t + u2 (t)(8 − 3t − t) + u3 (t)(t − 4 − (8 − 3t)) + u4 (t)(0 − (t − 4))
= t − 4u2 (t)(t − 2) + 4u3 (t)(t − 3) − u4 (t)(t − 4).
For example, for 3 < t ≤ 4, g(t) = t − 4(t − 2) + 4(t − 3) = t − 4.
The reason why we express piecewise functions in terms of Heaviside functions
uc (t) is to determine their Laplace transform. One common mistake is to believe
that this transform can be computed directly from the expression of the function
on each subinterval. We start with uc (t) itself using the only too available to do
17.1. Laplace transform of special functions
79
this, i.e., the definition (16.1):
Z
∞
L{uc (t)} =
0
Z
(split integral)
(substitute expression on each interval)
(assume s > 0)
uc (t)e−st dt
Z ∞
uc (t)e−st dt +
uc (t)e−st dt
Z0 c
Z ∞c
−st
=
0 · e dt +
1 · e−st dt
0
c
Z ∞
=
e−st dt
c
· −st ¸∞
e
=
−s c
µ
¶ µ −cs ¶
e−st
e
= lim
−
t→∞ −s
−s
−cs
e
=
.
s
c
=
This formula is recorded as entry #17 of Table 17.1. It can be used, together
with the linearity property #14 of Table 16.1, to determine the Laplace transform
of linear combinations of Heaviside functions, such as the function from Example
17.1.
Example 17.3
L{u1 (t) + u2 (t) − 3u3 (t) + u4 (t)} = L{u1 (t)} + L{u2 (t)} − 3L{u3 (t)} + L{u4 (t)}
= e−s + e−2s − 3e−3s + e4 s .
An additional formula, discussed in Section 17.2.1 is needed to determine the
Laplace transform of a function such as the one considered in Example 17.2.
17.1.2
Point distributions
Let c ≥ 0 fixed, h > 0, and consider the function fh (t):
80
Chapter 17. Advanced Laplace Transform Techniques
6¾
δc (t)
¾
fh (t) as h → 0
(
fh (t)
1
h
0 c
c+h
fh (t) =
0
1
h
if t ≤ c or t > c + h
if c < t ≤ c + h
-t
We are interested in the limit of the function fh (t) as h → 0. Call it δc (t)
(“delta-c”). The “function” δc (t) must vanish everywhere except t = c, where it
must be infinite. It is called the Dirac distribution, and is an example of a point
distribution.
The Dirac distribution is often used as a mathematical model for physical
phenomena which occur in a very short amount of time (h → 0) but with a finite
transmission of energy, momentum, etc (finite positive “area” under the “curve”).
Think for example of two cars colliding (ouch!).
The “function” δc (t) is, in some sense, the derivative of the piecewise function
uc (t) introduced in Section 17.1.1:
1. u0c (t) = 0 = δc (t) for all t 6= c
2. Since the area below fh (t) is 1 independently of h > 0, the area below δc (t)
must also be equal to 1 (this is an instance where 0 × ∞ = 1!).
3. The cumulative area below δc (t) thus remains 0 for t ≤ c and jumps to 1
whenever t > c. This is uc (t)!
Formally, we write
Z
δh (t) =
u0c (t)
½
t
in the sense that
δc (τ )dτ =
0
0 if t ≤ c
= uc (t).
1 if t > c
Useful properties of δc (t) include:
• δc (t)f (t) = δc (t)f (c) for any t and function f , since both sides vanish for t 6= 0
and are obviously equal for t = c.
Z ∞
Z ∞
Z ∞
•
δc (t)f (t)dt =
δc (t)f (c)dt = f (c)
δc (t)dt = f (c).
0
0
0
• With f (t) = e−st we obtain
L{δc (t)} = e−sc
(recorded as entry #18 in Table 17.1).
17.2. Transformation formulas
81
17.2
Transformation formulas
17.2.1
Shift in t
A shift in t is a transformation t → t − c for some constant c. If c < 0 the effect of
the shift is to move the graph of f (t) to the right. If c > 0 it moves the graph to the
left. A truncated version of f (t) is then obtained by multiplying the shifted function
f (t − c) by the Heaviside function uc (t) (to avoid introducing f -values which were
not taken into account in the Laplace transform of the original function f (which
only uses t ≥ 0, see (16.1)).
26
1
0
−1
←−
f (t + 1)
1
2
3
4
−1
5 -
26
1
0
1
2
3
26
+1
−→
f (t)
4
5 -
−1
f (t − 1)
1
0
1
2
3
4
5 -
−1
2 = f (2) at t = 1
2 = f (2) at t = 3
26
u1 (t)f (t − 1)
1
Truncate t < 1 part ⇒
0
1
2
3
4
5 -
−1
The Laplace transform of the shifted and truncated function uc (t)f (t − c) is
Z
∞
L{uc (t)f (t − c)} =
Z0 ∞
(uc (t) = 0 for t ≤ c, uc (t) = 1 for t > c)
=
c
Z
(change of variable t = τ + c)
∞
=
0
Z
uc (t)f (t − c)e−st dt
f (t − c)e−st dt
f (τ )e−s(τ +c) dτ
∞
f (τ )e−sτ e−sc dτ
Z ∞
= e−sc
f (τ )e−sτ dτ
=
0
0
(recognize the Laplace transform (16.1) of f (t))
= e−sc F (s).
This formula is recorded as entry #19 of Table 17.1. It is useful both ways: to compute Laplace transforms of piecewise functions, as well as find the inverse transform
of functions of s with exponential factors e−cs .
Example 17.4 The function g(t) from Example 17.2 has Laplace transform
82
Chapter 17. Advanced Laplace Transform Techniques
L{g(t)}
=
=
L{t + 4u2 (t)(t − 2) + 4u3 (t)(t − 3) − u4 (t)(t − 4)}
L{t} − 4 L{u2 (t)(t − 2)} + 4 L{u3 (t)(t − 3)} − L{u4 (t)(t − 4)}
Identify f (t − 2) = t − 2
↓
f (t) = t
↓
1
F (s) = 2
s
=
i.e., G(s) =
1
s2
−
?
1
s2
f (t − 3) = t − 3
↓
f (t) = t
↓
1
F (s) = 2
s
−2s
4e
+
4e
−3s
?
1
s2
f (t − 4) = t − 4
↓
f (t) = t
↓
1
F (s) = 2
s
−
e
?
1
.
s2
−4s
1 − 4e−2s + 4e−3s − e−4s
.
s2
½
Example 17.5 Determine u(t) = L
−1
¾
e−s
.
(s + 1)(s + 2)
e−s
1
= e−s ·
(s + 1)(s + 2)
(s + 1)(s + 2)
PFD
?
1
1
F (s) =
−
s+1 s+2
L−1
?
f (t) = e−t − e−2t
17.2.2
L−1 - u1 (t)f (t − 1)
= u1 (t)(e−(t−1) − e−(t−2) )
What is L−1 {F (s)G(s)}?
As pointed out in the Typical mistakes page 73 the Laplace transform of a product of
two functions f (t) and g(t) is NOT the product of their Laplace transform F (s)G(s).
So what function is F (s)G(s) the Laplace transform of? It turns out
F (s)G(s)
L−1
−−−−→
Z
t
f (t) ? g(t) =
f (τ )g(t − τ )dτ.
0
The quantity f (t) ? g(t) is the convolution product of f (t) and g(t).
(17.1)
17.2. Transformation formulas
83
Example 17.6
U (s) =
1
=
s2
1
s
1
s
↓ L−1
↓ L−1
1
1
(#1)
Z
⇒ u(t) = 1 ? 1 =
t
1 · 1 dτ = t.
0
(#1)
(compare with #2)
Example 17.7 Suppose we want to show that
L{tn } =
n!
sn+1
,
n = 1, 2, . . .
This is true for n = 1 (#2 in Table 16.1). Assume it is true up to some n − 1. We
show it must be true for n using the convolution formula:
U (s) =
n!
=
sn+1
(n − 1)!
sn
↓ L−1
t
n−1
n
s
↓ L−1
n
(induction)
Z
n−1
⇒ u(t) = t
t
?n=
τ n−1 · n dτ = tn .
0
(#1)
(compare with #2)
Two remarks:
1. f (t) ? g(t) = g(t) ? f (t) (the order is not important)
2. If g(t) = 1 then (17.1) reduces to the formula (16.21).
The convolution product formula offers an alternative to using the PFD for determining the inverse transform of a product of rational functions of s:
PFD *
U (s) = . . .
L−1
j
U (s) = F (s)G(s)
L−1 j f (t) and g(t)
u(t)
*
f (t) ? g(t)
Example 17.8
U (s) =
2
s+1
s2
1
+1
↓ L−1
↓ L−1
2e−t
(#3)
sin t
(#4)
⇒ u(t) = 2e−t ? sin t.
84
Chapter 17. Advanced Laplace Transform Techniques
Z
where 2e−t ? sin t =
t
2e−τ sin(t − τ )dτ . Here it is easier to use instead
0
Z
t
sin t ? 2e−t =
Z
(sin τ )2e−(t−τ ) dτ = 2e−t
0
t
eτ sin τ dτ
0
to avoid expanding sin(t − τ ). Using ... we obtain
u(t) = sin t ? 2e−t
¡
¢¯t
= 2e−t 12 eτ (sin τ − cos τ ) ¯0
¡
¢
= e−t et (sin t − cos t) + 1
= sin t − cos t + e−t .
which can be compared to the result
17.3
More transformation formulas
17.3.1
Shift in s
17.3.2
Dilation in t, contraction in s
17.3.3
What is L−1 {F 0 (s)}?
Example 17.9 Consider the IVP
½
u0 + u = δ2 (t) cos(πt),
u(0) = 2.
1. First note that δ2 (t) cos(πt) = δ2 (t) cos(2π) = δ2 (t).
2. Determine the Laplace transform U (s) of the solution u(t) of the IVP:
u0
+
u
L{u0 } + L{u}
?
= sL{u} − u(0)
= sU (s) − 2
Thus
sU (s) − 2 + U (s) = e−2s
=
↓L
=
?
= U (s)
⇒
δ2 (t)
L{δ2 (t)}
?
e−2s (#18)
U (s) =
e−2s
2
+
.
s+1 s+1