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Chapter 16
Basic Laplace Transform
Techniques
16
...
In contrast, the Laplace transform
transforms an ODE into an equivalent algebraic equation (without derivative)
...
16
...
1
Definition
Definition 16
...
The Laplace transform of f (t) is a function F (s) = L{f (t)}(s)
defined by
∞
F (s) =
f (t)e−st dt
...
1)
0
The Laplace transform is an integral transform
...
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
...
Values of the Laplace transform for complex values of s can however be
useful in practical applications
...
Note that the Laplace transform only uses values of f (t) for t ≥ 0
...
The Fourier transform and other integral transforms are used in numerous applications (stability of algorithms, medical imaging,
...
1
...
Basic Laplace Transform Techniques
Laplace transform of standard functions
Let’s compute the Laplace transform of a few simple functions
...
2 The Laplace transform of f (t) = 1 is
∞
L{1} =
1 · e−st dt
0
(assume s = 0)
=
e−st
−s
∞
(16
...
s
=
(assume s > 0)
(16
...
2c)
t=0
(16
...
2e)
Line (16
...
1) of the Laplace transform, substituting f (t) =
1
...
2b) the result of the integration is evaluated between t = 0 and
t = ∞ (understood in the limit sense)
...
2c)
...
2b)
assumes s = 0 because of a division by 0 (if s = 0 then (16
...
Some additional restriction must be placed on s for the limit appearing in (16
...
The result (16
...
1
...
3 The Laplace transform of f (t) = t (a given real number) is
∞
L{t} =
t · e−st dt
0
(assume s = 0)
= t·
∞
∞
−
0
−st
0
e−st
dt
−s
te−st
te
−
t→∞ −s
−s
1
= 0 − 0 + L{1}
s
1 1
=
s s
1
= 2
...
3a)
lim
(16
...
3c)
0
(16
...
3e)
(16
...
3a) uses the definition (16
...
Integration by parts yields (16
...
3c)
...
3d) note that the limit vanishes provided s > 0, and the integral was computed
in Example 16
...
The result (16
...
1
...
1
...
4 The Laplace transform of f (t) = eat (a given real number) is
∞
L{eat } =
0
=
∞
eat · e−st dt
(16
...
4b)
0
=
e(a−s)t
a−s
∞
(16
...
s−a
=
lim
−
e(a−s)t
a−s
(16
...
4e)
(16
...
4a) uses the definition (16
...
The property (1
...
4b), from which (16
...
4d) follow (assuming s = a)
...
The result (16
...
1
...
5 The Laplace transform of f (t) = sin(bt) is
∞
L{sin(bt)} =
sin(bt) · e−st dt
...
5a)
0
From (??) we obtain (with a → −s)
=
e−st
− b cos(bt) + (−s) sin(bt)
2 + b2
s
∞
(16
...
s + b2
=
lim
t→∞ s2
−
−b
s2 + b2
(16
...
5d)
(16
...
5c) exists as long as the exponential factor eat has a finite
limit, which is the case for s > 0
...
5e) utilizes Euler’s formula (1
...
Basic Laplace Transform Techniques
nizing sin(bt) as
eibt
...
4 we obtain, with a = ib,
L{sin(bt)} = L{ eibt }
= L{eibt }
1
=
s − ib
s + ib
=
(s − ib)(s + ib)
s + ib
=
s2 + b2
b
= 2
...
5e) is recorded as entry #4 in Table 16
...
The existence of the Laplace transform for certain values of s hinges upon the
convergence of the integral (16
...
e
...
2
Example 16
...
Thus the integral (16
...
e
...
0≤
1
1
t
then (assume s > 0)
∞
Example 16
...
Thus f (t) =
1
t
1
0
dt
= e−s [ln t]1 = e−s (0 − (−∞)) = ∞
0
t
does NOT have a Laplace transform
...
When a Laplace transform exists the limit at ∞ such as the ones appearing in (16
...
2a) or (16
...
16
...
16
...
Basic properties
16
...
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
...
(16
...
Example 16
...
1, entries #2 and #4 (with b = 1)
...
2
...
(16
...
7) is recorded in Table 16
...
Example 16
...
1, entries #1 and #2 (f (t) = t)
...
10 For f (t) = sin(bt) we have f (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
...
Using the linear property (16
...
e
...
1
...
11 For f (t) = eat we have f (t) = aeat and f (0) = 1, so that
L{aeat } = s L{eat } − 1
...
6) and Table 16
...
8)
66
Chapter 16
...
Note that we could have used (16
...
s−a
We include here a proof of the derivative formula (16
...
9)
0
= −f (0) + s L{f (t)}
...
9) is guaranteed to exist if |f (t)| ≤ Ceαt for all t ≥ 0 (exponential
order)
...
The derivative formula can be applied recursively to yield formulas for the
Laplace transform of higher order derivatives of f (t)
...
(16
...
1, entry #16
...
12 Take f (t) = sin(bt), so that f (0) = 0, f (t) = b cos(bt), f (0) = b
and f = −b2 sin(bt)
...
10) reduces to
L{−b2 sin(bt)} = s2 L{sin(bt)} − s · 0 − b,
(16
...
e
...
1, entries #4,
−b2
b
b
= s2 2
− b,
s2 + b2
s + b2
which is easily verified
...
11) could be used to determine L{sin(bt)}:
−b2 L{sin(bt)} = s2 L{sin(bt)} − s · 0 − b
⇒
L{sin(bt)} =
b
...
3
...
3
67
Solution of linear IPVs
Examples 16
...
12 show that the Laplace transform of certain functions can
easily be obtained by using the linearity property (16
...
7), (16
...
Here we formalize this idea by showing that the Laplace transform
of the solution of the second-order constant coefficients linear IVP
au + bu + cu = g(t),
u(0) = α,
u (0) = β
(if a = 0)
can be obtained this way
...
3
...
Collecting
terms yields
(as2 + bs + c)U (s) − aαs − (aβ + bα) = G(s),
i
...
,
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
Example 16
...
(16
...
Basic Laplace Transform Techniques
u
−
2u
=
↓L
=
L{u } −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
...
1
...
14 Find U (s) if u(t) is the solution of the IVP
u − 2u = eat ,
u(0) = 1,
(16
...
u
−
2u
L{u } −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
+
...
1, respectively
...
13)
...
15 Find U (s) if u(t) is the solution of the IVP
2u + 3u + u = 10 cos t,
u(0) = 1,
u (0) = −1
...
14)
16
...
Solution of linear IPVs
69
2u
+
3u
+
u
=
10 cos t
↓L
=
10 L{cos t}
2 L{u } +3 L{u } + L{u}
?
= s2 L{u} − su(0) − u (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
...
,
(2s2 + 3s + 1)U (s) − 2s − 1 =
or
U (s) =
10
,
+1
s2
10
,
+1
s2
10
2s + 1
+
...
15)
The second term in (16
...
1
...
15) cannot
be (as) easily written as the Laplace transform of a recognizable function in Table
16
...
16
...
2
Inverse transform
In order to determine the solution u(t) in Example (16
...
15)
...
e
...
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)}
...
1) for the Laplace transform:
L−1 {F (s)} =
1
2iπ
F (s)est ds
...
16)
70
Chapter 16
...
1):
exchange the roles of t and s, change the sign in the exponential
...
16))
...
s
The alternative to formula (16
...
Fortunately, many Laplace transform functions F (s) are rational functions (see Table
16
...
It is advisable at this point to review
how a PFD can be determined by consulting Section 3
...
Example 16
...
s2 + 3s + 2
1
1
1
=
−
(see Example 3
...
Inverse transform each term in the PFD of F (s) using Table 16
...
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
...
17 Determine f (t) = L−1
s+7
...
Factor denominator: cannot be done with real coefficients since
s2 + 2s + 10 = (s + 1)2 − 1 + 10 = (s + 1)2 + 9
...
+ 2s + 10
(s + 1)2 + 9
2
...
3
...
1: express the
numerator in the form s + 7 = (s + 1) + 2(3)
...
16
...
Solution of linear IPVs
71
Example 16
...
15 and consider
F (s) =
10
...
Factor denominator (real coefficients):
10
10
= 2
...
PFD setup:
(s2
10
as + b
c
d
= 2
+
+
...
17)
3
...
(16
...
s3 : 0 =
s2 :
0 =
s: 0 =
1 : 10 =
2a + c + 2d,
3a + 2b + c + d,
a + 3b + c + 2d,
b + c + d
...
We can set it up in matrix form
2
3
1
0
0
2
3
1
1
1
1
1
2
a
0
1 b 0
=
2 c 0
1
d
10
and use the row reduction method of Section ?? page ??:
(16
...
Basic Laplace Transform Techniques
r2 → r2 − 3r1
r3 → r3 − r1
−→
r3 → r3 − 3r2
r4 → r4 − r2
−→
r4 → r4 − 5 r3
4
−→
r1 → r1 − r4
r2 → r2 + r4
r3 → r3 − 16 r4
5
−→
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 → 1 r1
2
2 0
−→
1 10
1
0
−2 0
r2 → 1 r2
2
1
0
−→
1
10
0
1
r3 → 4 r3
−1 0
5
4
0
−→
2
10
1
0
−1 0 r4 → − 1 r4
2
16
0
−→
5
−2 10
0
5
r → r1 − 1 r3
2
0 −5 1
r2 → r2 + 1 r3
4
0 16
−→
1 −5
1
1 0 2
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
...
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
...
4
...
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+ 1
2
+
↓ L−1
e− 2 t
(#3)
−
−5
s+1
1
5
s+1
↓ L−1
sin t
(#4)
+
1
↓ L−1
e−t
(#3)
1
⇒ f (t) = L−1 {F (s)} = −3 cos t − sin t + 8e− 2 t − 5e−t
...
3
...
14) of Example 16
...
16
...
3
(16
...
1 includes the Laplace transform of standard functions
...
Typical mistakes and useful tips
• [mistake] t should NOT appear in the Laplace transform U (s) of u(t)
...
The confusion sometimes occur because the dt in the
∞
integral (16
...
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
...
2b)
...
• [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
...
s
A mistake is often made the other way around
...
(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
...
The correct function u(t) can be obtained
74
Chapter 16
...
From Example
...
• [mistake] The Laplace transform of f (t) is NOT F (s) (but is sF (s) − f (0))
...
Remember that the ICs are used early in the solution
process
...
Thus
check whether your solution satisfies the ICs! (at least the first one, u(0)
...
14) given by (16
...
• [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
...
18 the set-up (16
...
2
So you can still proceed if you get stuck in the system solution for a, b, c, d
...
This remark is sometimes useful to check whether a Laplace transform or inverse transform makes
sense
...
If s = iω, large s means large ω, i
...
, high frequency
...
L
1 −−→
−−
Exercises
75
Exercises
16
...
Use
a) either the expression of sinh(bt) and cosh(bt) in terms of ebt and e−bt
b) or Problem 1
...
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
...
3 determine the inverse Laplace transform of the given function
...
2
...
3
...
4
...
0
a) Verify that g (t) = f (t)
...
7) to g(t) and show that
1
L{g(t)} = F (s) ,
s
i
...
,
F (s)
1
s
t
L−1
−−→
−−
f (τ )dτ
...
21)
76
Chapter 16
...
1
...
1
Laplace transform of special functions
17
...
1
Piecewise functions
Let c ≥ 0 fixed and consider the function uc (t):
6
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
...
, a switch)
...
Example 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)
...
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
...
Whenever the function f (t) exhibits a discontinuity at some time t = c a
jump term uc (t)(right function − left function) is added
...
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
−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)
...
The reason why we express piecewise functions in terms of Heaviside functions
uc (t) is to determine their Laplace transform
...
We start with uc (t) itself using the only too available to do
17
...
Laplace transform of special functions
79
this, i
...
, the definition (16
...
s
This formula is recorded as entry #17 of Table 17
...
It can be used, together
with the linearity property #14 of Table 16
...
1
...
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
...
2
...
2
...
1
...
Advanced Laplace Transform Techniques
6
δc (t)
fh (t) as h → 0
fh (t)
1
h
0 c
c+h
0
1
h
fh (t) =
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
...
The “function” δc (t) must vanish everywhere except t = c, where it
must be infinite
...
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”)
...
The “function” δc (t) is, in some sense, the derivative of the piecewise function
uc (t) introduced in Section 17
...
1:
1
...
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!)
...
The cumulative area below δc (t) thus remains 0 for t ≤ c and jumps to 1
whenever t > c
...
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 = 0
and are obviously equal for t = c
...
0
• With f (t) = e−st we obtain
L{δc (t)} = e−sc
(recorded as entry #18 in Table 17
...
17
...
Transformation formulas
81
17
...
2
...
If c < 0 the effect of
the shift is to move the graph of f (t) to the right
...
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
...
26
1
0
−1
←−
f (t + 1)
1
2
3
4
−1
5 -
26
1
0
+1
−→
f (t)
1
2
3
4
5 -
−1
26
f (t − 1)
1
0
1
2
3
4
5 -
−1
2 = f (2) at t = 1
2 = f (2) at t = 3
26
Truncate t < 1 part ⇒
u1 (t)f (t − 1)
1
0
1
2
3
4
5 -
−1
The Laplace transform of the shifted and truncated function uc (t)f (t − c) is
∞
L{uc (t)f (t − c)} =
0
(uc (t) = 0 for t ≤ c, uc (t) = 1 for t > c)
c
(change of variable t = τ + c)
∞
=
∞
=
0
∞
=
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
0
∞
= e−sc
f (τ )e−sτ dτ
0
(recognize the Laplace transform (16
...
This formula is recorded as entry #19 of Table 17
...
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
...
4 The function g(t) from Example 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
...
, 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
Example 17
...
(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
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)
...
0
The quantity f (t) g(t) is the convolution product of f (t) and g(t)
...
1)
17
...
Transformation formulas
83
Example 17
...
0
(#1)
(compare with #2)
Example 17
...
This is true for n = 1 (#2 in Table 16
...
Assume it is true up to some n − 1
...
0
(#1)
(compare with #2)
Two remarks:
1
...
If g(t) = 1 then (17
...
21)
...
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
...
84
Chapter 17
...
Here it is easier to use instead
0
t
sin t 2e−t =
t
(sin τ )2e−(t−τ ) dτ = 2e−t
0
eτ sin τ dτ
0
to avoid expanding sin(t − τ )
...
we obtain
u(t) = sin t 2e−t
= 2e−t
1 τ
2 e (sin τ
− cos τ )
t
0
= e−t et (sin t − cos t) + 1
= sin t − cos t + e−t
...
3
More transformation formulas
17
...
1
Shift in s
17
...
2
Dilation in t, contraction in s
17
...
3
What is L−1 {F (s)}?
Example 17
...
1
...
2
...
s+1 s+1