Laplace quad

The Laplace Transform
Unit 2: Modeling in the Frequency Domain
Part 2: The Laplace Transform
Engineering 5821:
Control Systems I
Faculty of Engineering & Applied Science
Memorial University of Newfoundland
January 15, 2010
ENGI 5821 Unit 2, Part 2: The Laplace Transform
1 The Laplace Transform
The need for Laplace
The Inverse Laplace Transform
The Laplace Transform Pair Table
Laplace Transform Theorems (Part 1)
Partial-Fraction Expansion
In the previous section we saw that the responses to a series RL
circuit (which could be any other linear system) was composed of
constants, decaying exponentials, and sinusoids. The complex
frequency representation can handle all of these by utilizing the
following representation with different values for s:
x(t) = {Xest
}
However, if the input is not of this form it is more difficult to solve
for the system’s response. The Laplace transform allows almost
any realistic input to be described as an infinite sum of complex
exponentials. The output will also be of this form, and it becomes
much easier to describe how the system converts from input to
output.
The Laplace Transform is defined as,
L{f (t)} = F(s) =
∞
0−
f (t)e−st
dt
where f (t) is the time-domain signal that we wish to describe in
terms of an infinite sum of complex exponentials. In other words,
we want to transform f (t) into the frequency-domain.
The LT exists if the integral converges. The lower-limit is 0−
means that if there is a discontinuity at 0, we start before the
discontinuity. This allows impulse functions to be included.
Notice that we are only integrating for positive time (including 0).
We will be assuming that our time-domain functions are 0 for
negative time.

The Inverse Laplace Transform is deﬁned as,
L−1
{F(s)} =
1
2πj
σ+j∞
σ−j∞
F(s)est
ds = f (t)u(t)
The ILT can be viewed as reconstructing the original time-domain
signal f (t) from an inﬁnite sum of complex exponentials.
Notice the use of u(t) (the unit step function). If f (t) ever had a
negative-time part, it would not have been captured by the LT.
Therefore, the ILT can’t bring it back.
Also notice the limits of integration. The constant σ must be
chosen so that the path of integration stays clear of any singular
points of F(s).
Example
e.g. Find the LT of f (t) = Ae−atu(t).
The function does not contain an impulse, so integration can begin
at 0.
F(s) =
∞
0
f (t)e−st
dt
=
∞
0
Ae−at
e−st
dt
= A
∞
0
e−(s+a)t
dt
= −
A
s + a
e−(s+a)t
∞
t=0
=
A
s + a
Visualizing the LT
What does the frequency-domain signal F(s) look like for a
time-domain signal f (t)?
e.g. f (t) = u(t)e−2t
The result on the previous slide tells us that F(s) = 1
s+2. For every
point s = σ + jω on the complex plane we get a complex number
F(s). We will visualize these complex numbers in two ways...
1
s+2: Real and Imaginary Components

1
s+2: Magnitude and Phase
It is clear that |F(s)| → ∞ near s = −2. This is a pole of this
function.
We don’t often utilize the Laplace integral directly. The transforms
for a number of important functions appear below:
f (t) F(s)
1. δ(t) 1
2. u(t) 1
s
3. tu(t) 1
s2
4. tnu(t) n!
sn+1
5. e−atu(t) 1
s+a
6. (sin ωt)u(t) ω
s2+ω2
7. (cos ωt)u(t) s
s2+ω2
(Typo in book for item 4.)
s
s2+9: Magnitude and Phase
e.g. f (t) = u(t) cos 3t According to the table F(s) = s
s2+9
. There
are now two poles, corresponding to the roots of the denominator.
Theorem Name
1. L{f (t)} = F(s) =
∞
0− f (t)e−stdt Deﬁnition
2. L{kf (t)} = kF(s) Linearity theorem
3. L{f1(t) + f2(t)} = F1(s) + F2(s) Linearity theorem
4. L{e−atf (t)} = F(s + a) Frequency shift theorem
5. L{f (t − T)} = e−sT F(s) Time shift theorem
6. L{f (at)} = 1
a F(s
a ) Scaling theorem

Example
e.g. Find the inverse Laplace transform of F(s) = 1
(s+3)2 .
This F(s) does not appear directly in the LT table. However, we
see that it is a shifted version of 1
s2 which corresponds to tu(t)
(the ramp function). The frequency shift theorem is,
L{e−at
f (t)} = F(s + a)
Therefore, we can conclude that f (t)u(t) = e−3ttu(t).
Note: Unless otherwise specified, we will assume that the inputs
to the systems we are studying do not begin until t = 0. Hence,
we will leave off u(t) from our time-domain responses. Therefore,
for the example above the answer is f (t) = e−3tt.
Theorem Name
7. L{df
dt } = sF(s) − f (0−) Differentiation
8. L{d2f
dt2 } = s2F(s) − sf (0−) − f (0−) Differentiation
9. L{dnf
dtn } = snF(s) − n
k=1 sn−kf (k−1)(0−) Differentiation
10. L{
t
0− f (τ)dτ} = F(s)
s Integration theorem
11. f (∞) = lims→0 sF(s) Final value theorem
12. f (0+) = lims→∞ sF(s) Initial value theorem
See the textbook for special conditions on theorems 11 and 12.
(Typos in book for items 8 and 10.)
Example
e.g. What is the inverse Laplace transform of s? Assume initial
conditions are zero.
Use the first differentiation theorem (theorem 7):
L{
df
dt
} = sF(s) − f (0−)
with F(s) = 1. The ILT of 1 is δ(t). Therefore,
f (t) =
dδ(t)
dt
If F(s) is complicated it can be difficult to find the ILT. We will
often see rational functions which have the form,
F(s) =
N(s)
D(s)
Where N(s) and D(s) are polynomials. If the order of N(s) is less
than the order of D(s) then we can apply a partial-fraction
expansion. Consider the following function,
F(s) =
s3 + 3s2 + 2s − 5
s2 + 3s + 2
If we want a lower-order numerator we can actually carry out
polynomial division until the remainder has this property, or we can
find other ways to simplify. For the example above we can factor
out s from the first three terms of the numerator and get,
F(s) = s −
5
s2 + 3s + 2

F(s) = s −
5
s2 + 3s + 2
We can further factor the quadratic term,
F(s) = s −
5
(s + 1)(s + 2)
Functions like the second term can be expanded as follows,
5
(s + 1)(s + 2)
=
K1
s + 1
+
K2
s + 2
In general, there are three cases for partial fraction expansion. We
will use the current example to illustrate case 1...
Case 1: Real and Distinct Roots
5
(s + 1)(s + 2)
=
K1
s + 1
+
K2
s + 2
We must solve for K1 and K2. Multiply the equation by (s + 1),
5
s + 2
= K1 +
K2(s + 1)
s + 2
This should be valid for all s. Let s approach -1 to eliminate
everything else on the R.H.S.. We get K1 = 5. Apply the same
strategy to obtain K2 = −5. Returning to our full F(s) we have,
F(s) = s −
5
s + 1
+
5
s + 2
We can now apply known Laplace transforms and theorems,
f (t) =
dδ(t)
dt
− 5e−t
+ 5e−2t
Case 2: Real and Repeated Roots
e.g.
F(s) =
2
(s + 1)(s + 2)2
This can be expanded as follows,
2
(s + 1)(s + 2)2
=
K1
s + 1
+
K2
(s + 2)2
+
K3
s + 2
We can solve for K1 = 2 using the previously described method.
To find K2 we multiply by (s + 2)2 to isolate K2,
2
s + 1
=
2(s + 2)2
s + 1
+ K2 + (s + 2)K3
Letting s = −2 we get K2 = −2.
To find K3 we differentiate the equation above...
2
s + 1
=
2(s + 2)2
s + 1
+ −2 + (s + 2)K3
Differentiating w.r.t. s,
−2
(s + 1)2
=
2s(s + 2)
(s + 1)2
+ K3
Letting s = −2 we find K3 = −2. Thus, the whole expansion is,
2
(s + 1)(s + 2)2
=
2
s + 1
−
2
(s + 2)2
−
2
s + 2

Case 3: Complex Roots
Same procedure as for case 1, except we have to deal with complex
roots. (Book presents another alternative).
e.g.
F(s) =
3
s(s2 + 2s + 5)
=
3
s(s + 1 + j2)(s + 1 − j2)
=
K1
s
+
K2
s + 1 + j2
+
K3
s + 1 − j2
Using the same procedure as before we get K1 = 3
5. Likewise we
can solve for K2 and K3 only now we get complex numbers,
(workings not shown)
K2 = − 3
20(2 + j), K3 = − 3
20(2 − j).
F(s) =
3
5s
−
3
20
2 + j
s + 1 + 2j
+
2 − j
s + 1 − 2j
Applying the ILT we obtain,
f (t) =
3
5
−
3
20
(2 + j)e(−1−2j)t
+ (2 − j)e(−1+2j)t
Assuming that f (t) should be purely real, we can utilize Euler’s
formula to capture the complex numbers. Finally we arrive at,
f (t) =
3
5
−
3
5
e−t
cos 2t +
1
2
sin 2t

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