Why Laplace Transform?
Lecture 6
Frequency-domain analysis:
Laplace Transform
Laplace transform is the dual (or complement) of the time-domain
analysis.
In time-domain analysis, we break input x(t) into impulsive
component, and sum the system response to all these components.
In frequency-domain analysis, we break the input x(t) into
exponentials components of the form est, where s is the complex
frequency:
s = α + jω
(Lathi 4.1 – 4.2)
Peter Cheung
Department of Electrical & Electronic Engineering
Imperial College London
Laplace transform is the tool to map signals and system behaviour from
the time-domain into the frequency domain.
X (s)
Time-domain
analysis
h(t)
x (t )
y (t )
x (t )
L
Y (s)
frequency-domain
analysis
H(s)
L-1
y (t )
URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals
E-mail: p.cheung@imperial.ac.uk
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Lecture 6 Slide 1
E2.5 Signals & Linear Systems
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Definition of Two-sided Laplace Transform
For a signal x(t), its Laplace transform is defined by:
∞
− st
For the purpose of the 2nd year curriculum, let us assume that all
signals are causal. For this the Laplace transform is defined as:
−∞
∞
The signal x(t) is said to be the inverse Laplace transform of X(s). It
can be shown that
x(t ) =
1
c + j∞
2π j ∫c − j∞
X ( s)e st ds
X ( s) = L[ x(t )] = ∫ x(t )e− st dt
0
where c is a constant chosen to ensure the convergence of the first
integral.
Note that this definition is slightly more complex than what you have seen
in Dr Jaimouka’s 2nd year Control course. This general definition does not
assume casuality.
This is the same as that defined on the 2nd year Control course, and is
known as one-side (or unilateral) Laplace transform.
Remember that the Laplace transform is a linear tranform (see
Jamouka’s notes, p15):
This general definite is known as two-sided (or bilateral) Laplace
Transform.
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E2.5 Signals & Linear Systems
Lecture 6 Slide 3
Lecture 6 Slide 2
Definition of One-sided Laplace Transform
X ( s) = ∫ x(t )e dt
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Lecture 6 Slide 4
A few examples
A few examples (2)
Find the Laplace transform of δ(t) and u(t).
∞
∞
L[δ (t )] = ∫ δ (t )e− st dt = 1
L[δ (t )]
L[eat u (t )] = ∫ eat e− st dt
for all s
0
0
∞
= ∫ e− ( s −a )t dt =
⇔ 1
0
∞
∞
0
0
L[u (t )] = ∫ u (t )e− st dt = ∫ e − st dt
1
= − e− st
s
L[u (t )]
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Find the Laplace transform of eat u(t) and cos ω0t u(t).
∞
0
⇔
=
1
s
1
s
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L[cos ω0tu (t )]
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Lecture 6 Slide 5
⇔
⇔
1
s−a
s
s + ω0 2
2
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Laplace transform Pairs (1)
1
s−a
1
L[cos ω0t u (t )] = L[e jω0t u (t ) + e − jω0t u (t )]
2
⎡
s
1
1
1 ⎤
= ⎢
+
⎥=
2 ⎣ s − jω0 s + jω0 ⎦ s 2 + ω0 2
Re s > 0
L[eat u (t )]
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Lecture 6 Slide 6
Laplace transform Pairs (2)
Finding inverse Laplace transform requires integration in the complex
plane – beyond scope of this course.
So, use a Laplace transform table (analogous to the convolution table).
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Lecture 6 Slide 7
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Lecture 6 Slide 8
Laplace transform Pairs (3)
Examples of Inverse Laplace Transform (1)
Finding inverse Laplace transform of
7s − 6
.
s2 − s − 6
(use partial fraction)
To find k1 which corresponds to the term (s+2), cover up (s+2) in X(s),
and substitute s = -2 (i.e. s+2=0) in the remaining expression:
Similarly for k2:
Therefore
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Lecture 6 Slide 9
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Examples of Inverse Laplace Transform (2)
Easy to make mistake with partial fraction.
Method to check correctness of:
Substitute s = 0 into the equation (could use other values, but this is
most convenient):
−6
4 3
X (0) =
=1= +
(+2)(−3)
2 −3
Therefore, using Pair 5 from table:
Finding the inverse Laplace transform of
2s 2 − 5
.
( s + 1)( s + 2)
The partial fraction of this expression is less straight forward. If the
power of numerator polynomial (M) is the same as that of denominator
polynomial (N), we need to add the coefficient of the highest power in the
numerator to the normal partial fraction form:
Solve for k1 and k2 via “covering”:
Therefore
Using pairs 1 & 5:
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Lecture 6 Slide 11
Lecture 6 Slide 10
Examples of Inverse Laplace Transform (3)
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Lecture 6 Slide 12
Application of Time Shifting
Time Shifting Property of the Laplace transform
Time Shifting property:
[u (t − 2) − u (t − 4)]
Find the Laplace transform of x(t) as shown:
Delaying x(t) by t0 (i.e. time shifting) amounts to multiplying its transform
X(s) by e − st0 .
(t − 1)[u (t − 1) − u (t − 2)]
Remember that x(t) starts at t = 0, and x(t - t0) starts at t = t0.
Therefore, the more accurate statement of the time shifting property is:
x(t ) = (t − 1)[u (t − 1) − u (t − 2)] + [u (t − 2) − u (t − 4)]
= (t − 1)u(t − 1) − (t −1)u(t − 2)
= (t − 1)u (t − 1)
−
+
(t − 2)u(t − 2)
u(t − 2) − u(t − 4)
−
u(t − 4)
Time shift
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Lecture 6 Slide 13
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Frequency Shifting Property
Application of Frequency Shifting
Frequency Shifting property:
Frequency shifting the transform X(s) by s0 amounts to multiplying its time
signal by e s0t .
Given
Apply frequency-shifting property with frequency shift
Observe symmetry (or duality) between frequency-shift and time-shift
properties.
, show that
E2.5 Signals & Linear Systems
Lecture 6 Slide 15
.
.
Replace s with (s+a) means frequency shift by -a. This yields the RHS of
the equation. By frequency-shifting property, we need to multiply the
− at
LHS by e .
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Lecture 6 Slide 14
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Lecture 6 Slide 16
Time-Differentiation Property
Time-differentiation property:
Repeated application of this property yields:
Proof of Time-Differentiation Property
Integration by parts gives:
For the Laplace integral to converge, it is necessary that
Therefore we get:
x(t )e− st → 0 as t → ∞
Frequency-differentiation property:
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Lecture 6 Slide 17
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Application of Time-Differentiation
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Time-Integration Property
Find the Laplace transform of the signal x(t) using time differentiation and
time-shifting properties.
d/dt
Time-integration property:
The dual property of time-integration is the frequency-integration
property:
Time-differentiation &
time shifting properties
d/dt
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Lecture 6 Slide 18
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Lecture 6 Slide 19
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Lecture 6 Slide 20
Time-Convolution & Frequency-Convolution
Properties
Scaling Property
Scaling property:
Time compression of a signal by a factor a causes expansion of its
Laplace transform in s-scale by the same factor.
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Lecture 6 Slide 21
E2.5 Signals & Linear Systems
Time-convolution property:
Convolution in time domain is equivalent to multiplication in s (frequency)
domain.
Frequency-convolution property:
Convolution in s (frequency) domain is equivalent to multiplication in
time domain.
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Application of the convolution Properties
Use the time-convolution property of the Laplace transform to determine
at
c(t ) = e u (t )* e u(t ).
1
( s − a)
Since
Therefore
Perform inverse Laplace transform gives:
eat u (t ) ⇔
ebt u (t ) ⇔
eat u (t )* ebt u (t ) ⇔
1
( s − b)
1
( s − a)( s − b)
If h(t) is the impulse response of a LTI system, then we have see in
lectures 4 & 5 that the system response y(t) to an input x(t) is x(t)*h(t).
Assuming causality, and that h(t ) ⇔ H ( s) and x(t ) ⇔ X ( s )
then
The response y(t) is the zero-state response of the LTI system to the
input x(t). It follows that the transfer function H(s):
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Lecture 6 Slide 23
Lecture 6 Slide 22
Relationship with time-domain analysis
bt
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Lecture 6 Slide 24
Summary of Laplace Transform Properties (1)
Summary of Laplace Transform Properties (2)
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Lecture 6 Slide 25
Relating this lecture to other courses
You have done Laplace transform in maths and in control courses. This
lecture is mostly a revision, plus emphasis on the convolution –
multiplication properties for the two domains.
Many of the properties are deliberately stated without proofs. It is more
important on this course to understand the actual interpretations of
Laplace transform (and more importantly the duality of time and frequency
domains) than the mathematic proofs.
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