The Laplace Transform

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The Laplace Transform

• BEE2113 Signals Systems
• R. M. Taufika R. Ismail
• FKEE, UMP

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Introduction

• Laplace transform is another method to transform
  a signal from time domain to frequency domain
  (s-domain)
• The basic idea of Laplace transform comes from
  the Fourier transform
• As we have seen in the previous chapter, not many
  functions have their Fourier transform such as t,
  t2, et etc.

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• The Fourier transform formula
• The Laplace transform formula is the modification
  of the above formula, that is, the term jw is
  replaced by s
• s is equal to s jw, where s is a large positive
  real number
• The Laplace transform formula
• However, the Laplace transform only support the
  function f(t) which domain t 0

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Example 1
Using definition, find the Laplace transform
of (a) (b) (c) (d)
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Solution
(a)
(b)
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(c)
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(d)
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Properties of L-transform

• Linearity
• Laf(t) bg(t) aLf(t) bLg(t)

• First shift theorem
• Le-atf(t) F(s a)

• Second shift thorem
• Lf(t - d) u(t - d) e-dsF(s)

• Time scaling

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Properties of L-transform (cont.)

• Time derivatives

• Time integral

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Example 2

• Determine the Laplace transform of
• (a)
• (b)

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Solution
(a)
(b)
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Example 3

• Determine the Laplace transform of
• (a)
• (b)

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Solution
Let
(a)
then
Therefore
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(b)
Let
then
Also
Therefore
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Inverse Laplace transform (ILT)

• The inverse Laplace transform of F(s) is f(t),
  i.e.

where L-1 is the inverse Laplace transform
operator.
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Example 4

• Find the inverse Laplace transform of

(a)
(b)
(c)
(d)
(e)
(f)
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Solution

• From the table of Laplace transform,

(a)
(b)
(c)
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(d)
Write
(e)
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Since the ILT of the term cannot be found
directly from the table, we need to rewrite it
as the following
(f)
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Example 5

• Find the inverse Laplace transform of

(a)
(b)
(c)
(d)
(e)
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Solution

• We use the partial fractions technique

(a)
L
L
(b)
L
L
L
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(c)
L
L
L
where, if we let
, then
Hence,
L
L
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(d)
L
L
L
L
L
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(e)
L
L
L
L
L
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The convolution theorem
where
is called as the convolution of
f(t) and g(t),
defined by
Convolution property
Therefore,
Sometimes,
denoted as
or simply
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Example 6
Use the convolution theorem to find the
inverse Laplace transforms of the following
(a)
(b)
(c)
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Solution
(a)
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(b)
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(c)
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Circuit applications

• 1. Transfer functions
• 2. Convolution integrals
• 3. RLC circuit with initial conditions

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Transfer function
h(t)
y(t)
x(t)
Network
System
In time domain,
In s-domain,
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Example 7 (Pb.13.64, pg.751)
For the following circuit, find H(s)Vo(s)/Vi(s).
Assume zero initial conditions.
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Solution
Transform the circuit into s-domain with zero
i.c.
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Using voltage divider
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Example 8 (Pb.13.65, pg.751)
Obtain the transfer function H(s)Vo(s)/Vi(s),
for the following circuit.
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Solution
Transform the circuit into s-domain (We can
assume zero i.c. unless stated in the question)
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We found that
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Example 9 (P.P.15.14, pg.705)
Use convolution to find vo(t) in the circuit
of Fig.(a) when the excitation (input) is
the signal shown in Fig.(b).
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Solution
Step 1 Transform the circuit into s-domain
(assume zero i.c.)
Step 2 Find the TF
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Step 4 Find vo(t)
For t lt 0
For t gt 0
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Circuit element models

• Apart from the transformations
• we must model the s-domain equivalents of the
  circuit elements when there is involving initial
  condition (i.c.)
• Unlike resistor, both inductor and capacitor are
  able to store energy

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• Therefore, it is important to consider the
  initial current of an inductor and the initial
  voltage of a capacitor
• For an inductor
• Taking the Laplace transform on both sides of eqn
  gives
• or

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• For a capacitor
• Taking the Laplace transform on both sides of eqn
  gives
• or

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Example 10 (Pb.16.23, pg.750)
Consider the parallel RLC circuit of the
following. Find v(t) and i(t) given that v(0)
5 V and i(0) -2 A.
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Solution
Transform the circuit into s-domain (use the
given i.c. to get the equivalents of L and C)
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Then, using nodal analysis
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Since the denominator cannot be factorized, we
may write it as a completion of square
Finding i(t),
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Using partial fractions,
It can be shown that
Hence,
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Example 11
The switch in the following circuit moves from
position a to position b at t 0 second.
Compute io(t) for t gt 0.
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Solution
The i.c. are not given directly. Hence, at
first we need to find the i.c. by analyzing the
circuit when t 0
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Then, we can analyze the circuit for t gt 0 by
considering the i.c.
Let
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Using current divider rule, we find that
Using partial fraction we have