Laplace Transform

1
Laplace Transform

• Chapter 5

2
General Discussion

• The French mathematician Pierre-Simon Laplace
  (1749-1827) developed a mathematical method for
  solving differential equations by the use of
  algebraic techniques. The Laplace method uses a
  so-called transform or operational procedure.
• The British physicist Oliver Heaviside
  (1850-1925) is credited with early usage of
  operational methods in circuit analysis.

3
Overview

• The Laplace transform permits the solution of
  complex integrodifferential equations using
  algebraic methods. In particular, actual models
  can be represented directly using Laplace
  techniques.
• Both the transforms of time functions and the
  inverse transforms of Laplace functions are
  considered. The waveforms of Chapter 2 appear
  extensively in these developments, since they are
  widely employed in circuit applications.

4
Objectives

• State the definition of the Laplace transform.
• Discuss the advantages of the Laplace transform
  approach in solving differential equations.
• Derive the Laplace transforms of some of the
  common functions of Table 5-1.
• Use Table 5-1 to determine the Laplace
  transforms of common waveforms encountered in
  circuit analysis.
• Apply the operation pairs of the Table 5-2
  (derivative of a time function, integral of a
  function, shifting theorem, multiplication by
  e-at) for appropriate waveforms.

5
Objectives

• Write the form of a polynomial, identify its
  degree (or order), and indicate the number of
  roots.
• Express a polynomial in factored form.
• State the various classifications of roots.
• Apply Table 5-1 to determine inverse Transforms
  of easily identifiable transform functions.
• Define the terms zero and pole as they relate
  to Laplace- functions.

6
Objectives

• State the four ways of classifying poles for
  the purposes of inverse transformation.
• Perform a partial fraction expansion for
  first-order real poles.
• Determine inverse transforms corresponding to
  first-order complex poles.
• Determine inverse transforms corresponding to
  multiple-order poles.

7
General Discussion
As an aid in understanding transform analysis,
let us consider the process of multiplication by
logarithms as illustrated in the next slide in
Figure 5-1. The first step in performing the
multiplication is to replace the original numbers
by their logarithms. Next, the logarithms are
added, and the result, of course, is the
logarithm of the quantity desired. As a final
step, the antilogarithm (or inverse logarithm) is
found, and the desired result is obtained. Thus,
the process of multiplication has been replaced
by transformation, addition, and inverse
transformation.
Numbers to be multiplied
Product of Original numbers
Determine logarithms Of numbers
Determine Inverse logarithms
Add logarithms
Figure 5-1 Steps involved in multiplication by
logarithms
8
General Discussion

• An analogous situation exists with transform
  circuit methods. A general description of the
  process is shown in Figure 5-2.
• First we begin with a circuit or system that is
  described by a set of differential equations.
• We then replace the system by a so-called
  transform system. We speak of the original
  system as a representation in the time domain and
  the new system as a representation in the
  transform domain. (often referred to as the
  complex-frequency domain).
• The transform system has the property that any
  solution may be obtained by algebraic procedures.
  
• After the desired response is obtained in the
  transform domain, we invert or calculate the
  inverse transform to obtain the desired response
  in the time domain.

Differential Equation To be solved
Determine Inverse Transform Of result
Desired response
Determine Laplace transform of equation
Solve equation by algebra
Figure 5-2 Steps involved in Laplace transform
analysis
9
General Discussion

• In performing the transformation to the
  transform domain, two approaches are possible
• 1. The differential equation may be written in
  the time domain, followed by transformation of
  the differential equation.
• 2. The circuit or system may first be transformed
  directly, followed by direct expression of the
  equations in the transform domain.
• The engineering analyst will have occasion to
  follow both procedures, but the second method is
  usually simpler for circuit applications.

10
Laplace Transform

• The Laplace transform of a function of time f(t)
  is a function of a new variable s and is
  designated as F(s). The process of performing the
  Laplace transformation is designated as

(5-1)
The process of inverse transformation is
designated as
(5-2)
11
Laplace Transform

• Note that we have used a lowercase f for f(t) and
  a capital F for F(s). It is necessary to make
  this distinction, because f(t) and F(s) are not
  equal. In this context, we emphatically state that

(5-3)
It is common for beginners to write expressions
of the form of Equation (5-3) as an equality. The
pair of functions f(t) and F(s) correspond to
each other in a transform sense, but are not
equal. The correct way to express an equality is
by Equations (5-1) and (5-2).
12
Laplace Transform

• The definition of the Laplace transformation is

(5-4)
We thus multiply f(t) by e -st and integrate from
t 0 to t 8 to obtain F(s). This means that
the given function times e -st is integrated over
all positive time. This may require breaking up
the function into several different integrals in
cases where f(t) is defined in a different sense
over different intervals, or it may require
integrating over only a short period of time in
cases where f(t) is identically zero over most of
positive time.
13
Laplace Transform
1. The Laplace transform of a constant times a
given function of time is equal to the constant
times the transform of the original function of
time. Thus,
(5-5)
2. The Laplace transform of the sum of several
functions is equal to the sum of the individual
transforms. Thus,
(5-6)
The Laplace transform of the product of two
functions is not the product of the two
individual transforms. Thus,
(5-7)
14
Laplace Transform
15
Laplace Transform
Example Derive the Laplace transform of the
unit step function.
Solution By definition, we have
We interpret the s to be positive in the
foregoing equation, so that lim ?-st 0. Thus,
t?8
16
Laplace Transform
Example Derive the Laplace transform pair (T-5).
We may use "simpler" transform pairs to
accomplish the task. Solution Although we could
integrate ?-st sin wt if we cared to do so, we
can accomplish the same task by accepting, for
the moment, pair (T-4) and employing the
exponential definition of the sine function. This
definition reads
(5-10)
Thus, by Equations (5-5), (5-6), and (T -4),
(5-11)
17
Laplace Transform
Exercise The narrow voltage pulse shown in
Figure 5-3 is used to excite a network whose time
constants are very long compared with the width
of the pulse. Define an impulse that approximates
the given pulse, and find its Laplace transform.
FIGURE 5-3 Voltage pulse for Example 5-3.
18
Laplace Transform
Solution The area of the pulse is 100 V x 10-3 s
0.1 Vs. Thus, according to the methods of
Chapter 2, we may approximate v(t) as
The transform of v(t) is V(s). From (T-l),
We might note in passing that, while the impulse
function may be somewhat strange in the time
domain, its transform is the simplest of all
functions because it is merely a constant!
19
Laplace Transform
Exercise Determine the Laplace transform of the
current
Solution In order to use the table, we must
decompose i(t) into the sum of a sine function
and a cosine function. The reader can verify that
Letting I(s) represent the transform of i(t), we
have from (T -5) and (T -6),
20
Laplace Transform
21
Some Definitions And Concepts
Classification of Roots 1. A root may be a real
root, imaginary root (e.g., j5) or a complex root
(e.g., 4 j5). A complex or imaginary root is
always accompanied by its complex conjugate.
Thus, the complex conjugate of a jb is a - jb,
and if a jb is a root of a polynomial, a - jb
is a root also. Similarly, -jc is the conjugate
of jc, and if one is a root, so is the other. 2.
Roots may be classified according to their order,
which is the number of times a root is repeated
in a given solution. The most common root is the
first-order or simple-order root. Higher-order
roots are referred to as multiple-order roots.
22
Inverse Transforms By Identification
Example
23
Inverse Transforms By Partial Fraction Expansion
Denominator with real poles of first order
Our basic problem at present is the determination
of the Ak coefficients. Multiply both sides of
the equation by (s - pk) and rearrange some terms
(5-84)
The inverse transform of the portion of the
equation (excluding the remaining portion R(s))
involving the real poles of simple order is
24
Inverse Transforms By Partial Fraction Expansion
Denominator with real poles of first order
Example
25
Complex Poles of First Order
There are several different approaches that may
be employed. We consider three different methods.
Two methods are approximately equal in
complexity, whereas the third method involves a
quick "plug-in" formula. The first two methods
will provide a better understanding of the
inversion process. The third method, much
shorter, is somewhat mechanical. Let us assume
for the moment that a given F(s) contains a
single pair of complex conjugate poles. In most
cases of interest at present, the real part is
negative, and thus the rectangular form of the
poles is
(5-103) (5-104)
where a is usually a positive number representing
a damping constant in the time domain, and w
represents an angular frequency in the time
domain.
26
Complex Poles of First OrderExpansion as a
quadraticMethod 1
In observing transform pairs (T -5) through (T
-8), we note that to use those transform pairs we
must be able to separate quadratic factors
containing complex or imaginary roots from the
remainder of the expression. Thus, assume that
F(s) contains a denominator factor of the form
(s2 bs c) whose roots are given by Equations
(5-103) and (5-104). According to the theory of
partial fraction expansion, we may expand the
function as follows
(5-105)
where A and B are constants to be determined, and
R(s) represents everything else in the expansion.
Since the denominator is of second degree, two
constants are required.
27
Complex Poles of First OrderExpansion as a
quadraticMethod 1
First, determine as much of the expansion for
R(s) as possible by other methods. Since Equation
(5-105) expresses an equality, select some
convenient values of s to substitute into the
equation and hence produce as many simultaneous
equations as are necessary to solve for the
constants. Complex poles of simple order with
negative real parts correspond to a damped
sinusoidal response in the time domain. As a
special case, purely imaginary poles correspond
to an undamped sinusoidal response in the time
domain.
28
Complex Poles of First OrderExpansion in terms
of the polesMethod 2
The second method is that of expanding the
function in terms of the poles directly, as we
did for the case of real poles. The main
differences are the facts that we have to
manipulate complex numbers, and some adjustments
must be made on the final time function to put it
in the best form. The key to this method is the
exponential definition of the cosine function
(5-121)
In determining the inverse transform by this
method, we factor the quadratic into two factors
as in the case of real poles, the essential
difference being that the factors involve complex
numbers. We then expand the function as in the
case of real poles, so the procedure of Equation
(5-84) serves equally well for this case. A
coefficient is required for both the terms
involving the complex roots. An important point
to remember is that the coefficients of the
conjugate poles are themselves complex
conjugates. After, the inverse transform can be
manipulated in form of Equation (5-121).
29
Complex Poles of First OrderA Trick
FormulaMethod 3
We now wish to turn our attention to a reasonably
simple formula for determining the portion of the
inverse transform of a given function that
corresponds to a pair of complex poles of simple
order. Like the second method, it allows us to
determine the damped or undamped sinusoidal part
of a time function without considering the
remainder of the function. Assume that F(s) is
written in the form
(5-132)
Q(s) represents everything else in F(s) besides
the quadratic factor.
(5-133)
_
_
The complex poles are p -a jw and p -a -
jw, of course.
30
Complex Poles of First OrderA Trick
FormulaMethod 3
_
We now evaluate Q(s) for the particular value of
s p -a jw . The resulting complex number is
expressed in polar form as a magnitude M and an
angle
(5-134)
We designate the portion of the time response of
interest as f1(t). It can be shown that f1(t)
is given by
(5-135)
31
Complex Poles of First OrderA Trick
FormulaMethod 3
An important special case Suppose the pair of
poles are purely imaginary (i.e., a 0). Then
We then substitute s jw into Q(s) and obtain
The time function is then
32
Multiple-order poles
First, let us consider the case of multiple-order
real poles. Assume that the denominator of F(s)
contains a factor of the form (s - p )r,
indicating that s p is a real pole of r-th
order. We may thus write F(s) in the form
or
where Q(s) represents everything else in F(s)
besides the (s - p )r factor. The partial
fraction expansion of F(s) requires the following
form
where R(s) is the expansion due to all other
poles.
33
Multiple-order poles
Let F1 (s) represent the expansion of interest
at present
It can be shown that the general k-th coefficient
is given by the formula
Once the coefficients are known, the inverse
transform can be determined by means of (T-l0).
The general form is
Multiple-order complex poles are best handled by
combining the method of this section with the
second technique described in the section on
complex poles of simple order.