UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Chapter 4 : Laplace Transform
The Laplace transform is an important tool for solving certain kinds of initial values problem, particularly those involving discontinuous forcing function, as occur frequently in areas such as electrical engineering. It is also used to solve boundary value problems involving partial differential equations to analyze wave and diffusion phenomena. The Laplace transform is an example of a class called integral transforms, and it takes a function (which we shall refer to as time) into a function frequency).
Definition:
( ) of one variable t
( ) of another variable s (the complex
ℒ ∞ ℒ −
)]( ) defined by The Laplace transform of a function f is is a function [ ( )](
[ ( )]( ) =
( )
= ( )
0
The improper integration is with respect to t and and defines a function of the new variables s for all s such that this integral converges.
Note- we call it improper integral because the upper limit in the integral is infinite, so the domain of integration is infinite. And because the lower limit in the integral is zero, it follows that the
≥
negative values of t for
( ) only for
( ) is ignored or suppressed that means
( ) contains information of
0. In general, however, unless the domain is clearly specified, a function ( )
is normally intepreted as being defined for all real values, both positive and negative, of t .
≥ � ≥ ( ), where
Making use of the Heaviside unit step function
( )=
We have
( ) ( )=
0 1
<0 0
0
( )
<0 0
Thus the effect of multiplying f (t) by H(t) is to co nvert it into a causal function. Graphically, the relationship between f (t) and f (t)H(t) is 1
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I
It follows that the corresponding Laplace transform
rather than { ( ),
Notation:
( ) contains full information on the
( ) ( ). Consequently, strictly speaking one should refer to { ( ) ( ),
behavior of
drop the
( )}
( )} as being a Laplace transform pair. However, it is common practice to
( ) and assume that we are dealing with causal functions.
Because the symbol
ℒ
[ ( )]( ) may be awkward to write in computations, we will
make the following convention. We will use lowercase letters for a function we put into the transform and the corresponding uppercase letters for the transformed function.
ℒ ℒ ℒ ℎ ℒ ∞ ∞ − ℒ ℒ − − − →∞ →∞ − →∞ − − ∎ [ ( )] = ( )
[ ( )] =
( )
[ ( )] =
( )
[ ]=
Example 1:
Let c be any real number, and ( ) = . The Laplace transform of f is the function defined by
[ ( )] = [ ] =
=
0
0
= lim
=
1
lim
0
0
=
lim
1
+
2
1
=
1
=
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 2:
Let a be any real number, and ( ) = by
. The Laplace transform of f is the function defined
Example 3:
Determine The Laplace transform of the ramp function ( ) =
.
Solution:
Exercise 1:
−
Determine The Laplace transform of the ramp function ( ) =
3
.
Properties of Laplace Transform
Now we consider some of the properties of the Laplace transform that will enable us to find
further transform pairs { ( ), definition.
( )} without having to compute them directly using the
The Linear properties of Laplace Transform:
A fundamental property of the Laplace transform is its linearity, which may be stated as follows:
ℒ ℒ ℒ
Let f and g be functions where Laplace transforms exist, and let and be constants. Then
[
( )+
( )] =
[ ( )] + ( )+
=
3
[ ( )]
( ).
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Laplace Transform Table f (t )
L( f )
f (t )
1
1
cos ω t
L( f )
s
s t
2
s t n
s
t a
sinh a t
a t
e a t sin ω t
1
at
ω
t sin (ω t )
2ω s
( s − a ) n +1
(s t cos (ω t )
a−b
(s − a )(s − b )
Γ
Note that: ( + 1) =
2
(s − a )2 + ω 2
n!
e at − e bt
−a
(s − a )2 + ω 2
s−a n
2
s−a
cos ω t
s a +1
t e
−a
a
n +1
e
2
s
3
Γ (a + 1)
e a t
s
2
+ ω
cosh a t
n!
n = 0,1,...
s
2
2
ω
2
2!
t
s
2
+ ω
sin ω t
1 s
s
2
s2
(s
!
Example 4:
Determine The Laplace transform of the ramp function ( ) = 3 + 2
Solution: 4
2
4
.
2
+ ω
)
2 2
− ω + ω
2
)
2 2
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Exercise 2:
Determine the Laplace transform of the following function −4t
(i)
k (t ) = −5t e
(ii)
w(t ) = cos(5t ) − 2t cos(3t )
(iii)
f (t ) = 4 + 3t sin( 2t ) − 5t + 2e
(iv)
g (t ) = e
(v)
h(t ) = 14t e
2
+ sin(3t )
3
−4t 3
−7 t
sin(8t ) 2 t
− 4 sin(7t )
The Inverse Laplace Transform
ℒ ℒ− ℒ ⟹ ℒ−
If [ ( )]( ) =
1
( ), then
[ ( )] = ( ) denotes a function ( ) whose Laplace transform
is ( ) and is called the inverse Laplace transform of ( ) . That is
1
[ ( )] = ( )
Example 5:
ℒ− − ℒ − 1
1
1 1
=
[ ( )] = ( )
Since
= 1
Since
ℒ − ℒ [
[1] =
1
]= 1
■
Remark:
The inverse Laplace transform also has linearity property
ℒ− ℒ − ℒ− 1
[
( )+
1
( )] =
=
[ ( )] +
( )+
5
( ).
1
[ ( )]
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I The most obvious way of finding the inverse transform of the function
( ) is using a table of
transforms. Sometimes it is possible to write down the inverse transform directly from the table,
but more often than not. It is first necessary to carry out some algebraic manipulation of ( ). In particular, we frequently need to determine the inverse transform of a rational function of the form
( )
, where
( )
( ) and
( ) are polynomials in s. in such cases the procedure is first to
resolve the function into partial fractions and then to use the table of transforms.
Example 6:
Find the following inverse Laplace transforms. (i) (ii) (iii) (iv) (v)
ℒ− − ℒ− − ℒ− − ℒ− − ℒ− − 1
3
6
3
1
6)3
(
1
1
3
1
2
2
9
1
2+
1
4
+
6
2
2 +3
28
6
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Exercise 3:
Find the inverse Laplace transform of the following functions. (i)
G( s) =
(ii)
P( s ) =
5s s + 11 2
1 s + 35
(iii)
F ( s ) = −
(iv)
Q( s ) =
−
−
4 s +8 2
1 ( s − 4) 5
3s 2 − 27 ( s 2 + 4) 2 s
s + 64 2
The First Shifting Theorem (s – shifting)
Exponential modulation theorem The shifting theorems of this section will enable us to solve problems involving pulses and other discontinuous forcing functions. We will show that the Laplace transform of
−
( ) is the
transform of ( ), shifted a units to the right. This shift is achieved by replacing s by s-a in
( ) to obtain
(
).
Theorem:
If ( ) is a function having Laplace transform ( ), then for any number a,
L[e a t f (t )] = F ( s − a ) L −1 [ F ( s − a )] = e a t f (t )
or
This conclusion is also called shifting in the s variable. Proof:
L[e f (t )](s) = at
∫
∞
0
− st at
∫
∞
e e f (t )dt = e − 0
( s − a ) t
f (t )dt = F ( s − a) ■
7
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 7:
Given L[cos(bt )] =
s s +b 2
at
= F ( s ) . Find the Laplace Transform of e cos(bt ) .
2
Solution: Use First shifting theorem,
L[e at cos(bt )] = F ( s − a ) And L[cos(bt )] =
s s 2 + b2
= F ( s )
So
L[e at cos(bt )] = F ( s − a ) =
( s − a) ( s − a) 2 + b 2
Example 8:
Given L[t ] = 4
24 s
5
4 5t
. Find the Laplace Transform of t e .
Solution:
Example 9:
ℒ− ⁄ 1
Find
[1 ( + 2)2 ]
Solution:
Exercise 4:
Determine 1. 2.
ℒ−− ℒ [
3
sin2 ]
(
+1)2 ( 2 +4)
1
1
8
■
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Derivative-of-transform property:
This property relates operations in the time domain to those in the transformed s domain, but initially we shall simply look upon it as a method of increasing our repertoire of Laplace transform pairs. The property is also sometimes referred to as the multiplication-by-t property. A statement of the property is contained in the following theorem.
Derivative of transform: Theorem:
If ( ) is a function having Laplace transform of
ℒ ℒ − [ ( )] = ( )
then the functions
( )(
= 1, 2, . . . ) also have Laplace transforms, given by
[
( )
( )] = ( 1)
Example 10:
Find
ℒ
[ sinh(2 )].
Solution: Using derivative of transform
Example 11:
ℒ
Find [
2
sin(3 )]
Solution:
9
,
= 1, 2, …
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Exercise 5:
1. Find 2. Find
ℒℒ − [
3
2
[ cosh
]
]
Heaviside function (Unit step function)
Functions having jump discontinuities are efficiently treated by using the unit step function or Heaviside function H , defined by
0 H (t ) = 1
, t < 0 , t ≥ 0
This is the Heaviside function shifted a units to the right:
0 H (t − a ) = 1
, t < a , t ≥ a
In particular,
0
H (t − a ) g (t ) =
, t < a
g (t ) , t ≥ a
10
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I so the function
=
− (
) may be interpreted as a device for ‘switching on’ the function ( ) at
. In this way the unit step function may be used to write a concise formulation of
piecewise-continuous functions. Consider the piecewise continuous function ( )
≤ ≤ ≥ − − − − − − ≤ 1(
) 2( ) 3( )
( )=
0
< <
In terms of the unit step function, the function ( ) may thus be expressed as
( )=
Alternatively,
1(
) ( )+ [
2(
)–
1(
)] ( –
)+ [
3(
)–
2(
)] ( –
( ) may be constructed using the top hat function
Clearly, p393
(
which gives
)
(
)=
1 0
< otherwise
− − − ≤ ( )[ (
)
(
)] =
11
( ) 0
< . otherwise
(
) .
)
(
).
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 12:
Express in terms of unit step functions the piecewise-continuous causal fun ction
≤ ≤ ≥ ( )=
2 2 +4 9
0 3
< 3 < 5 5
Solution:
Second Shifting Theorem (shifting in the t variable)
If ( ) is a function having Laplace transform ( ), then for a positive constant a,
L[ H (t − a ) f (t − a )]( s ) = e − a s F ( s ) or
L −1 [e − a s F ( s )] = H (t − a ) f (t − a )
Example 13:
Find the Laplace transform L[ H (t )].
Solution:
∞ −
∞ −
∞ − ∙ ∎ [ ( )] =
( )
0
=
0
Example 14:
Find the Laplace transform L[ H (t − a )].
Solution: 12
1
=
=
0
1
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 15:
Find the Laplace transform L[ g ] , where
, t < 2 0 g (t ) = 2 t + 1 , t ≥ 2 Solution:
Example 16:
se −3 2 s + 4 s
Find
−1
L
Solution:
Exercise 6:
≤ ℎ
1- Find the Laplace transform of ( ).
(i)
( )=
cos2 0
1
≤ ≤ ≥ 2
< 3
(ii) ( ) =
2- Find the inverse Laplace transform of (i)
(ii)
ℒ− − ℒ− − 1
4
4
( +2)
( +3)
1
(
2 +1)
13
2 + 3 7
0 3
< 3 < 5 5
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Convolution
If f (t ) and g(t ) are defined for t
≥
0 , then the convolution f ∗ g of f with g is a function
defined by t
∫
( f ∗ g )(t ) = f (t − τ ) g (τ )d τ for t ≥ 0 such that this integral converges. 0
t
Remark:
∫
( f ∗ g )(t ) = ( g ∗ f )(t ) = g (t − τ ) f (τ )d τ 0
Example 17:
Find the convolution of
f (t ) = sin t and g (t )
= t .
Solution:
The Convolution Theorem:
If ( ) and ( ) have Laplace transforms ( ) and
( ) respectively, then,
L[ f ∗ g ] = L[ f ] L[ g ] Equivalently,
L[ f ∗ g ]( s ) = F ( s )G ( s ) And the inverse convolution theorem is
L −1 [ FG ] = f ∗ g
14
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 18:
by using the convolution theorem 2 s ( s 4 ) − 1
− Find L 1
Solution:
Example 19:
Find
L[t ∗ cos t ]
Solution:
Exercise 7:
Determine
ℒ − 1
1
2(
+2)2
Solution of Initial value problem
To apply the Laplace Transform to the solution of an initial value problem, we must able to transform a derivative. This involves the concept of piecewise continuous function.
Suppose ( ) is defined at least on [ , ]. Then f is piecewise continuous on [ , ] if : 1.
f is continuous at all but perhaps finitely many points of [ , ]
2.
If f is not continuous at t 0 in ( , ), then ( ) has finite limits form both sides at
3.
0.
( ) has finite limits as t approaches a and as t approaches b from within the interval. 15
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Transform of a derivative Theorem:
Let f be continuous for t ≥ 0, and suppose Suppose also that lim e k →∞
− sk
′
is piecewise continuous on [0, k ] for every k > 0.
f ( k ) = 0 if s > 0. Then
L[ f ′]( s ) = sF ( s ) − f (0) Transform of a Higher derivative Theorem:
′
Let ,
,…,
( +1)
be continuous for
t ≥ 0, and suppose
[0, ] for every k > 0. Suppose also that
( )
is piecewise continuous on
lim e − sk f ( j ) ( k ) = 0 k →∞
For s > 0 and j = 1,2, …, n-1. Then
L[f ( n ) ](s) = s n F(s) − s n −1f ( 0) − s n − 2 f ′(0) − ... − sf ( n − 2 ) ( 0) − f ( n −1) ( 0) For second derivative case
L[ f ′′]( s ) = s 2 F ( s ) − sf (0) − f ′(0)
Example 20:
Solve
y ′ − 4 y = 1 ; y (0) = 1
Solution:
16
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Example 21:
Solve y ′′ + 4 y ′ + 3 y = e
t
; y (0) = 0 , y ′(0) = 2
Solution:
Example 22:
Solve the differential equation
− 2
2
subject to the initial conditions
+ 5
= 1 and
+ 6 =2
= 0 at = 0.
Exercise 8:
1. Solve 2. Solve
′′ ′ ′ − − + 4 = cos
+ 9 = sin
+ 2
+ 5 =1;
3
+ 2 =2
2
2
4.
Solve
(0) = 0
+ 6
2
3. Solve
;
2
(0) = 0 ,
;
= 1 and
4
;
(0) = 0
= 0 at = 0
= 1 and
17
= 1 at
=0
UECM1653 Mathematics for Engineering I UECM1683 Mathematics for Physics I Laplace Transform
1.
2.
Solve the initial value problem y ′ + 2 y = e
− t
; y (0) = 2 by
(i)
the method of integrating factor
(ii)
the Laplace transform
(a)
Find a particular solution of
(b)
Find the inverse Laplace transform of the following function using the Laplace
Sept 2011 2
y ′′ + 3 y ′ + 2 y = 3e x .
transform table provided.
(i)
2 s 2 + 3s + 5
e2 (ii)
3.
s
Sept 2011
s ( s − 2)
Use Laplace transform to solve the following initial value problem
y ′′ − 5 y ′ + 6 y
=
e t , y (0) = 0 , y ′ (0) = 0 Jan 2012
4.
−1
Obtain the inverse Laplace transform of L
2s + 3 2 . s 4 s 13 + + Jan 2012
5.
Use Laplace transform to solve the following system
x ′′ − 2 x ′ + 3 y ′ + 2 y
=
4
2 y ′ − x ′ + 3 y
=
0
=
0
x (0) = x ′(0) = y (0) 18
