MATHEMATICAL METHODS FOR ENGINEERS II
(EE203)
Chapter 3
LAPLACE
1.1 Introduction
1.2 Laplace Transforms
1.3 Inverse Laplace Transform
1.4 Solution of Differential Equations
1.5 Application to Step Functions and Delta
Functions
Lecture #3
1.50 Application to Step Functions and Delta Functions
INTRODUCTION
This section is important since we shall reach the point where
Laplace Transform method shows its superiority in engineering
applications.
We will introduce two functions:
A. Unit step function: u(t)
B. Unit impulse function (dirac’s delta):δ(t)
These functions are suitable for solving ODEs with complicated
right sides of considerably engineering interest such as single
waves, input (driving forces) that are discontinuous or act for some
time only.
1.51 Application to Step Functions
A. UNIT STEP FUNCTION: Definition
1. Unit step function, u(t-a) is a right shifted or delayed version of
u(t) by amount of a (unit time).
0,
u (t )  
1,
t0
t 0
0,
u (t  a )  
1,
ta
ta
A. UNIT STEP FUNCTION: Definition
2. A step function is a piecewise continuous function of the
form
 g1 0  t  a1

 g 2 a1  t  a 2
 g 3 a 2  t  a3

g (t )   
g
a  t  an2
 n  2 n 3
 g n 1 a n  2  t  a n 1

 g n t  a n 1
 g1  [ g 2  g1 ] H (t  a1 )  [ g 3  g 2 ] H (t  a 2 )  ....
 [ g n 1  g n  2 ] H (t  a n  2 )  [ g n  g n 1 ] H (t  a n 1 )
For n=3 the step function of the form is
 g1 0  t  a

g (t )   g 2 a  t  b
g t  b
 3
 g1  [ g 2  g1 ]H (t  a)  [ g3  g 2 ]H (t  b)
1.51 Application to Step Functions
Laplace Transform of Unit Step Function
Laplace transform of u(t) is given by:


1
Lu (t )   u (t )e dt   1.e dt 
s
0
0
 st
 st
For unit step function u(t-a):

Lu (t  a )   u (t  a )e st dt
0
a

0
a
  u (t  a )e st dt   u (t  a )e st dt



 as

1
e
Lu (t  a )  0   1.e st dt 
e   e as 
s
s
a
1.51 Application to Step Functions
Second Shifting Theorem
The first shifting Theorem (s-shifting) is:
L e f (t )  F ( s  a )


e f (t )  L F ( s  a )
Laplace
Inverse Laplace
at
at
1
Now, we have second shifting Theorem (t-shifting):
L f (t  a ) H (t  a )  e  as F ( s )
Laplace


f (t  a ) H (t  a )  L1 e  as F ( s )
Inverse Laplace
1.51 Application to Step Functions
EXAMPLE:
1. Find the Laplace Transform of the function:
 1 t  4
f (t )  
t4
1
2. Describe this function in term of unit step function
a)
1
 (2  t ) 0  t  2
f (t )   2
0
t2
b)
1

f (t )  2
 1

0t 4
4t 7
t7
 g1 0  t  a
g (t )  
g 2 t  b
 g1  [ g 2  g1 ]H (t  a )
 g1 0  t  a

g (t )   g 2 a  t  b
g t  b
 3
 g1  [ g 2  g1 ]H (t  a)  [ g3  g 2 ]H (t  b)
3. Find the Laplace transform of the expression:
a) 2(t  1) H (t  1)  (t  2) H (t  2)
 (t 2)
e
H (t  2)
b)
c) sin 2(t   ) H (t   )
Answer
2e  s e 2 s
a)
 2
2
s
s
b)
e 2 s
s 1
2e s
c) 2
s 4
L f (t  a ) H (t  a )  e  as F ( s )
4. Find the Laplace transform of the expression:
1) f (t )  (2t 2  3t  1) H (t  3)
2) f (t )  e t H (t  2)
Answer
1) e
3 s
 4 15 26 
 3 2 
s 
s s
e 2 ( s 1)
2)
s 1
1.52 Application to Delta Functions
B. UNIT IMPULSE (Dirac’s Delta) FUNCTION: Definition
The Dirac’s delta can be loosely thought of as a function on the
real line which is zero everywhere except at the
δ(t)
origin, where it is infinite.
,
 (t )  
0,
t 0
t0

and
  (t )dt  1

t
Similarly, for the shifted case:
,
 (t  a)  
0,
δ(t-a)
ta
ta
a
t
1.52 Application to Delta Functions
Laplace Transform of Delta Function

If f(t) is a continuous function at t=a, then
 f (t ) (t  a ) dt 
f (a)
0
This result is known as the shift of an impulse function, because all
values of f(t) are shifted except at t=a
 st
1)If f ( t )  e , then the Laplace transform of the unit impulse

function is
L{ (t  a)}   e st  (t  a)dt  e as
When a=0
0
L { ( t )}  1
1.52 Application to Delta Functions
Laplace Transform of Delta Function
2) By replacing f(t) with est f (t) the Laplace transform of f (t ) (t  a )
is


0
0
L{ f (t ) (t  a)}   e st [ f (t ) (t  a)dt   [e st f (t )] (t  a)dt e as f (a)
Therefore
L{ f (t ) (t  a)}  f (a)e
When a=0
L{ f (t ) (t )}  f (0)
 as
Laplace transform of Step and Impulse
function
Example
1)  (t   )   (t  3 )
1
2) (t  )
2
3) 7 (t  3)
1
3
4) t  (t   )
3
5)t 2 (t  2)
Answer
1) e s  e 3s
2)e
1
 s
2
3) 7e 3 s
3
 
4)   e
3
5)4e  2 s
1
 s
3