Subject Title:
Technical Mathematics 6
Total Contact Hours:
Lecture:
26
Tutorial:
13
Practical:
13
Other
Subject Aims:
1. To support the Semester 6 module in Control by introducing Laplace Transforms
and their use in solving linear differential equations.
2. To deepen the students knowledge of geometry, by introducing the formulation
and solution of 3D problems involving lines and planes in vector notation. To be
able to calculate the divergence, gradient and curl and relate these to ideas in
geometry and physics.
3. To use iterative techniques to solve equations numerically as an introduction to
material they may cover in year 4.
1
2
3
4
5
6
7
8
9
to find and invert Laplace transforms
directly and indirectly using Tables and
partial fractions.
to use Laplace Transforms in the solution of
linear first and second order differential
equations with constant coefficients
to calculate the power spectrum of a wave
given its Fourier series (periodic function)
to write the coordinates of points in i, j, k
notation from a diagram
to find the equations of lines and planes and
solve problems involving these
to find the divergence, gradient and curl for
polynomially defined functions
to use the gradient to find the directional
derivative, maximum rate of change,
tangent planes and normal lines
Use Newton’s Method and the Bisection
Method in one variable to numerically solve
an equation
To use the first order Euler method to
numerically solve a first order differential
equation.
Overall Assessment Breakdown (% of
Marks)
Final
Exam
Programme
outcomes
reference
CA
On successful completion of the module,
students will be able to
Practical
Assessment Type
Project
Learning Outcomes:
Learning outcome
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














15
15
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MODULE/SYLLABUS CONTENT
Component / Topic
The Laplace Transform: Definition. The Laplace transform of
common functions. Linearity rules. Transform of derivatives. Use of
partial fractions. Application: Solution of first and second order
linear ODE’s with constant coefficients.
Fourier Series: Periodic functions. Even and odd functions.
Fundamental Period and Frequency . Parseval’s Theorem and the
Power spectrum.
Vector Calculus: i, j, k notation. Equations of lines and planes in 3D.
Intersection point of line with a plane. Gradient, divergence, curl.
Directional derivatives and maximum rate of change. Tangent
planes and normal lines to a surface.
Numerical Methods: Newton-Raphson and Bisection methods. First
order Euler method for a first order differential equation
Hours
12
2
8
4
PRACTICALS, PROJECTS AND CONTINUOUS ASSESSMENT
Description
CA Test covering the Laplace Transform and 3D lines
and planes
CA to use a spreadsheet to implement bisection,
Newton or Euler methods and high threshold Keyskills
test
Final Exam
Due date
Week 7
% of Marks
15
Week 12
15
Semester
Exam
70
BIBLIOGRAPHY
Essential and/or recommended reading
Engineering Mathematics, 4th edition, K.A. Stroud, MacMillan, 1995.
Additional or supplementary reading/reference material
Further Engineering Mathematics, K.A. Stroud (Third Edition 1996)
Laplace Transforms
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Semester 6 Maths
Section 1: Laplace Transforms
The Laplace transform F(s) of a function of the function f(t) at s, specified for t>0
is defined by the expression
F (s )  Lf t    e  st f (t )dt

0
The transform exists at s if the value of F(s) is finite. The transform may exist for
all, some or no values of s. The most common case is some s values, and that is
enough.
e.g. 1 (Transform does not exist for s<a but does for s  a )
Suppose f(t) = eat, where a is a constant. Then,



0
0
0
F (s )   e  st e at dt   e  st  at dt   e
s  a t
Now, if s  a then –(s-a) is positive. In this case
dt

0
e s a t dt will be   i.e. F(s)
does not exist if s<a.
If s  a ,

 e (s a )t 
 e 0 
  F (s )  1
F (s )  
  0  

s a
  (s  a )  0
  (s  a ) 
since, e   0, e 0  1
Normally, we don’t worry about values for s for which F(s) is  as long as we can
find F(s) for some s values, as in the above example.
e.g. 2 Let u(t) be the unit constant function for t  0 (also called the Heaviside
Function).
1
u (t )  
0
if t  0
if t  0
u(t)
1
t
Laplace Transforms
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Then,

U(s)   e  stu(t)dt
0

  e  st .1dt
0
 0   1s e  st 
   0  e
1

s
1
s

0
 s0

The Heaviside function is used to “switch on” inputs. If f(t) is defined for all t
values then u(t)f(t) = 0 if t<0 and u(t)f(t) = f(t) if t  0 i.e. f(t) has been ignored if
t<0 and switched on for t  0 . This is a very handy function for looking at the
response of a system to an input.
Thankfully, we don’t have to work out Laplace Transforms from scratch, but can
look them up in tables ( see next page).
Some Notation
We will write functions of time t in lower case letters e.g. f(t), u(t) etc
We will write the Laplace Transform in upper case letters e.g. F(s), U(s)
We will say that F is the Laplace transform of f. Sometimes this will be abbreviated
as Ff and you will also see F = L(f).
Laplace Transforms
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Table of Laplace Transforms:
Function , f(t)
Laplace Transform, F(s)
1
t
1/ s
1/ s2
2 / s3
n!/ sn1
1/(s  a)
t2
tn
e  at
tn e  at
sin(bt)
n!/(s  a)n1

s / s

b 
b / s2  b2
cos(bt)
2
2

eat sin(bt)
b /  s  a   b2
2
s  a  / s  a 
eat cos(bt)
2

 b2



s / s  b 
2bs /  s  b 
s  b  / s  b 
sinh(bt)
b / s2  b 2
cosh(bt)
2
t sin(bt)
2
2
t cos(bt)
2
u(t) unit step function
u(t-d)
(t)
(t  d)
2
2
2
2
2
2
1/ s
e /s
1
 sd
e  sd
Using the Laplace Table
The Laplace Table is used left to right and right to left. Using it right to left is
called “finding the Inverse Laplace Transform”. This is why we use the notation
Ff because if we know f we can find F (the Laplace Transform) and if we know F
we can find f (the Inverse Laplace Transform).
Result 1
F  Gs   F s   Gs 
This says that the Laplace Transform of the sum of two functions is the sum of the
two Laplace Transforms.
Example
Find the Laplace Transform of t 2  3t 4
Laplace Transforms
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Solution
From the table,
2
t2  3
s
So using Result 1
t 2  3t 4 
t4 
and
2
s
3
 3
24
s
5

4!
s 4 1
2
s
3


24
s5
72
s5
Example
Find the Laplace Transform of 2t 2e 3t  3 sin(3t )  3
Solution
From the table,
t 2e  3t 
2
(s  3)3
and
sin( 3t ) 
3
s 2  32
and
1
1
s
So using Result 1
 2   3  1
3
2t 2 e  3t  3 sin( 3t )  3  2
3
 (s  3)3   s 2  3 2   s 


4
9
3

 2

3
2
s
(s  3)
s 3
Example
Find the Laplace Transform of 2te 2t  3 cos(3t )  3t
Example
Find the Laplace Transform of u(t )  2 (t  2)  3 (t  3)
Laplace Transforms
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Example (right to left)
Find the inverse Laplace Transform of
72
s5
Solution
The solution from right to left in the Table is not quite as straightforward. Start by
examining the denominator, which is s5 in this case. Look for the line in the table
on the right hand side which gives this denominator, and write down this line. The
only possibility is
n!
where n must be 4
t n  n 1
s
4!
 t4 
s
5

24
s5
Now we turn our attention to the numerator, which is 72 in our original expression
4! 24
1 4
1
72 4
72
t4  5  5

t  5 
t  5
24
24
s
s
s
s
72 4
t  3t 4
So our inverse transform is
24
Example
Find the inverse transform of
2
3
 3
(s  1) s
Solution
Identify the line in the table giving a denominator of (s  1) and the line giving a
denominator of s 3 .
Writing down the lines from the Table
1
2
2
3
3
e t 
and t 2  3
 2e  t 
and t 2  3
(s  1)
(s  1)
2
s
s
2
3
3

 3  2e  t  t 2
(s  1) s
2
Laplace Transforms
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Example
Find the inverse transform of
3
(s  1) 2  4
Solution
Appropriate line from Table:
Adjust numerator:
Example
Find the inverse transform of
7
(s  1)
4

1
(s  1) 4
Solution
Appropriate lines from Table:
Adjust numerators:
Do More Examples in Class
Laplace Transforms
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Worksheet on using the tables:
Find the Laplace transforms of
t 12
1.
 (t  4 )
2.
3.
e 2t sin( 6t )
u (t  4 )
4.
sinh( 2t )
5.
4t 2
6.
7 sin( 2t )
7.
t 5  4 cos(3t )  6t 2
1
2t 2 e 5t  e t sin( t )
9.
2
e 2t
7
10. 14e 3t 
11
Find the inverse Laplace Transform of:
4
1.
2
s  16
6
2.
s4
s3
3.
(s  3) 2  16
1
4.
s 7
1
5.
s8
s2  4
6.
(s 2  4 ) 2
8.
7.
8.
9.
10.
11.
12.
13.
e 3 s
s
7
s3
5
s2
4s
2
s  25
9
2
s  16
6s  6
(s  1) 2  81
6e 2s
13s
2
s (s  16) 2
Laplace Transforms
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Using laplace transform techniques to solve differential equations
As far as we are concerned, the main use of Laplace Transforms is in solving
differential equations.
A bit more notation:
We will use F  to represent the Laplace Transform of
Result 2
df
df
, so F  
dt
dt
F s   sF s   f 0 for a suitable function f(t).
[Suitable means that F (s ) and F(s) must exist for some s values and also “f must
be continuous from the right at t = 0”. We won’t worry about this at all.]
Proof: [Not examined]
F (s ) 





0
0

e st
df
dt
dt


 
d st
 d st

 dt e f (t )  dt e f (t )dt





 e st f (t ) 0    se st f (t )dt
0

 e  F ( )  e 0 F (0)   e st f (t )dt
0
 F (s )  sF (s )  F (0)
Normally we will drop the explicit reference to s and write
F   sF  f (0)
Results 1 and 2 can be used to turn differential equations into algebra.
Example 1 - a simple example using Results 1 and 2
Solve the first order differential equation,
dy
 3y  1
dt
y (0 )  1
[We saw equations like this last semester].
If Y and Y  are the Laplace Transforms of y and
Results 1 and 2,
Y   sY  y (0)  sY (s )  1
dy
respectively then, using
dt
(Result 2 using Y not F)
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and Laplace Transforming the original differential equation
Y   3Y  0
 sY  1  3Y  0
We can rearrange this to find Y;
s  3Y  1  0
 s  3Y  1
Y 
Looking at the Table of Transforms, 1
1
s  3
s  3 is the transform of
e 3t .
i.e. y(t) = solution of the DE = e 3t
This example shows the method for solving differential equations,
Original
DE
Laplace Transform
Algebraic Equation
for Y in terms of s
DE
Rearrange
Look up y(t) in table
Y=
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Example 2
Changing the previous example slightly, suppose we want to find y where,
dy
 3 y  e t ,
dt
y (0 )  1
….(1)
This is an example of an inhomogeneous differential equation. The term e t is
called a driving term, and represents a time varying input to the system described
by
dy
 3y  0
dt
To solve (1) we again take Laplace Transforms.
On the left hand side of (1),
On the right hand side,
dy
 3y  Y   3Y  sY  y (0)  3Y
dt
 sY  1  3Y
 (s  3)Y  1
e t  1
So we have
s  1
s  3Y  1  1s  1
 s  3 Y 
1
1
s  1
1  s  1

s  1
s2

s  1
s  2
Y 
s  1s  3
This is the Laplace Transform of the solution y(t) we require. There is a problem
however, and that is that this Y is not in our table. To proceed, we need the
method of partial fractions, which will allow us to rewrite Y in a form we can use.
Laplace Transforms 12
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RATIONAL FUNCTIONS TECHNIQUE:PARTIAL FRACTIONS
Often it is necessary to break down a complicated rational function of the form
P(s)
Q(s)
(where P(s) and Q(s) are polynomials in s, and the degree of the top polynomial is
less than the degree of the bottom polynomial), into the sum of simpler fractions
called PARTIAL fractions.

The type of partial fraction that you use depends on the FACTORS OF THE
BOTTOM POLYNOMIAL.
We will look at 3 cases:
Case 1:
All the factors of the bottom Q(s) are linear and nonrepeating.
Case 2:
Q(s) has some repeated linear factors.
Case 3
Q(s) has some irreducible quadratic factors.
Examples of case 1:
s
(s  2)(s 1)
Examples of case 2:
2s  1

(s  8) 2
Examples of case 3:

2s  1

or
3  2s
s(s  5)(s  3)
or
2

s(s  3)2
or


s s2  4
or
3
s 1
 
(s  3)  1(s  2)
2
2
s  1 (s  2)
s 2 1
or
s  3 (s 1)
2

or
3
4s 2  3s 1
2 
2
2
(s 1) s 2  3  s  1  9



 
We will start with the most common case, case 1:


 linear and non-repeating.
Case 1: All the factors of the bottom Q(s) are
CASE 1: All the bottom factors are linear (i.e. of the form x  some number) and
then is no repeated factor (i.e. there is no factor which is squared or cubed, etc.)
and there are no irreducible quadratic terms (don’t worry about this!). In this case
therefore, we are talking about rational functions of the form:
top polynomial
( x  a )( x  b )...( x  g )
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In this case we can rewrite the rational function as follows:
top polynomial
A
B
G

 .... 
=
( x  a )( x  b )...( x  g )
x a x b
xg
For example :
5s 1
A
B


(s  2)(s 1) s  2
s 1
Example

1
 14
1
Show that

 4
(s  7)(s  3) s  7 s  3
Solution
This is a case 1 partial fraction so we start with
1
A
B


(s  7)(s  3) s  7 s  3
Step1: Remove Fractions
Multiply both sides of the equation by the denominator on the left hand side
1  (s  7 )(s  3) A(s  7 )(s  3) B(s  7 )(s  3)


(s  7 )(s  3)
s7
s3
 1  A(s  3)  B(s  7 )
Step2: Choose s values to find A and B
The equation above is true for all values of s. We can choose s values to make
things simple:
Choose s = -3 so that (s+3) = 0 and we have
1
1  A(0)  B( 4)  1  4B  B 
4
Choose s = -7 so that (s+7) = 0 and we have
1
1  A( 4)  B(0)  1  4 A  A  
4
Step3: Substitute A and B into the original expression
1
1
1
 4  4
(s  7)(s  3) s  7 s  3
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Example
Write
3s 1
in partial fraction form
(s  1)(s  2)
Solution
 Here P(s) = 3s-1 has degree 1 and Q(s) = (s+1)(s+2) has degree2
3s 1
A
B


Therefore we can write
(s  1)(s  2) s  1
s2
Multiply both sides by (s+1)(s+2):

3s 1
A
B
(s  1)(s  2) 
(s  1)(s  2) 
(s  1)(s  2)
(s  1)(s  2)
s 1
s2
 3s 1 A(s  2)  B(s 1)

If choose s = -2
 3 2 1 A(2  2)  B(2 1)

 7  A.0  B(1)
 7  B
B7


 s = -1
If choose

 3 1 1 A(1 2)  B(11)
 4  A(1)  B(0)
 4  A
 A  4

Therefore

3s 1
4
7



(s  1)(s  2) s  1
s2

 
 
Example

Show that
s4

5
4

1
4
s 2  8s  12 s  6 s  2
(Hint: Factorise the left hand side denominator first)
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Example
The transfer function as part of an aircraft control system is
5s  3
.
s  5s  6
Use the technique of partial fractions to find the values A and B needed to write
A
B

the transfer function in the form
.
s3 s2
2
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Example
Find the fixed constants A, B, C so that the partial fraction decomposition can be
4s  3
A
B
C



completed:
s(s 1)(s  3) s
s 1 s  3
Solution:
4s  3
A
B
C



s(s 1)(s  3) s
s 1 s  3

Multiplying both sides of this equation by the left hand side denominator we
deduce that,

4s  3
A
B
C
s(s  1)(s  3)
s(s  1)(s  3)  s(s  1)(s  3)
 s(s  1)(s  3)
s(s  1)(s  3)
s
s 1
s3
 4s  3  A(s 1)(s  3)  Bs(s  3)  Cs(s 1)
Now in turn, put , s= 0 , s=1, s = -3. You will find at each stage that 2 of the A, B, C
terms will vanish:

If choose s = 0
4(0)  3  A(0  1)( 0  3)  B(0 )  C(0)
 3  3A
 A 1
If choose s = 1

4(1)  3  A(0)  B(1)(1  3)  C(0)

 7  4B
7
B
4
If choose s = -3

4 3  3A(0 ) B(0 )  C(3)( 3  1)
 9  12C
9
B
12
Therefore
:

 1  74   43 
4s  3

s(s  1)(s  3) s s  1 s  3
1
7
3
 

s 4s  1
4s  3 
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Example
Find the fixed constants A, B, C so that the partial fraction decomposition can be
completed:
 2s  1
A B
C
 

s(s  1)(s  2) s s  1 s  2
Solution
 2s  1
A B
C
 

s(s  1)(s  2) s s  1 s  2
Multiplying both sides of this equation by the denominator s(s –1)(s + 2) we deduce
that,
 2s  1 A(s  1)(s  2)Bs(s  2)Cs(s  1)
The latter equation holds for all values of the variable s. Hence, by choosing
appropriate values of s we may easily determine A, B and C. For s  0 , s  1 and
s  2 we get,
If choose s  0  1  A( 1)( 2)  A  
If choose s  1   1  B(3)  B  
1
,
2
1
,
3
If choose s  2  5  C( 2)( 3)  C 
5
.
6
Thus,
 2s  1
11 1 1  5 1 
   
 

s(s  1)(s  2) 2  s  3  s  1  6  s  2 
Example
Find the fixed constants A, B, C so that the partial fraction decomposition can be
completed:
s 2  4s  1
A B
C



s(s  1)(s  3)
s s 1 s  3
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Case 2:
Denominator Q(s) has some repeated linear factors.
Sometimes a linear factor is repeated twice or three times or four times, etc. This
means that we will have an expression like  s  somenumber  or
2
 s  somenumber 
3
or ... below the line. When this occurs, you have to be careful
as we have to use a partial fraction for EACH of the powers. If you want, you can
use the following table:
Factor in given rational
function
top polynomial
s  somenumber
Corresponding partial fraction
A
s  somenumber
A
s  somenumber
top polynomial
(s  somenumber ) 2
top polynomial
(s  somenumber ) 3
+
B
(s  somenumber ) 2
B
C
A
+
+
2
s  somenumber
(s  somenumber ) (s  somenumber ) 3
Examples of case 2:
4s 2  3s  2
A

(s  1) (s  3) (s  1)
2
3s 4  7s 3  2s 2  5
s (s  1) (s  5)
3
2
A

s
3

B
s
2

2

B
C
, A, B,C  constants ;

s 1 s  3
C
D
E
F



2
s (s  1) (s  1) (s  5)
A, B,C, D, E, F  constants
Example:
Write in terms of partial fractions
8s  1
 s  1
2
Solution:
In this case
8s  1
 s  1
2

A
B

s  1 (s  1)2
Step1 we multiply both sides by (s  1) 2 to remove fractions
8s  1
A
B
(s  1) 2 
(s  1) 2
2
s 1
(s  1)
(s  1)
 8s  1  A(s  1)  B
(s  1) 2 
2
Step2: we can choose s = 1 as before so that (s-1) = 0 and we get
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8 1  B  B  7
We cannot choose another s value to directly find A however. There is more than
one approach to finding A but the easiest method is called “equating
coefficients”. In this case, we note that there must be the same “amount of s” on
both sides of the equation. On the left hand side we have 8s and on the right we
have As, so that A must be 8.
Step3; don’t forget to write the answer down
8s  1
s  12

8
7

s  1 (s  1) 2
Example
Write
2s  1
in partial fraction form.
s 1 (s  2)2
 
Solution:
As the linear factor (s+2) repeats (i.e. we have (s+2) and (s+2)2 in the denominator)
the required partial fraction is of the form,

2s  1
A
B
C



2
s  1(s  2) s  1 s  2 s  22
where A, B and C are fixed constants which are to be found.
Step1: multiply both sides by the denominator (s – 1)(s+2)2
s  1(s  2) 2
2s  1
s  1(s  2)
2
 s  1(s  2) 2
A
B
C
 s  1(s  2) 2
 s  1(s  2) 2
s 1
s2
s  22
 2s  1  (s  2)2 A  s  1(s  2)B  s  1C
…(1)
The latter equation holds for all values of the variable s.
Step2: Choose good s values
Choose s = -2
 2 2  1  (0) 2 A  (0)B   2  1C
 3  3C  C  1
Choose s = 1
 21  1  (1  2) 2 A  (0)B  (0)C
 3  9A  A 
1
3
We again have the problem of not being able to choose a good s value to find B. We
again “equate coefficients”. The best strategy is to equate the highest power of s
on both sides, which is s2 . On the left hand side of (1) we have 0 lots of s2. On the
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right hand side of (1) the (s  2) 2 A term will contribute As2 if you multiply it out
and the (s  1)(s  2)B term will contribute Bs2. So we have
1
 B  A  
0  AB
3
An alternative method of finding B is simply to choose another s value, like s = 0

 2 0 1 (0  2)2 A  (0 1)(0  2)B  (0 1)C
1 4A2B1C
But as we already know C = 1 and A 


1
 1  4  2B 11
3

 1
1
 2B
3
1
then
3
B
1
3

This method can involve more calculation though. Either way



2s  1
1 1
1 B
1
 .
 .

2
3 s 1 3 s  2 s  2 2
s 1 (s  2)
 


Example :

Show that
2s  1
(s  2)s  32
3
3
7
5
 25  25 
s  2 s  3 s  32

Solution
2s  1


(s  2) s  3
2

A
B
C


s 2 s 3 s 3 2



Laplace Transforms 21

IT Tallaght Maths Group: CC BY-NC-SA 3.0
Case 3
Denominator Q(s) has some irreducible quadratic factors.
Irreducible means that the quadratic term in the denominator cannot be factorized
into two brackets.
3s 2  2s  5 A Bs  C
Examples of case 3:
, A, B,C  constants
  2
s s 4
s(s 2  4 )
s3  s  2
and

A
s(s  1)(s  2) s
2

B Cs  D

s 1 s2  2
Example:
5s 2  7s  8
Write
in partial fractions.
s 1 (s 2  2s  5)
 
Solution
The denominator contains an irreducible quadratic term (i.e, it cannot be easily
factored into two linear terms. – If it did factor into linear factors then we would
 be back to cases 1 or 2.) As it does not in this example we must write :
5s 2  7s  8
A
Bs  C


2
(s  2s  5)
s 1 (s 2  2s  5)
s 1
 
 
Note: Linear term on bottom means constant A on top. Quadratic term on bottom
means linear term Bs + C on top.

As before to find these constants multiply both sides by the denominator
( s – 1) (s2-2s+5).
 
s 1 (s 2  2s  5)
5s 2  7s  8
A
Bs  C
 s 1 (s2  2s  5)
 s 1 (s2  2s  5) 2
2
(s  2s  5)
s 1 (s  2s  5)
s 1
 
which gives:
 
 
 
5s 2  7s  8 (s 2  2s  5)A s  1Bs  C 
The latter equation holds for all values of the variable s. So we can choose any
values for s and set up three simultaneous equations for A, B and C.
By putting s = 1 we can get one value easily:
Choose s = 1
 512  71  8 (12  21  5)A 0B(1)  C 
 6  4 A 0
3
 A
2
We cannot make any more brackets = 0 by a good choice of s. As for case 2
however, we can equate coefficients. Start with the highest power s2 first
Laplace Transforms 22
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equate s2
equate s
 5  A  B  B  5  A  B  5  32 
  7  2 A  B  C
7
2
  7  2( 32 )  72  C   7   13
 C  C   12
2
[Alternatively, we could choose different s values, though this normally involves
more fractions and hence more work for you!
3
 502  70  8  (0 2  0  5)  0  1B 0  C 
2
15
C
2
1
  C
2
 8
Now choose another value for s, s = 2 and using A = 1.5 and C = -0.5

3
  1 
 522  72  8  ( 2 2  2( 2)  5)   2  1 B( 2)    
2
 2 

2
1
 14   2B 
2
2
 14  7  2B
 7  2B
7
 B
2
]
7 s 1
2  2
2
2
2


s

1
s  1(s  2s  5)
(s  2s  5)
5s 2  7s  8
So

3
Examples:
5s 2  2s  7
Write
in partial fractions.
s  1 (s 2  4)
 
5s 2  2s  7
2
3s  1

 2
Solution: Check the solution is
2
s  1(s  4) s  1 (s  4)

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The point of partial fractions is that the fractions we get are all in the Laplace
Table and we can find their inverse transform. This lets us solve complex
differential equations.
Note: There are many “partial fraction calculators” on the web. Just Google on
“partial fraction calculator” and let the web find partial fractions for you! You can
also Google “Laplace Transform calculator”…
Example
An electrical supply, with a time varying current te 2t is applied to a circuit with a
capacitor C. The current i at time t in the circuit satisfies,
di
 3i  te  2t
dt
, i ( 0)  0
(a) If I is the Laplace Transform of i show that,
I
1
s  3s  22
(b) Write I in partial fractions and hence find i(t).
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Inverse Laplace in this Example
In a previous example we got Y 
Using the Laplace Table,
3s  1 .
2
 2
s  1 s  4


2
 2e t .
s  1


The second fraction is not directly in the table. We look at the s 2  4 term on
the bottom and find it in the Laplace Table. We find,


s 4 
2
 both have s  4 on bottom.
s 
cos2t   2
s  4 
sin2t  

2
2





Now we want to write the top line in the problem, 3s  1 , in terms of the two top
lines “2” and “s” from the table and put.
3s  1  H (2)  K (s )
This is like our partial fractions problems:
  1  H (2)  K   21
Choose s = 0
3K
Equate s
So that
3s  1 H (2)  K (s )
2
s

  12 . 2
 3. 2
2
2
s 4
s 4
s 4
s 4








  12 sin( 2t )  3 cos(2t )
So the inverse of Y is,
y (t )  2e t  1 2 sin( 2t ) 3 cos( 2t )
Laplace Transforms 25
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Example 3
Find the inverse Laplace Transform of Y 
Solution

s2
s
2
9


We look at the s 2  9 term on the bottom and find it in the Laplace Table. We
find,


s 9 
2
 both have s  9 on bottom.
s 
cos3t   2
s  9 
sin3t  

3

2




Now we want to write the top line in the problem, s  2 , in terms of the two top
lines “3” and “s” from the table and put
s  2  H (3 )  K (s )
Choose s = 0
Equate s
So that
 2  H (3 )  H 
 1 K
2
3
s2
H (3 )  K (s ) 2
3
s

 3. 2
 1. 2
2
2
s 9
s 4
s 9
s 9
 2 3 sin( 3t )  cos(3t )




Example 4
Find the inverse Laplace Transform of Y 

2s  1
s
2
4




(Hint – head towards line 11, 12 in the Table like you did for lines 7, 8 in e.g. 3).
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Example 5
Find the inverse Laplace Transform of Y 
Solution

5s  2
s  22  9

We look at the (s  2)2  9 term on the bottom and find it in the Laplace Table.
We find,


(s  2 )  9 
2
 both have (s  2)  9 on bottom.
s2

e 2t cos3t  
2
(s  2)  9 
e 2t sin3t  
3


2




Now we want to write the top line in the problem, 5s  2 , in terms of the two top
lines “3” and “s - 2” from the table and put
5s  2  H (3)  K (s  2)
This is like our partial fractions problems:
 8  H (3)  H  83
Choose s = 2
5K
Equate s
So that
5s  2
H (3)  K (s  2) 8
3
(s  2)
Y


5
2
2
2
3 s  2  9
s  2  9
s  2  9
s  22  9
 8 3 e 2t sin( 3t )  5e 2t cos(3t )
Example 6
Find the inverse Laplace Transform of Y 
s
3s  2
2

 6s  13
Solution
In this example, there is one quadratic factor on the bottom line. The first thing to
do is complete the square for this factor,
 s  3  3 2  13
2
i.e. s 2  6 s  13  s  3  9  13
2
 s  3  4
2
 Y (s) 
3s  2
s  3
2
4

This can now be done like the previous example (get H = 3 and K   72 ).
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Example 7
Find the inverse Laplace Transform of Y (s ) 
s
4s  1
2

 2s  3
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Example 8
Find the inverse Laplace Transform of Y 
5s 2  7s  8
s  1s 2  2s  5
Solution
Using partial fractions we get Y 
3
The first term is easy:

2
s  1
3
3
2
s  1
2e
t

7 2 s  12 
s
2
 2s  5

from the Table.


For the second term, we again look for the denominator s 2  2s  5 in the table.
Unfortunately it is not there, and we must convert it using completing the square.
s 2  2s  5  s  12  12  5  s  12  1  5  s  12  4
Now,
s
7
2
2s
 12
 2s  5


7
2s
 12
s  12  4
As in the previous example, we can find this denominator in the Table and find,
e t sin( 2t ) 
2
e t cos( 2t ) 
s  12  4
As before, we need to write our top line
“2” and “(s - 1)”.
7
2
7
s
s  1
s  12  4
s  1 2 in terms of the top lines above,
s  12  H (2)  K (s  1)
3 ( 2)
s  1
2



2
2
s  1  4 s  1  4 s  12  4
 s  1  3
2
 72 
 .

2
2
 s  1  2 2  2 s  1  2 2
7
2s

1
2
7
2
3
 7 2 e t cos( 2t )  e t sin( 2t )
2
So inverse of Y is y (t ) 
3 t 7 t
3
e  e cos2t   e t sin2t 
2
2
2
Laplace Transforms 29
IT Tallaght Maths Group: CC BY-NC-SA 3.0
More Laplace Transform Examples
Example 1
What is the inverse Laplace Transform of Y 
2s
.
s  4
Solution
Again, this is not in the tables directly due to the s term on the top line. for this
kind of problem we must write the top line as “a multiple of the bottom line plus a
number”
2s  A(s  4)  B
i.e.
We have seen this kind of thing before so
  8  A(0)  B  B  8
Choose s = -4
Equate s
2 A
So we get
2(s  4)  8 2(s  4)
8
8
Y


 2
s  4
(s  4)
(s  4 ) ( s  4 )
Now we can use the table, Y  2 (t )  8e 4t
Examples
Find the inverse Laplace Transforms of,
(a) Y 
3s  1
s  2
(b) Y 
s 3
s  4
Laplace Transforms 30