INSTITUT TEKNOLOGI BRUNEI
DEPARTMENT OF ELECTRICAL AND COMMUNICATION
ENGINEERING
HIGHER NATIONAL DIPLOMAS IN
ELECTRICAL POWER AND CONTROL ENGINEERING
AND
COMMUNICATIONS AND COMPUTER SYSTEMS ENGINEERING
Course Notes
ENGINEERING MATHEMATICS B
1
Engineering Mathematics B
10201
Summary of Aims
To enable students to progress and gain skills in the mathematical tools and approaches required by the
Engineering application areas including the formulation and solution of first and second order
differential equations, the manipulation of complex numbers in cartesian, polar, and exponential
forms, the definition and application of the Normal distribution, and an introduction to Bessel
functions, matrices and vectors.
Principal Objectives
The expected learning outcome is that the student:
A
First Order Differential Equations
1
Formulates and solves problems that can be represented by first order differential equations and
investigates their application to the transient behaviour of linear circuits.
a.
b.
c.
B
Determines and sketches a family of curves given their derivative (simple functions) and specifies
particular curves from boundary conditions. Derives equations of the type
dQ
 kQ
dt
and solves them.
Uses the appropriate terminology and formulates differential equations from simple physical
problems.
Solves first order differential equations by using techniques of direct integration, separation of
variables and Laplace Transform.
Second Order Differential Equations and Laplace Transforms
2
Formulates problems that can be represented by second order linear differential equations with
constant coefficients. Solves them using Laplace Transforms. Investigates their application to the
transient behaviour of linear circuits.
a.
b.
c.
d.
e.
Defines the Laplace transform of f(t).
Uses tables of Laplace Transforms.
Determines the inverse Laplace Transforms of simple forms using a list of transforms and partial
fractions.
Uses Laplace transforms for the solution of differential equations with constant coefficients and given
initial conditions.
Applies Laplace transforms to the transient behaviour of linear circuits.
2
C
Algebra
3
Extends his knowledge of the applicability of Complex Numbers, Hyperbolic Functions, Matrices
and Bessel Functions.
a.
b.
c.
d.
e.
f.
g.
h.
D
Vectors
4
Defines vectors and scalars and applies vector algebra in an Engineering context.
a.
b.
c.
d.
e.
E
Determines roots of a complex number.
Quotes and uses the exponential form of complex numbers. Relates it to transmission line
applications.
Expresses hyperbolic functions in exponential form and determines functions and inverses of real and
complex numbers.
Relates Bessel functions of the first kind to the Fourier coefficients in the Fourier series expansion of
a frequency modulated wave.
Determines values of J0 , J1, etc using graphs, tables and a computer algebra package for different
values of the modulation index
Recognizes and knows how to apply matrix notation, using standard electronics examples, to the
solution of systems of linear equations with two unknowns.
Solves linear equations using the Gauss-Seidel method.
Solves equations using the Newton-Raphson method.
Defines a vector and a scalar.
Applies vector addition and subtraction.
Defines unit vector and rectangular unit vectors i, j, and k.
Distinguishes between vector and scalar fields.
Defines scalar and vector products.
Statistics
5
Defines the Normal distribution and applies it in Engineering situations.
a.
b.
c.
d.
e.
Relates mean and standard deviation to the Normal distribution.
Defines the standard Normal distribution.
Converts non-standard distributions into standard form.
Uses tables of the standard Normal distribution to solve problems in an Engineering context.
Distinguishes between the standard Normal distribution and the normal error function.
3
A
Differential Equations
Introduction
A differential equation is any equation involving an unknown function and its
di
derivatives. Very often the derivatives are taken with respect to time, as in
, or with
dt
dy
respect to distance, as in
.
dx
What do differential equations look like? Here are two typical examples of differential
equations
di
L  iR  E
(1)
dt
(2)
LC
d 2v
dv
 RC  v  E
2
dt
dt
Differential equations must contain derivatives and may contain any other functions of
the variables, or their derivatives and constants.
In equation 1 for example the unknown function (i.e. the dependent variable) is the
current i, the independent variable is the time t and L, E and R are given constants.
1.1 Order of a differential equation
di
 iR  E involves the first derivative with respect to time of an
dt
unknown time-dependent current i. This equation (1) is said to be a first order
differential equation.
The equation L
d 2v
dv
 RC  v  E involves both the first and second derivatives of
2
dt
dt
the unknown time-dependent voltage v. This equation is a second order differential
equation
The equation LC
In general the order of a differential equation is the order of the highest derivative of
the unknown function involved in the equation. In Engineering Mathematics B we shall
only consider first and second order differential equations as these are the ones most
frequently encountered.
Exercise 1
State the order of the following differential equations
(a)
d 3v
dv
4 3 2 v 5
dt
dt
(b)
dv
 10v
dt
(c)
4 y ' '  2 y '  xy 2
(d)
v' '  v  sin(t )
4
Answers: (a) third order (b) first order (c) second order (note in this equation the
dy
d2y
independent variable is x so y ' ' stands for
and y ' stands for
2
dx
dx
(d) second order but this time v ' ' stands for
d 2v
since the independent variable is now t
dt 2
not x
1.2 Linear and non-linear differential equations
Equations (1) and (2) on page 4 are called ‘linear’ equations because the unknown
dependent variable and its derivatives only appear in the equation to power 1 with
coefficients that are either constant or functions of the independent variable only (i.e.
products of derivatives are not allowed) . Equations where this is not so are called
‘non-linear’. Non-linear equations are difficult to solve symbolically and in this course
we shall only deal with linear equations.
For example both of the equations below are non-linear:
2
 di 
L    iR  E (first order non-linear)
 dt 
d 2i 2
L 2 i R  E
dt
(second order non-linear)
What’s so special about linear equations? The crucial thing about linear equations is
that you can add two solutions together to a third solution. i.e if i1 (t ) and i2 (t ) are both
solutions of a certain differential equations then a i1 (t )  b i2 (t ) is also a solution (a and b
are any constants). This is not the case for non-linear equations.
Exercise 2.
Which of the equations in Exercise 1 are linear equations?
Answer (a), (b) and (d)
1.3
Formulation of differential equations
1.3.1 Charging an inductor using a direct voltage supply (LR circuit)
The differential equation (1) on page 4 may be obtained by applying Kirchoff’s law to
the circuit shown in Figure 1 below: We assume that the switch is closed at time t = 0,
at which time the current flowing in the circuit will begin to increase, driven by the
di
applied voltage source. . At time t the voltage drop across the inductor is L
and
dt
across the resistor it is iR and together these voltage drops must add up to balance the
applied emf E, hence we obtain the differential equation describing the circuit
di
behaviour viz. L  iR  E
dt
5
i
E
Figure 1
You will need know how to solve the differential equation (1) to find an expression for
the current in the circuit as a function of time. The precise method of solution will be
covered in detail in a subsequent lecture but for now we may simply observe that one
E
particular solution to Equation (1) is i  (1  e Rt / L ) . You should check that this is
R
indeed a valid solution by directly substituting it back into equation (1).
1.3.2
Series LCR circuit with a direct voltage supply
In Equation (2) on page 4 v is the voltage across the capacitor at time t in a series LCR
circuit connected to a dc supply E volts,as shown in Figure 2 on the next page. The
switch will be assumed to be closed at time t = 0
E
Figure 2
The differential equation (2) on page 4 governing the circuit behaviour will now be
derived using Kirchoff’s law as follows:
6
Taking q to be the charge on the capacitor at time t and v to be the corresponding voltage
across it we know from the definition of capacitance that q  Cv and from the definition
dq
of current we also know that i 
.
dt
Combining these two expressions we can see that the current in the circuit may be
dq d (Cv)
dv

C
expressed in terms of the capacitor voltage by i 
dt
dt
dt
We can now substitute this expression for i into the resistor voltage drop iR to find that
dv
iR  RC
.
dt
di
dv
The inductor voltage drop is L and again by using i  C
we can write
dt
dt
di
d
dv
d 2v
L  L (C )  LC 2 .
dt
dt
dt
dt
Finally we now apply Kirchoff’s laws to balance the applied EMF and the three voltage
drops to obtain
LC
d 2v
dv
 RC  v  E
2
dt
dt
You will need to know how to solve the differential equation (2). The precise method of
solution will be covered in detail in a subsequent lecture and involves the use of Laplace
Transforms.
1.4
General Solutions and Particular Solutions of Differential Equations.
dy
 2 , This tells us that any
Consider the very simple first order differential equation
dx
function of x that has a constant slope equal to 2 will be a solution of this equation.
Therefore there will be many solutions to this equation, here are just a few of them:
y  2 x , y  2 x  1 , y  2 x  2 y  2 x  3 etc….,
in fact any function of the form y  2 x  c will be a solution, where c is an arbitrary
constant ( and c can be any number – not just a whole number). A few solutions are
plotted below in Figure 3 and we can see they form an infinite ‘family’ of solutions that
are all parallel straight lines with the same slope or gradient equal to 2.
7
10
8
2x
6
2x 1
2x 2
4
2
0
0
1
2
3
4
x
Figure 3
dy
 2 is y  2 x  c . The
dx
general solution of a first order differential equation will always contain one arbitrary
constant and for an nth order differential equation the general solution will always
contain n arbitrary constants.
We say that the General solution of the differential equation
The individual solutions drawn in Figure 3 are called Particular solutions. Whenever
we decide to choose a particular value for the arbitrary constant c we will obtain a new
particular solution. A Particular solution doesn’t contain any arbitrary constants.
To see how this works in practice suppose we have solved the first order differential
equation (1) to find the current in the circuit and we have obtained the general solution
containing one arbitrary constant:
i (t ) 
E
 ce  Rt / L
R
(general solution - c is the arbitrary constant)
Suppose we also know that in this particular case the current was zero when the switch
was closed at time t = 0. We say that the initial condition is given by
i (0)  0 .
Knowing the initial value of the current is zero enables us to select the value of the
arbitrary constant and determine the particular solution satisfying the initial condition as
follows:
Substituting t = 0 into the general solution i (t ) 
c
E
E
 ce  Rt / L we find 0   c so that
R
R
E
. The particular solution will therefore be given by
R
i (t ) 
E
(1  e  Rt / L )
R
(particular solution – contains no arbitrary constants)
8
This is the only solution that satisfies both the differential equation and the initial
condition.
Summary:
i (t ) 
i (t ) 
E
 ce  Rt / L = general solution
R
E
(1  e  Rt / L ) = particular solution satisfying i (0)  0
R
Exercise 3: Given E = 9 volts, R = 1 ohm and L = 100 mH find the particular solution
if we know the initial condition is i (0)  1 A.
Answer:
General solution: i (t ) 
E
 ce  Rt / L  9  ce 10t
R
Initial condition: i (0)  1
Therefore 1  9  ce0  9  c , hence c  8 and the particular solution is given by
i(t )  9  8e10t . The particular solution is plotted below in Figure 4 on the next page
(notice the graph starts at t = 0 and i = 1 as expected.)
9
6
 10t
9 8e
3
0
0
0.13
0.25
0.38
0.5
t
Figure 4
Summary:
i (t ) 
E
 ce  Rt / L  9  ce 10t = General Solution
R
i(t )  9  8e10t = Particular Solution satisfying i (0)  1
9
2. Methods for solving first order differential equations
2.1
Direct Integration
Some first order differential equations may be solved by directly integrating both sides
of the equation with respect to the independent variable. If the independent and
dy
dependent variables are x and y respectively this method only works if
can be
dx
expressed directly in terms of x and constants only with no occurrences of y.
In general it may be possible to use direct integration for first order equations of the
form
dy
 f ( x) .
(3)
dx
It will of course only be possible to use the direct integration method if the function
f ( x ) on the right-hand side of equation (3) can be integrated ( remember not every
function can be integrated symbolically)
By integrating both sides of equation (3) we obtain
dy
 dx dx   f ( x)dx
but since integration is the reverse of differentiation it means that
dy
 dx dx  y and the
solution of the differential equation (3) is thus y   f ( x)dx
Example 1. Solve the differential equation
dy
 2x
dx
Answer. By direct integration of both sides we find y   2 xdx  x 2  c .
This is the general solution of the first order differential equation and contains one
arbitrary constant, c.
dy
 4 cos( x) . Find (i) the general solution
dx
and (ii) the particular solution satisfying the boundary condition that y (0)  1
Example 2. Solve the differential equation
(note when a differential equation is expressed in terms of some distance variable x
instead of a time variable t we usually call the given condition a ‘boundary condition’
rather than an ‘initial condition’ because the word ‘initial’ usually means ‘at the
beginning time’)
Answer: (a) General solution by direct integration
y  4sin( x)  c
(b) Particular solution
y  4sin( x)  1
10
2.2
Separation of Variables
dy
is expressed in terms of y as long as the x’s and y’s can
dx
dy
 f ( x) g ( y )
be ‘separated’ in the sense that
dx
dy
y
Example 3
dx
This method can be used even if
Direct integration is not possible here because although the LHS would integrate to give y we
would still be unable to integrate the RHS until we know y in terms of x. i.e. we would need
the answer in order to find the answer !
Instead we treat dy and dx as separate terms which may be treated algebraically. We then
rearrange the equation so that all the y terms are on the left and all the x terms are on the right
obtaining the equation:dy
 dx
y
We now integrate both sides

giving
so that
dy
 dx
y 
ln( y )  x  c , where c is an arbitrary constant
y  e x C  e x ec  Ae x
where ec  A is also an arbitrary constant
dv 2
 v  0 and v = 5 when t = 0
dt
Example 4
dv
 v 2
dt
Rearranging

Separating variables
dv
 dt
v2
dv
Integrating
 v
which gives
1
t c
v
2
v
so that
  dt
1
(General Solution)
t c
Substituting the boundary condition 5 
case
v
1
t
1
5

1
so that c = 1/5 and in this particular
0c
5
(Particular Solution)
5t  1
11
3.
Further Examples on the Formulation of Differential Equations
3.1
RC Circuit: Discharge of Capacitor through a Resistor
Let the initial charge on the capacitor be Q0 when the discharge commences.(i.e. at time
t = 0) and at a later time t the charge is q.
q
C
R
i
Applying Kirchhoff’s Voltage Law to the circuit gives:
q
 iR  0
C
and as i 
dq
we have
dt
q
dq
R
0
C
dt
Rearranging
q
dq
 R
C
dt
Separating variables
dq
1

dt
q
( RC )

Integrating both sides
dq
1

dt
q
( RC ) 
which gives
ln(q)  
or, by taking antilogs
qe
or

t
c
( RC )
t
c
( RC )
q  Ae

e

t
( RC ) c
e
t
( RC )
Substituting the initial condition we find q  Q0e
(General Solution)

t
( RC )
(Particular Solution)
Note:
This circuit has been described by a differential equation which has been solved for q.
Other variables of the circuit (i, vC and vR , ) may be determined from their relation to q.
i
Q
dq
  0 et /( RC )
dt
( RC )
12
vC 
q Q0  t /( RC )

e
C C
vR  iR  
alternatively
Q0  t /( RC )
e
C
vC  vR  0 therefore vR  vC  
Q0 t /( RC )
e
C
In practice the easiest and most common measurement made on a circuit is that of voltage,
very often with the voltage displayed on an oscilloscope.
Measurements on the discharge of a capacitor would start with the initial voltage on the
capacitor, V0 from which the initial charge, Q0 0, could be calculated if required.
Let us repeat the analysis of this discharge, this time in terms of the most likely variable to be
measured, vC .
vC  iR  0
Applying Kirchhoff’s Voltage Law
Substituting
The equation becomes
i
dv
dq d
 (CvC )  C C
dt dt
dt
vC  RC
dvC
0
dt
Rearranging and separating variables
dvC
1

dt
vC
( RC )
Integrating
ln(vC )  
Applying initial conditions
ln(V0 )  c ln(V0) = c
Substituting
1
t c
( RC )
ln(vC )  ln(V0 )  
1
t
( RC )
vC
 e  t /( RC )
V0
Taking antilogs
vC  V0 e  t /( RC )
As before any other variable of the circuit may be found from its relation to vC .
13
3.2
RC Circuit: Charging of Capacitor through a Resistor
At time t = 0 let the capacitor be connected to a direct voltage supply, V volts through the
resistor, R. Let the initial charge on the capacitor be zero when the charging commences (i.e.
at time t = 0).
q
C
R
i
V
We know from experiment what should happen: A current will flow in the circuit to charge
the capacitor and the capacitor will begin to charge up. As the capacitor charges up it
produces an opposing voltage and the current gradually reduces. When the voltage across the
capacitor finally reaches V volts the charging current has dropped to zero.
It takes a finite time to charge the capacitor. The time taken depends on the size of the
resistor and the size of the capacitor. We shall see below by solving the differential equation
that it takes approximately a time 5RC seconds to fully charge.
The quantity RC is called the time constant of the circuit and has units of time in seconds. A
useful rule of thumb is that it takes about 5 time constants to discharge (or charge) a capacitor.
Applying Kirchhoff’s Voltage Law to the circuit gives:
and as i 
dq
we have
dt
Rearranging
Separating variables
Integrating
q
 iR  V
C
q
dq
R
V
C
dt
q
dq
V  R
C
dt
dq

1
dt
R
q
V
C
dq
1
 q   R  dt
V
C
which gives
C ln(
or
ln(
q
t
V )    c
C
R
q
t
c
V )  

C
RC C
14
t
t
t

 c'


q
c'
RC
RC
RC
( V )  e
e e
 Ae
C
c
Let c  and take antilogs
C
'
t

q
 V  Ae RC
C
or
(General Solution)
Substituting the initial conditions 0 = V + A
so that A = -V
t
t


q
RC
RC
 V  Ve
 V (1  e )
C
Therefore
q  VC (1  e
or

t
RC
)
(Particular Solution)
As before the other variables may be found from their relation to q.
vC 
t

q
 V (1  e RC )
C
vR  V  vC  V  V (1  e

t
RC
)  Ve

t
RC
t

vR V
RC
i
 (1  e )
R R
The figures below show graphs for the case V = 10 volts, R = 1 M  and C = 1  F. for this
case the time constant RC = 1 second so it takes about 5 seconds to charge up. The first graph
below shows the voltage across the capacitor rising with time and the second graph shows the
charging current slowly disappearing.
15
t


 ( R C) 
V  1 e

10
5
0
0
1
2
3
4
5
t
Capacitor Voltage
15
6
15
10
t
V e
( R C)
5
0
0
1
2
3
4
5
6
t
Charging current
3.3
3.4
RL Circuit : Switching out a dc supply at t = 0 when the series current is I0.
Applying Kirchhoff’s Voltage Law
vL  vR  0
or
L
di
 iR  0
dt
Rearranging and Separating Variables
di
R
  dt
i
L
Integrating
R
ln(i )     t  c
L
Taking antilogs
ie
 R
 t c
 L
i  I 0e
Applying Initial Conditions
16
 ec e
R
 t
L
 R
 t
 L
 Ae
 R
 t
 L
3.4 RL Circuit : Switching on a dc supply at t = 0 when the series current is
zero.
Consider connecting the RL circuit to a dc supply, V volts, at time t = 0.
vL  vR  V
This time
L
or
di
 iR  V
dt
Rearranging, etc
di
1
   dt
V  iR  L 
Integrating
 V  iR   L   dt
1
di
t
1
   ln V  iR    c
L
R
R
ln V  iR      t  cR
L
Rearranging
V  iR  e
Taking antilogs
Applying Initial conditions
R
 t cR
 L
 Ae
R
  t
 L
V = A
R
so that
or
 t
1
i    (V  Ve  ( R / L )t  Ae  L 
R
V 
i    (1  e  ( R / L )t )
R
The time constant for the LR circuit is thus L/R seconds and again it takes about 5 time
constants to stabilize the system, whether switching on or off.
The graph below shows the rise of current with time for the case where R = 100 ohms
and L = 100 mH. The time constant L/R = 1 millisecond = 0.001 seconds. Note that
after about 5 time constants the current has stabilised.
0.15

 R
V 
 1 e
R
t
L



0.1
0.05
0
0
0.002
0.004
0.006
t
current in the coil
17
3.5
Use of Laplace Transforms
The previous methods for the solution of first order differential equations, while very useful in
the special cases considered, cannot be applied generally and can only rarely be applied to
second order equations, Neither of these restrictions apply to the method using Laplace
Transforms that will be discussed in the next section
4.
Laplace Transforms.
Before defining these it is useful to remember that Algebraic Equations are generally much
easier to solve than Differential Equations and to note that the main feature of Laplace
Transforms, as far as we are interested, is that they convert Differential Equations into
Algebraic Equations.
(Compare this with the process of ‘taking logs’ which allows multiplication to be replaced
with addition by the appropriate use of log tables)
4.1
Definition of the Laplace transform
The Laplace transform of a function of time f (t ) is defined by the integral

 f (t )  0
f (t )e st dt
Carrying out the integration produces a function of the Laplace variable s. In symbols we
write
 f (t )  F (s)
In practice whenever we need to find the Laplace transform of a given function of time we
don’t need to use this integral definition because for all the important standard functions that
we will need this has already been done and the results presented in a table of Laplace
transforms. You will be provided with a table of Laplace transforms in end- and phase- tests.
1
. We obtained this answer from the
s
table but we will obtain the same answer by integration to convince you that the integration
method and the tables produce the same answer:
As a simple example the table shows that
1 
Using the integration definition for the case f (t )  1 we have


1
1
1
1
1  0 1e st dt   e st    e    e0  
 s
0  s
  s  s
18
4.2 The Inverse Laplace transform
Suppose we happen to know the Laplace transform of a function of time is some given
function of s, say F(s), and we want to know what the corresponding function of time is. We
say that to find the function of time is to find the Inverse Laplace transform of F(s). There
is a complicated integral expression for F(s) but we don’t need to know what it is because to
find either Laplace transforms or Inverse Laplace transforms we always use the same tables.
As a simple illustration the Laplace transform (using tables) of e t is
Laplace transform of
1
so the inverse
s 1
1
is just e  t
s 1
In symbols
L e  t  
1
s 1
for the forward transform. In words this says that the Laplace transform of et is
1
s 1
For the Inverse transform we can write
 1  t
L 1 
e
 s  1
In words this says that is e  t is the function of time whose Laplace transform is
4.3
1
s 1
Rules of Laplace transforms
There are 4 main features (which will be called the Rules of Laplace Transforms) that make
them useful:
Rule 1. L kf (t )  k L  f (t )
i.e. constants may be ‘taken outside’ the LT
Rule 2. L  f (t )  g (t )  L  f (t )  L g (t )
Rule 3. L  f ' (t )  s L  f (t )  f (0) .
Rule 4. L  f ' ' (t )  s 2 L  f (t )  sf (0)  f ' (0)
In rule 3. and rule 4. we have used the following notation:
f (0) means the value of f(t) when t = 0, i.e. f(t = 0)
df
f ' (0) means the value of
when t = 0 i.e. f ' (t  0)
dt
d2 f
''
f (0) means the value of
when t = 0 i.e. f '' (t  0)
2
dt
19
For instance if we had the function of time f (t )  3  2t  3t 2 then we would have
f (0)  3,
f ' (t )  2  6t
f ' (0)  2
f ' ' (t )  6
f ' ' (0)  6
4.4 Special Cases when all initial conditions are zero.
Rules 3 and 4 are particularly simple if all initial ( t = 0) conditions are zero i.e. if
f (0)  0, f ' (0)  0 and f ' ' (0)  0 because then we have as a special case:
Rule 3. L  f ' (t )  s L  f (t ) .
Rule 4. L  f ' ' (t )  s 2 L  f (t )
and it is seen that to find the Laplace transform of a first derivative of a function of time we
need only multiply the Lapalace transform of the function by s and multiplying by the square
of s gives the second derivative.
4.5 Examples of Laplace Transforms:
L 1 
1
(a unit step)
s
L k  
k
, where k is any constant
s
1
(unit ramp)
s2
n!
L t n   n 1
s
L t 
L e  at  
1
sa
An extensive list of Laplace Transform pairs is given in Appendix A
20
4.6 Application Example:Using Laplace Transforms
At time t = 0 a 6V dc supply is applied to an inductor of 4H in series with a resistor of 2
We are told that no current was flowing immediately before the voltage supply was connected
so we know the initial condition is that i = 0 at t = 0
4
The differential equation obtained is:
di
 2i  6
dt
 di

L 4  2i   L 6
 dt

Take Laplace Transforms of both sides
Apply the rules stated above
 di 
L 4   L 2i  L 6
 dt 
(using Rule 2)
 di 
4 L    2 L i  L 6 (using Rule 1)
 dt 
4s L i  2 L i 
Rearrange to find L i
L i 
6
s(4s  2)
L i 
3
1
2 s(s  1 )
6
s
(using Rule 3 and tables)
2
i  1.5 x 2(1  e0.5t )
Taking Inverse Transforms (Tables)
i  3(1  e0.5t )
or
This particular example could have been solved using separation of variables but if the
voltage source had been a function of t (e.g. sin(t) ) then it would not have been possible to
separate the variables. The Laplace transform method would still work though.
Let us now see how the Laplace transform method works for the case where the right-handside is sin(t):
The differential equation f or the case where the right-hand-side is sin(t) is
4
Take Laplace Transforms:
di
 2i  sin(t )
dt
 di 
4  +
 dt 
2i = sin(t)
21
 di 
4  +
 dt 
1
s 1
1
4s i + 2 i = 2
s 1
Apply the rules stated above
i =
2i =
2
1
1
4( s  )( s 2  1)
2
Taking the Inverse Transform will give the expression for i.
The solution of differential equations by this method relies on:
the application of the rules of Laplace Transforms,
the accurate application of the rules of conventional algebra,
proficiency in the use of the table of Laplace Transform pairs.(see Appendix A)
4.7
More on the Rules of Laplace Transforms
In Section 4.1 we saw the Laplace transform of a function of time f (t ) is defined by the
integral

 f (t )  0
f (t )e st dt
And in Section 4.3 we noted some of the rules of Laplace transforms. In this section we use
the definition of the Laplace transform to explain where these rules come from.
The first two arise because the Laplace Transform of a function of time is defined in terms of
an integration and are the same as the rules governing the process of integration.
Rule 1
If the function of time includes multiplying constants (e.g. 3t, 5sin(t), 2 e-t) these constants
may be ‘taken outide’ the integral and hence ‘outside’ of the Laplace Transform.

kf (t )  0

kf (t )e st dt  k  f (t )e st dt  k
0
Examples:using Rule 1:
[3t]
= 3
[t]
[5sin(t)] = 5
[sin(t)]
[2 e-t]
[ e-t]
= 2
22
 f (t )
Rule 2
The rules of integration tell us that the integral of a sum of two or more functions is equal to
the sum of the integrals of each function separately.
For example  t + sin(t) + e-t ) dt = t dt +sin(t) dte-t dt
Therefore because the Laplace Transform is defined as an integral the same rule applies to
Laplace Transforms. We can use this idea to prove Rule 2 as follows:
.we have

 f (t )  g (t )  0  f (t )  g (t )  e st dt


0
0
  f (t )e st dt    g (t )e st dt
 f (t )
=
hence
g (t )
 t + sin(t) + e-t ) =
.
Rules 3and 4
By definition
+
[t] +
[sin(t)]

 f (t )e
[f(t)] =

 st
'
dt and therefore
[ f (t ) ] =
0
du/dt = -se-st
then

'
[ f (t ) ] =
 f ' (t )e
 st
dt
0
Integrating by parts with u = e-st
so that
[e-t]
and dv/dt = f ' (t )
and v = f(t)
 f ' (t )e
 st
-st
dt = [e f(t)]
0

0

+ s  f ( t )e  st dt = -f(0) + s
[f(t)]
0
where f(0) = [ the value of f(t) when t = 0 ].
For the special case when f(0) = 0 this simplifies to
f ' (t ) ] = s
[f(t)]
Rule 3 can be used to obtain Rule 4 for the transform of the second derivative viz.


L f ' ' (t )  s 2 L  f (t )  sf (0)  f ' (0)
4.8
Hints on Using the Laplace Transform Tables
1
The tables show general cases and include algebraic symbols (a, b, c,  ,etc) in
place of the actual numbers for a particular transform or inverse transform. Thus
when using the tables look for the general form of the function.
Question:
Find the Laplace Transform of e -4t.
Answer:
e -4t. will not be found in the tables but e -at will.
Therefore since
(e -at ) = 1/(s + a) ,
(e -4t ) = 1/(s + 4)
23
substituting a = 4 gives
2
Multiplying constants may be ‘taken outside’ the transform or inverse transform.
Question:
Answer:
3
. Find the Laplace Transform of 3 e -4t
(3 e -4t) = 3 ( e -4t ) = 3/(s + 4)
The tables do not include any multiplying constants with s or s2 but in practice the
transform for which the inverse is required may contain such constants. These must be
converted into multiplying constants for the whole transform (not just the s or s2 term)
and may then be treated as in note 2 above.
Question:
Find the inverse transform of 1/(2s + 3)
Answer:
Take out the factor 2 and rewrite as
Then f(t) =
4
1
2
( inverse transform of
1
3
2( s  )
2
1
3
(s  )
2
=
1

2
)=
1
3
(s  )
2
1
2
 e -(3/2)t
In some cases the tables give the transform as a general expression and it is then
necessary to do some preliminary work to find what the tables do give in the particular
case required.
Question:
Find the Laplace Transform of t2
Answer:
Powers of t are given by the expression
1
t n-1
(n  1)!
Thus for t2 we take must n = 3 and the expression reduces to
The tables therefore show that the Laplace Transform of
1
2
t2 is
1
2
t2
1
s3
Thus
(t2) =
(2 
1
2
t2 ) = 2
1
( t2 )
B
Second Order Differential Equations
5.1
Introduction
2
=
2
s3
Second order equations arise from the analysis of circuits containing both capacitance
and inductance. The formulation of these equations again makes use of the application
of Kirchhoff’s Voltage Law.
24
5.2
LCR Series Circuit : Switching on with i = 0 and q = 0 at t = 0
Applying Kirchhoff’s Voltage Law we have L
di q
  iR  V
dt C
In terms of the voltage on the capacitor, v , we know that q = Cv
Thus the current i = dq/dt = C dv/dt and di/dt = C d2v/dt2
On substitution we obtain the second order differential equation in terms of v
LC
d 2v
dv
 RC  v  V
2
dt
dt
To solve this equation we take Laplace Transforms of both sides and apply Rules 1-3
LCs 2 L v  RCs L v  L v  L V  
L v  LCs 2  RCs  1)  
L v 
L v 
V
s
V
s
V
s  LCs  RCs  1) 
2
V
R
1 

LCs  s 2  s 
)
L
LC 

What follows depends on the form of the quadratic obtained when the values of L, C and R are
R
1
substituted into the quadratic term: s 2  s 
. We pause here to revise the three types of
L
LC
roots that a quadratic equation can have.
25
5.3 Revision of Quadratic Equations
The quadratic equation
has the solutions
ax2 + bx + c = 0
b  b 2  4ac
x=
2a
If we denote the two solutions by x1 and x2 we can then express the quadratic term
ax2 + bx + c in the form of a product
a(x - x1 )(x - x2 )
x1 and x2 are called the roots of the quadratic and are given by
x1 =
b  b 2  4ac
2a
and
x2 =
b  b 2  4ac
2a
There are three possibilities:
The roots may be real and unequal, real and equal, or pure imaginary. The type depends on
the value of b 2 - 4ac.
If b 2 - 4ac is positive the roots are real and unequal:
If b 2 - 4ac is zero the roots are real and equal:
If b 2 - 4ac is negative the roots are imaginary.
Question:
Express the quadratic 2x2 +10x +12 in terms of its roots.
Answer:
b 2 - 4ac is positive (100 -96 = 4 ) and therefore the roots are real and unequal
10  100  96
Applying the formula for the roots x =
4
so that x = -3 or -2
Therefore
2x2 +10x +12 = 2(x + 3)(x + 2)
You can check the answer by removing the brackets – both sides are the same
5,4 Application to the LCR Series Circuit analysis
Example 5.4.1 Overdamping
The overdamped case occurs when the quadratic expression has two real and different
roots.
R
1  1 2

)  = s(s + 4s + 3)
For example suppose LCs  s 2  s 
L
LC  3

1
1
Then (v) = V/[ s(s2 + 4s + 3)] = V/[ s(s + 3 )(s + 1)]
3
3
3V
=
s ( s  3)( s  1)
26
And from the tables the Inverse Laplace transform of
1
is
s( s  a)( s  b)
1 
b  at
a  bt 
e 
e 
1 
ab  b  a
ba

We need to take b = 3 and a = 1 and find
1
3
1
3 -t
1
v = 3V( )( 1 - ( ) e-t + ( e 3t ) e-2t = V( 1 e + e-3t )
3
2
2
2
2
The solution is graphed below for the case V = 10 volts. Note that the capacitor
charges to its final value of 10 volts with no oscillations
15
10
v( t )
5
0
0
2
4
6
t
overdamping.
Example 5.4.2:
Critical Damping
The critically damped case occurs when the roots are real and equal.
1
R
1 

)  = [ s(s2 + 4s + 4)]
Suppose we had the case LCs  s 2  s 
L
LC  4

Then
1
(v) =V/[ s(s2 + 4s + 4)]
4
We thus need to express the quadratic
x2 + 4x + 4 in terms of its roots.
This time b 2 – 4ac is zero (16 – 16 = 0) and therefore the roots are real and equal. This means
the quadratic is a perfect square and by inspection
x2 + 4x + 4 = (x + 2)2
Hence
1
(v) = V/[ s(s + 2)2]
4
4V
=
s( s  2 ) 2
And from the tables the Inverse Laplace transform of
We need to take a = 2 and obtain
27
1
1
is 2 1  e  at  ate  at ) 
2
a
s(s  a)
v = 4V(1/4)(1 – e-2t – 2t e-2t ) = V(1 – e-2t – 2t e-2t )
The graph below shows the solution for the case V = 10 volts.
Notice that again the voltage rises to its final value with no oscillation
15
10
v2( t)
5
0
0
2
4
6
t
critical damping
Example 5.4.3:
Under Damping
The underdamped case occurs when the roots are complex conjugates. With underdamping
we find oscillatory behaviour.
Suppose we had the case
R
1  1

LCs  s 2  s 
) =
s(s2 + 4s + 10)]
L
LC  10

1
s(s2 + 4s + 10)]
10
We observe that this time b2 < 4ac so the quantity under the square root is negative and the
roots are complex.
Then
(v) =V/
In this case we proceed by “completing the square”. A completed square is a term that looks
like ( s  a)2  b2 and this is useful because it is easy to find from Tables the Inverse transform
of 1/( ( s  a)2  b2 )
We need to consider the quadratic:
x2 + 4x + 10
To complete the square take the 4 and halve it to get 2. Then note that the perfect square of
(x+2) is
(x + 4/2)2 = x2 +4x + 4
This is not quite what we want because our expression is x2 + 4x + 10 not x2 + 4x + 10 but
we can complete the square x2 +4x + 4 by adding 6 to it to get the desired 10.
i.e.
x2 + 4x + 10 = (x + 4/2)2 +6 = = (x+2)2 + 6
28
Thus if we have a term containing a quadratic such as
1
it may be converted into
[ s  4 s  10 ]
2
1
[( s  2) 2  6]
the standard completed square form
This matches the form of a number of transforms in Appendix A * which are shown as:
1
[( s  a ) 2 b 2 ]
In this case a = 2 and b = 6.
(v) 
For our differential equation we have
From Tables the Inverse of
10V
10V

s( s  4s  10)
s  ( s  2) 2  6 
2
1
1
1

e  at sin(bt   )
is 2
2
2
2
2
2
(
a

b
)
s  ( s  a)  b 
b a b
b
where   tan 1  
a
1
1

Taking V = 10 volts, a = 2 and b  6 we obtain v  100  
e2t sin( 6t   ) 
6 10
 10

 6
And   tan 1 
  0.886 radians
 2 
The graph is plotted below for V = 10 volts. Notice this time the voltage overshoots its final
value and oscillates before settling down.
15
10
v( t )
5
0
0
2
4
6
t
underdamping
5.5
Partial Fractions
The previous discussion on the application of Laplace Transforms assumes the availability of
an extended list of Laplace Transform Pairs. If however only a limited list is available it is
still possible to find a solution by first simplifying the Laplace Transform using the method of
partial fractions.
29
Consider the expression:
1
3

x2 x4
This can be expressed as a single fraction using a common denominator:
1
3
1( x  4)  3( x  2)
4 x  10



x2 x4
( x  2)( x  4)
( x  2)( x  4)
The fractions
4 x  10
1
3
and
are called the Partial Fractions of the fraction
( x  2)( x  4)
x2
x4
Conversely, starting with the fraction
4 x  10
it is possible to reverse the process and
( x  2)( x  4)
find its partial fractions as follows:
Let
4 x  10
A
B

=
x2 x4
( x  2)( x  4)
Equation 1
Multiplying both sides by (x+2)(x+4) gives
4x + 10 = A(x+4) + B(x+2)
Equation 2
Equation 2 must be true for all values of x and suitable values for x may be chosen to
eliminate one of the constants A and B.
If we substitute x = -2 in Equation 2 we make the B term disappear and we have
-8 + 10 = A(2)
therefore
A= 1
Now let x = -4 in Equation 2 to make the A term disappear so that
-16 + 10 = B(-2) therefore
B=3
This approach gives rise to the Cover-up Rule.
Consider Equation 2: Putting x = -2 means that B will be eliminated from the equation
leaving
A=
4 x  10
x4
with x =-2
Now consider starting from Equation 1:
4 x  10
A
B

=
x2 x4
( x  2)( x  4)
The constant, A, is ‘over’ the term (x+2) and it can be seen that the expression for A starting
from Equation 2 is the same as the LHS of Equation 1 with the (x+2) term removed (or
Covered-up) and x put equal to -2.
30
In short: to find A we first look on the RHS of Equation 1 to see what A is sitting ‘over’ – we
see that this is (x+2) so now we cover up the x +2 on the LHS of Equation 1 and substitute
8  10
x = -2 in what is left uncovered. This gives us A =
1
(2  4)
Similarly to find B we first look on the RHS of Equation 1 and see that B is sitting on top of
(x+4). Now we cover up the (x+4) on the LHS of Equation 1 and substitute x = -4 in the
uncovered part of the LHS of Equation 1 i.e with the (x+4) term removed (or Covered-up) .
16  10
This gives B =
= 3.
(4  2)
5.5.1 Recap of Solution by the Cover-up Rule.
Look at the RHS of Equation 1
A is ‘over’ the term ‘x+2’.Assign x the value that will make this term zero (i.e. put x = -2)
Now look at the LHS of Equation 1
A = LHS of Equation 1 with the ‘x+2’ term Covered-up and with x = -2 substituted
Therefore
A=
 8  10
=1
24
Question: Apply the Cover-up Rule to the following single fraction to find the partial
fractions.
5x  1
2
3

(Answer:
)
x 1 x 1
( x  1)( x  1)
Answer:
Let
5x  1
A
B

=
( x  1)( x  1) x  1 x  1
 51
=2
11
51
B is ‘over’ x - 1: Put x = 1: Cover-up ‘x -1’ in the fraction on LHS and B=
=3
11
A is ‘over’ x +1: Put x = -1: Cover-up ‘x+1’ in the fraction on LHS and A=
A
B

if a and b are
( x  a ) ( x  b)
A
B
different numbers. You cannot apply the rule for example to
as we shall

( x  a) ( x  a)2
see in the next case.
Notice you can only apply the cover-up rule to things like
5.5.2 Other Important Cases
1
If the denominator contains a squared factor then the expansion into partial
fractions includes a constant over the squared factor and a constant over the
factor itself.
31
2x
A
B
2 
2 
x 1
( x  1)
( x  1)
e.g.
Solution
We cannot use the cover-up rule here
Multiplying by (x+1)2 gives 2x =A + B(x+1)
Let x = -1
then
therefore
-2 = A
A = -2
Let x = 0
then
therefore
0 = A+B
B=2
so that
2
2x
2
2


2
2
( x  1)
( x  1)
x 1
If the denominator includes a factor of the form (x2 + a) then the numerator of this
partial fraction must be of the form Ax+B
e.g.
1
Ax  B
C
 2

2
( x  1)( x  2) x  2 x  1
Solution
By Cover-up Rule
Multiplying by (x+1)(x2+2)
C=
1
1

2
( 1)  2 3
1=(Ax+B)(x+1)+C(x2+2)
Putting x = 0 and substituting for C, 1 = B+
therefore
Putting x = 1
B=
2
3
1
3
1
1
1 = (A+ )(2)+ (3)
3
3
2
-1
3
1
A =3
rearranging 2A= 1 therefore
so that finally
1
 x 1
1


2
2
( x  1)( x  2) 3( x  2) 3( x  1)
32
5.6
Examples of second order differential equations using Laplace
transforms
Example 1: Solve using Laplace Transform
Answer: First, apply the Laplace Transform
Knowing that
,
and
we get
After easy algebraic manipulations we get
,
which implies
Next, we need to use the inverse Laplace.
We have (see the table)
For the second term we need to perform the partial decomposition technique first.
.
We get
Hence, we have
33
Since (see the table)
and (see the table)
Finally, we have
Example 2
Solve for i(t) for the circuit, given that V(t) = 10 sin5t V, R = 4 Ω and L = 2 H.
Answer
34
So
Equating coefficients of
:
gives
Equating coefficients of :
gives
So
So we have
35
A
Example 3
The Laplace Transform of the current in a certain series circuit is:
Invert the Laplace Transform to find the expresion for the current I(t), not by partial
fraction expansion, but by completing the square in the denominator, dividing the
expression for I ( s ) in two and using two entries from your table of Laplace
Transforms.
SOLUTION :
Using:
with a=1 and
6.
: we have
Voltage Transfer Function
The Voltage Transfer Function (VTF) is a useful mathematical device to separate out the
fixed properties of the circuit from the input and output voltages. If we know the VTF of a
circuit we can quickly find its output for any input by using Laplace Transform tables.
Consider a circuit having an input voltage, Vin, and an output voltage, Vout, these voltages
will generally be functions of time. Keeping the circuit always the same we can choose to
apply different input voltages. Different input voltages will of course give rise to different
output voltages
36
Vin
circuit
Vout
The Voltage Transfer Function (VTF) of a network is defined by
VTF =
LaplaceTransform(Vout )
LaplaceTransform(Vin )
with all initial conditions being zero.
By definition of the VTF the Laplace Transforms must always be calculated with all initial
conditions set to zero (i.e the initial charge, initial current, initial voltage and their rates of
change are all zero).
Note: With all initial conditions being zero we can use the simplified forms of Section 4.4 i.e.
(di/dt)= s (i), etc.
For linear circuits we shall see that the VTF
LaplaceTransform(Vout )
LaplaceTransform(Vin )
always works out to be the same function of the Laplace variable s no matter what input
voltage we choose to apply. If we know the VTF then we can easily find the Lapalce
Transform of the output voltage for a given input voltage by using the equation
LaplaceTransform(Vout )  VTF * LaplaceTransform(Vin )
We can then find the output voltage in the time domain by finding the inverse transform of
the LaplaceTransform(Vout ) .
As an example consider the simple network:
37
Vin  R1i 
We have that
q
 R2i
C1
Vout  R2i
 dq 
L Vout   L  R2i  R2 L    R2 s L q
 dt 


1
q
L q  R2 s L q
L Vin   L  R1i   R2i  = R1s L q 
C1
C1


Therefore VTF =
R2 s
LaplaceTransform(Vout )
=
s( R1  R2 )  1 / C1
LaplaceTransform(Vin )
VTF =
R2
R1  R2  1 / sC1
Notice the L q ha cancelled out and the VTF depends only on the circuit components. It is
the same expression whatever the input voltage. It is a property of the circuit itself – not a
property of the input voltage.
Compare the VTF with the voltage divider phasor ratio
V±
R2
out

V±in R  R 
1
2
V±
out
for an ac input:
±
Vin
1
jC1
Note that this would be exactly the same as the VTF if j is replaced by s. This idea gives
us a quick way for writing down VTFs. All you do is work out the phasor ratio using
impedances in the usual way and then change every j to s
38
Example:
If the input to the above network is a 10 volts step with R1=3k  , R2=1k  and C1=1mF then
10
(Vin) =
hence (Vout) is given by
s
(Vout) =
=
1000s
10

s(1000  3000)  1000 s
10
2.5

4 s  1 s  0.25
By inverting we find the output voltage is therefore Vout=2.5 e-0.25t
In a similar way the output voltage for this circuit can be found for any input providing it has
a Laplace Transform.
39
C
Algebra
7.
Complex Numbers
We have seen in Semester 1 that a complex number may be expressed in co-ordinate and in
polar form.
z  x  jy .
In co-ordinate (or rectangular) form
z  r cos( )  jr sin( )
In polar form
 r  cos( )  j sin( ) 
 r
We shall now describe yet another form for complex numbers called exponential form.
7.1
The exponential form of a complex number
It i8s shown in what follows that the complex number z  r can be written as
z = re j .
This is called the exponential form of complex number.
Important:  MUST be in radians.
For example
4e
j

2
 4



 4(cos( )  j sin( ))  j 4
2
2
2
Proof:
From work on infinite series it can be shown that for a number, x, (or angle which
must be in radians)
and
so that
sin x  x 
x3 x5 x7


 ……
3! 5! 7!
cos x  1 
x2 x4 x6


 ……
2! 4! 6!
x2 x3 x4 x5
e  1 x 



 ……
2! 3! 4! 5!
x
e jx  1  jx 
( jx ) 2 ( jx ) 3 ( jx ) 4 ( jx ) 5



 ……
2!
3!
4!
5!
40
x2
x3 x4
x5
 1  jx 
j

j
 ……
2!
3! 4!
5!
x2 x4
x3 x5
 (1 

......)  j ( x 
 ......)
2! 4!
3! 5!
= 1 x
= cos(x) + j sin(x)
Therefore a complex number z= r x is also equal to z = re jx .
We often use  instead of x and then we can write
r  re j  r (cos( )  j sin( ))
7.2 Powers and Roots of a complex number
The exponential form is particularly useful for establishing the rules for finding powers and
roots of complex numbers.
Take any complex number z in exponential form and raise it to the power of n, where the real
number n is either a whole number (giving a power of z) or a fraction (giving the roots* of z).
i.e.
z n   re j   r n  e j   r n e jn  r n  cos(n )  j sin(n ) 
n
n
This relation is known as De Moivre’s Theorem and we have already met it in Semester 1 in
polar form .
 r 
i.e.
n
 r n n
In polar form  may be in degrees or radians but in exponential form  must be a number
(or angle in radians).
7.2.1 Whole number Powers of a complex number
To find the whole number powers of a complex number it is useful to use exponential form
z n  r ne jn  r n  cos(n )  j sin(n ) 
Question: Find (3  j3) 4
Answer:
Then we have
first convert (3  j 3) to exponential form: (3  j 3)  18e j /4
(3  j3)4 

18e j /4

4
 324e j  324
41
This method is much quicker than working out the 4th power by removing the brackets.
*
Notice that when we take a whole number power of a complex number z n we will
always only get one answer. However this is not true when we take roots (i.e. fractional
1
powers). For example if we take square roots as in z 2 we will get two different answers, for
1
1
cube roots z 3 there are three different answers, for fourth roots z 4 there are four different
answers and so on. In the next section we will see that the extra answers come from the fact
that
sin( )  sin(  2 )  sin(  4 )  sin(  6 )...
cos( )  cos(  2 )  cos(  4 )  cos(  6 )...
r  r(  2 )  r(  4 )  r(  6 )...
therefore
We see in the next section that these alternatives mean that more than one root will be found
7.2.2 Square Roots of a complex number
1
2

1
2
(r )  r ( ) = the first square root.
2
1
2
  360o
1

= r (

1
2
) = r (  180o ) = the second square root.
2
2
Attempts to add a further 360o and obtain another square root merely gives an expanded form
of the first square root:
Or
1
That is r 2 (
  720 o
2

1
)  r 2 (  360 o )  r 2 ( ) = the first square root again.
2
2
7.2.3 Cube Roots of a complex number
In this case there will be 3 roots.
The first cube root is found from
1
1

(r ) 3  r 3 ( )
3
the next is
1
3
 r (
  360
3

1
3
)  r (  120 o )
3
and the third and final root is
1
3
 r (
  720 o
3

1
3
)  r (  240 o )
3
42
If we continue adding the 360 degrees we simply cycle through the same three answers
already found.
7.3
Application to transmission lines
The characteristic impedance,z0, of a correctly terminated transmission line is given by
z0 
R  j L
G  j C
and the p
  ( R  jL)(G  jC)
These expressions include the square root of complex numbers but the subject of the equation
is a physical quantity for which the ‘second root’ has no meaning since the real part would be
negative. In the case of characteristic impedance the second root would imply a negative
resistance and in the case of the propagation coefficient it would imply negative
attenuation/negative losses.
8
The Hyperbolic Functions
In this section we shall meet some new functions sinh(  ), cosh(  ) and tanh(  ). These
functions are called hyperbolic functions. Normally the functions are pronounced as shine
cosh and than. They are useful in transmission line applications and in other areas of
electrical engineering.
A comprehensive list of the properties and graphs of the hyperbolic functions may be found in
Appendix B.
8.1 Definition of the hyperbolic functions
The hyperbolic functions are defined in a slightly similar way to the circular functions
sin(  ), cos(  ) and tan(  ) . We begin by defining sin(  ) and cos(  ) in terms of complex
exponentials:
WE know that
e j  cos( )  j sin( )
(a)
Hence It follows that
e j  cos( )  j sin( )  cos( )  j sin( )
(b)
From these equations the trigonometric or circular functions of sin( ) and cos( ) may be
obtained:
Adding (a) and (b) we get
e j  e  j  2cos .
Therefore
cos( ) 
e j  e j
2
43
(c)
e j  e j  2 j sin( )
subtracting (a) – (b) gives
e j  e j
(d)
2j
sin( ) and cos( ) are called circular functions and may be considered to be defined by (c)
and (d) above. Another useful pair of functions can be obtained from these definitions simply
by dropping the j . The new functions are called hyperbolic functions
sin( ) 
Therefore
Hyperbolic functions are defined using similar exponential expressions
e x  e x
cosh( x) 
2
.
sinh( x) 
e x  e x
2
sinh( x) e x  e x
tanh( x) 

cosh( x) e x  e x
The hyperbolic functions are graphed below. Notice that the graph of cosh(x) has the same
shape as a string hanging under gravity.
10
5
cosh( x)
3
2
1
0
1
2
3
1
2
3
x
cosh(x)
10
5
sinh( x)
3
2
1
0
5
10
x
sinh(x)
44
2
1
tanh( x)
3
2
1
0
1
2
3
1
2
x
tanh(x)
The hyperbolic functions and their inverse are easily evaluated using the special keys on your
scientific calculator. Make sure you can use your calculator to do this.
Here are some examples for you to check using your calculator:
cosh(1.5) = 2.352…
sinh(-0.6) = -0.637…
tanh(1.2) = 0.834….
cosh 1 (2)  1.317…
sinh 1 (0.85)  0.771…
tanh 1 (0.5)  0.549
Question: A parallel-wire transmission line with conductors of diameter d spaced a distance s

between center-lines has capacitance per unit length C 
. . Rearrange this
1  s 
cosh  
d 
formula to find a formula for s.
Answer:
s
 s  
  
  
cosh 1   
therefore  cosh   and s  d cosh  
d
d  C
C 
C 
45
9.
Bessel Functions and their relation to Frequency Modulation
The treatment of frequency modulation in communication systems requires a knowledge of
some mathematical functions called Bessel functions. No in-depth knowledge of Bessel
functions is necessary at HND level but you do need to know what the graphs look like and
the places where the Bessel functions are equal to zero. For those interested in knowing more
about the application to FM there is a further discussion in Appendix C.
Thre are various kinds of Bessel function but we are only interested in Bessel functions of the
First Kind. Bessel functions are written as J n ( x) . The subscript n is an integer.
Let us look first at some graphs.
9.1
Graphs of J n ( x) J n ( x)
The graphs of Bessel functions look roughly like oscillating sine or cosine functions that
decay proportionally to 1/√x.
Here are the graphs of J 0 ( x) , J1 ( x) and
J 2 ( x)
46
.
2
Zeros of Jn(x)
The table below shows those values of x where Jn(x) is zero.
s
n=0
n=1
n=2
n=3
n=4
n=5
1 2.405
3.832
5.135
6.379
7.586
8.780
2 5.520
7.016
8.147
9.760
11.064 12.339
3 8.654
10.173 11.620 13.017 14.373 15.700
4 11.792 13.323 14.796 16.224 17.616 18.982
5 14.931 16.470 17.960 19.410 20.827 22.220
6 18.071 19.616 21.117 22.583 24.018 25.431
7 21.212 22.760 24.270 25.749 27.200 28.628
8 24.353 25.903 27.421 28.909 30.371 31.813
9 27.494 29.047 30.571 32.050 33.512 34.983
As you may see from Appendix C the reason why the zeros are important is that by choosing
the FM modulation index to be one of the x values in the first column of the table the
transmitted signal concentrates all of its power into the information-carrying sideband
frequencies. Not doing so results in some power being sent at the carrier frequency, which
doesn’t contain any signal information.
9.3 Evaluation of Bessel functions
Most hand calculators do not have Bessel function keys. To find the value of a Bessel
function one can use Mathcad.
Bessel functions using Mathcad:
J0(x) Returns J0(x).
J1(x) Returns J1(x).
Jn(m, x) Returns Jm(x).
Arguments:
x must be a real scalar.
x cannot have units.
m must be an integer between 1 and 100 inclusive.
47
10
Matrices
Matrices are mathematical objects that have numerous applications. We shall be using them
to represent and solve linear equations. We need to know what a matrix looks like and how to
manipulate matrices. A matrix is defined as a rectangular array of numbers.
1 5


 3 4
2 6
e.g.
or
 1 5


 3 4


 2 6
The convention for describing a matrix is to give the number of rows first, then the number of
columns.
Thus the matrix above is described as a ‘3 by 2 matrix’(i.e. 3 rows by 2 columns).
Similarly for identifying an element in the matrix. Thus the number ‘5’ in the matrix is
identified as element ‘1,2’ (i.e. 1st row, 2nd column)
10.1
Addition and Subtraction of matrices
Addition of matrices is defined as addition of corresponding elements. Thus for addition to be
possible the matrices must have the same number of rows and must have the same number of
columns.
e.g.
1 5
2 3
1 + 2 5 + 3
3 8








 3 4 + 4 5 = 3 + 4 4 + 5 = 7 9
2 6
6 1
2 + 6 6 + 1
8 7
Subtraction is similarly defined as subtraction of corresponding elements.
e.g.
10.2
1 5
2 3
1- 2 5 - 3
-1 2 








 3 4 - 4 5 = 3 - 4 4 - 5 =  - 1 - 1
2 6
6 1
2 - 6 6 - 1
- 4 5 
Multiplication
10.2.1 Multiplication of a matrix by a number or algebraic variable (scalar multiplication)
is achieved by multiplying each element of the matrix by the number.
e.g.
3 
2 3
3  2 3  3
6 9






4 5 = 3  4 3  5 = 12 15
6 1
3  6 3  1
18 3 
Any matrix can always be multiplied by a number
Viewed in reverse it can be seen that a common factor of the elements may be ‘taken outside’
the matrix.
48
10.2.2 Multiplication of a matrix by another matrix results in a ‘product’ matrix in which
each element of the product matrix is the sum of products of the row elements of the first
matrix and the corresponding column elements of the second matrix. This therefore requires
that the number of row elements of the first matrix is equal to the number of column
elements of the second matrix.
Note:
The number of row elements = the number of columns and
the number of column elements = the number of rows.
1 5


 3 4
2 6
e.g. Consider a 3 by 2 matrix:
Each row contains 2 elements, one for each column.
So 2 row elements means 2 columns.
Similarly each column contains 3 elements, one for each row.
So 3 column elements means 3 rows.
This rule for multiplication is usually expressed as ‘the number of columns of the first
matrix must equal the number of rows of the second’.
e.g.
[3 by 2]  [2 by 1] is possible because the number of columns of the first (2) is the
same as the number of rows of the second (also 2) and would result in a product matrix
[3 by 1]
but [2 by 1]  [3 by 2] is not possible.
In general [m by n]  [n by p] is possible with a product matrix [m by p]
a11

Consider the product A  B where A = a 21
a 31
a12 
b11 b12

a 22  and B = 
b21 b22
a 32 
b13 
b23 
[3 by 2]  [2 by 3] gives a product matrix [3 by 3]. Let this be the matrix C where
c11

C = c21
c31
c12
c22
c32
c13 

c23 
c33 
and where c11 is the element in the first row, first column and is equal to the sum of products
of elements in the first row of matrix A and corresponding elements in the
first column of matrix B
49
c11  a11b11  a12b21
i.e
The element c12 c12 is ‘first row’, ‘second column’ and is given by
c12  a11b12  a12b22 c12
10.2.3
Summary:
1.
Test to see if multiplication is possible and if so the size of the product
matrix. i.e. use [m by n]  [n by p] is possible with product matrix [m by p]
2.
In the product matrix identify the element required.
e.g. say c23 in the previous example is at row 2, column 3.
3.
Identify row 2 in matrix A and column 3 in matrix B:
4.
a11 a12 
b11 b12 b13 


A = a21 a22 
and B = 

b21 b22 b23 
a 31 a 32 
Sum the products of corresponding row elements in A and column elements
in B.
c23  a21b13  a22b23
i.e.
Note: In general the commutative law of conventional algebra does not apply to
matrices
i.e. in general A  B  B  A
(An important exception involves the Identity (or Unit) Matrix: see below)
10.3 Identity (or Unit) Matrix (I)
This is a square matrix (2 by 2 ,3 by 3, etc) which corresponds to 1 (unity) in conventional
algebra. Any matrix multiplied by its unit matrix will remain unchanged.
2 by 2:
 1 0 0


3 by 3: 0 1 0
0 0 1
 1 0
 0 1


Example
1 5
 (1  1) + (5  0) (1  0) + (5  1) 
 1 0




 3 4  0 1 =  (3  1)  (4  0) (3  0) + (4  1) 


(2  1) + (6  0) (2  0) + (6  1) 
2 6
50
1 5


=  3 4
2 6
10.4 The Determinant of a Matrix
A square matrix has the same number of rows and columns.
A square matrix always has a determinant. If a matrix is not square it does not have a
determinant.
You have met determinants in semester 1 with Cramer’s Rule. Here is a brief reminder on
how to evaluate determinants:
The determinant is defined as a function of the ‘cross products’ of the elements of a matrix.
a b 
For the 2 by 2 matrix 
 the determinant is defined as (ad - bc).
c d 
It is written as
a b
c d
a b

For the 3 by 3 matrix d e
 g h
c

f
i 
the determinant is defined as
a(ei - fh) - b(di - fg) + c(dh - eg)
It is written as
a b
d e
c
f
g h
i
10.5 The Inverse of a Matrix
The inverse, A-1, of a matrix A is defined such that A-1  A = I = the Identity Matrix.
For the rigorous determination of an inverse it is necessary to know the definition and use of
co-factors, the transpose of a matrix and the adjoint (Adj) of a matrix. Then the inverse is
Adj ( A)
found as
.
Det ( A)
a b 
However for the 2 by 2 matrix 
 the inverse is given by
c d 
A-1
 d  b
 c a 


=
=
Det ( A)
 d  b
 c a 


a b
c d
51
10.6 Systems of Equations
Matrices may be applied to represent and solve systems of equations.
e.g. the two simultaneous equations :
3x + 2y = 13
and x - y = 1
may be written as one equation with the left hand side as a product of matrices and the right
hand side as the product matrix.
3 2 
1  1


or
x
13
=
 y
1
 
 
x
A  S = B where   = S = the Solution Matrix
 y
The procedure to find S is similar to that of conventional algebra.
i.e.
Multiply both sides by the inverse of A
A-1  A  S = A-1  B
I

S = A-1  B
S = A-1  B
Example
3x + 2y = 13
x - y= 1
3 2   x 
13
1  1  y  =  1 

 
 
3 2 
Now the inverse of 

1  1
is
Equation 1
 1  2
 1  2
 1 3 
 1 3 
1  1  2 




=
= - 
3 2
5  1 3 
5
1 1
Multiplying both sides of Equation 1 by this inverse gives
x
1  1  2  13
1 ( 13  2)   3
 y  = - 5  1 3   1  =  5 ( 13  3)  = 2 

 

  
 
Therefore
x = 3 and y = 2
52
11
Numerical Methods
11.1
Newton-Raphson method for the solution of non-linear equations.
Some non-linear equations can be solved exactly but many cannot. In fact it is fair to
say that most equations cannot be solved exactly and no formula exists to find their
solutions.
(a) Some non-linear equations that can be solved exactly are the quadratics,
cubics and quartics.
For example the quadratic equation ax 2  bx  c  0 has two solutions given by
the well-known formula x 
b  b2  4ac
.
2a
There are similar (but more complicated) formulae for the three solutions of
any cubic (third power) equation and for the four solutions of any quartic
(fourth power) equation.
(b) Some of the non-linear equations that cannot be solved exactly are the higher
order equations, from the fifth power upwards. For these equations there is
no general formula for the solution and they cannot be solved exactly. Other
non-linear equations that cannot be solved exactly may have logarithmic,
exponential or trigonometric terms. Here are some examples of equations that
cannot be solved exactly:
et  t  2  0
cos(t )  1  t
ln(t )  1  t
3x5  4 x 4  x3  2 x 2  x  0
When an equation cannot be solved exactly all is not lost because we can still solve it
approximately using numerical methods. We can actually obtain the solutions very
accurately indeed, to as many decimal places as we choose, even though we cannot
find the exact answer.
There are many different numerical methods available for solving non-linear
equations and in this section we shall study one of them, the Newton-Raphson
Method. We shall demonstrate this in what follows by means of three examples.
53
Example 1
Solve the equation e  t  t  2  0
This is a non-linear equation for which there is no formula but an approximate
solution may be estimated graphically. Below we draw the graph of the function
f (t )  et  t  2 and look to see the value of t where it becomes zero. Any solution of
the equation f (t )  0 is called a root of the function f (t )
f (t )  e t  t  2
t=1.83
0
t
It can be seen from the graph that the function has only one root, i.e. there is only one
solution of the equation f (t )  0
The approximate value of the root is estimated from the graph to be about t = 1.83.
This approximate value may be improved using the Newton -Raphson method. To
see how this can be done consider a general function f(t) plotted against values of t
near a root
f(t)
ttrue
t1
t0
t
Say the value of t where the curve crosses the t -axis (i.e. where f(t) = 0 ) is estimated
(very approximately) as t0.
It can be seen that a better approximation would be t1 which is where the tangent to
the curve when t = t0 cuts the t axis.
f (t 0 )
The slope of this tangent is f (t 0 ) 
t 0  t1
rearranging
t 0  t1 
f (t 0 )
f (t 0 )
54
t1  t 0 
so that
f (t 0 )
f (t 0 )
This is essentially the Newton-Raphson method; it says that if t 0 is the approximate
value of a real root of the equation f (t )  0 then a closer approximation to the root is
f (t 0 )
t1  t 0 
.
f (t 0 )
We need not stop the process here however. We know that t1 is a better
approximation than t 0 but it appears from the graph that an even better one may be
found by repeating the process in terms of the tangent at t1 to give t2 :and again to give
f (t n )
t3 and so on until the
term is negligible in terms of the accuracy required.
f (t n )
In summary the Newton-Raphson method requires us to start with a guess value, t0
and repeat the iterative process given by
f (t )
tn1  tn  ' n
f (tn )
Until the solutions no longer change to the number of decimal places required.
For the example above t0 = 1.83 and
f (t )  e  t  t  2 so that f (t )  e  t + 1
e  tn  t n  2
and therefore tn 1  tn 
e  t n  1
For the first iteration we find
t1 = 1.83 -
e
-1.83
+183
. -2
= 1.83+0.01142 = 1.84142
 e -1.83  1
For the second iteration we obtain
t2 = 1.84142 -
e -1.84142 +1.84142 - 2
= 1.84141
e-1.84142  1
For the third iteration we have
t3 = 1.84141 -
e -1.84141 +1.84141 - 2
= 1.84141
e-1.84141  1
Since the third iteration gives the same answer to 6 decimal places as the second
iteration we have solved the equation correct to 6 decimal places.
The final solution is t = 1.84141.
55
In the example above we started off the process with a fairly good guess t0 = 1.83.
What would have happened if we had started with not such a good guess? Well we
have eventually got the required accuracy but it would probably have needed more
turns round the loop.
For example suppose we had taken as a starting guess t0 = 1.8 then you may show
yourself that t1 is found to be 1.84157 and repeating the process using this value for t1
again gives 1.84141 for t2 and t3 etc.
If we had started even further away with t0 = 1 we would find t1= 2, t2= 1.84348 and
t3= 1.84141.
Advantages and drawbacks of the Newton-Raphson method
The advantages of the Newton-Raphson method over other methods of successive
approximations is that it can be used for any type of mathematical equation (i.e.ones
containing trigonometric, exponential, logarithmic,hyperbolic and algebraic
functions), and it is usually easier to apply than other methods.
The main drawback of the method is that it doers not always converge to the root you
want if the initial guess is not good enough. Also the derivative of the function is
necessary and in some cases this might be difficult to obtain.
Example 2
Use the Newton-Raphson method to find the positive root of the quadratic equation
5 x 2  11x  17  0 correct to 5 decimal places. Check your answer against the exact
answer using the quadratic formula.
A graph of the function f ( x)  5 x 2  11x  17 is shown below:
200
100
2
5x  11x 17
4
3
2
1
0
1
2
3
4
100
x
From the graph we see the positive root is close to x = 1 so we take this as our starting
guess value.
By differentiation we find f ' ( x)  10 x  11 so the Newton-Raphson method has
f (x )
xn1  xn  ' n
f ( xn )
56
5 xn 2  11xn  17
Hence xn 1  xn 
with x0  1 . Since we want to be accurate to 5
10 xn  11
decimal places we shall show the working to 5 decimal places.
For the first iteration we find
5 x0 2  11x0  17
5 x 12  11 x 1  17
1
x1  x0 
 1
 1
 1.04762
10 x0  11
10 x 1  11
21
For the second iteration we obtain
x2  x1 
5 x12  11x1  17
5 x 1.047622  11 x 1.04762  17
 1
 1.04709
10 x1  11
10 x 1.04762  11
For the third iteration we have
x3  x2 
5 x2 2  11x2  17
5 x 1.047092  11 x 1.04709  17
 1
 1.04709
10 x2  11
10 x 1.04709  11
Since the third iteration gives the same answer to 5 decimal places as the second
iteration we have solved the equation correct to 5 decimal places.
The final solution is x = 1.04709
For comparison the exact solution using the quadratic formula is
11  461
 1.047091015.......
10
Example 3
The transient voltage v  10te  t is graphed below. Find any time where the voltage is
equal to 2 volts. Give your answer correct to 3 decimal places.
4
3
t
10 t e
2
1
0
2
4
6
t
57
8
10
Solution. First we write the equation to solve in the form required to use the NewtonRaphson method i.e. in the form f (t )  0
f (t )  10tet  2  0
The derivative of f (t ) may be found using the product rule:
f ' (t )  10et (1  t )
Take as a starting value t0  2 then
f (t0 )
 .52219
f ' (t0 )
t1  t0 
For the first iteration we thus find that
f (t0 )
 2  0.52219  2.52219
f ' (t0 )
For the second iteration we find that
f (t )
t2  t1  ' 1  2.52219  0.02038  2.54300
f (t1 )
For the third iteration we thus find that
f (t )
t3  t2  ' 2  2.54300  0.000358665=2.54264
f (t2 )
The final solution is t = 2.543
The working for Newton-Raphson can be tedious if done by hand, as we have just
done, and it is usually more efficient to write a small computer program to carry out
the calculations. In Mathcad the root function is one example of such a program,
PSPICE also used Newton-Raphson.
58
1. By graph it is estimated that f (t )  e 2t  t  0 when t = 0.4
Use the Newton -Raphson method to find t correct to 4 decimal places.
(0.4263)
2. Similarly f (t )  sin( t )  t  0.5  0 when t = 1.5
(1.4973)
59
11.2
Gauss-Seidel Method for Simultaneous Linear Equations
Simultaneous linear equations may be solved graphically by plotting both lines and
finding the point of intersection.
Consider the two lines shown in graph A
y
line 1
line 2
0
x
Graph A
The point of intersection could be reached following the arrows.
i.e.
Find y for line 1 when x = 0
Use this value of y to find x for line 2
Use this value of x to find y for line 1
Use this value of y to find x for line 2
Continue until the values of x and y are constant which are the co-ordinates of
the point of intersection.
At each step in this process the form of the calculation is repeated ,each time
producing a value which is a closer approximation to the final value. Each step in this
repetitive process is termed an iteration and the method is said to be an iterative
method.
This is the basis of the Gauss-Seidel method for the solution of simultaneous
equations without drawing the graphs.
1
2
3
4
5
Put x equal to 0 in the equation of ‘line 1’ and find the value of y
Put this value of y into equation of ‘line 2’ and find the value of x
Put this new value of x into the equation of ‘line 1’ and find the new
value of y
Put this new value of y into equation of ‘line 2’ and find the new value
of x
Continue this process until the x and y values reach a constant value.
With the line equations of graph A this process would involve many calculations
before the x and y values became constant. This is because the magnitudes of the
slopes of the lines are similar. (In fact the method will not give a result if the
magnitudes of the slopes are equal)
This method is best applied when the slopes of the lines are very different as shown in
Graph B. In this case constant values would be obtained after only a few iterations
provided that the correct equation is chosen as ‘line 1’. Choose the wrong one and
it would be the dotted path that would be followed with the x and y values moving
further and further away from the point of intersection.
60
y
line 2
line 1
0
x
Graph B
How to choose the correct equations?
Choose the equation whose x coefficient has the largest magnitude to find x values
and the
and the equation whose y coefficient has the largest magnitude to find y values.
e.g.
5x + y = 8
x
8 y
5
2x +20y = 13
y
13  2 x
20
Start with x = 0 and work to two places of decimals
x
y
0
0.65
1.47
0.50
1.50
0.50
At which stage y is same as before and x would be as well (if we bothered to
calculate it!)
61
D
Vectors
(See Chapters 1 and 2 of ‘Vector Analysis’ by Murray R Spiegel in the Schaum
Outline Series).
12.1 Vectors and scalars
12.1.1 A VECTOR is a quantity having both magnitude and direction, such as
displacement, velocity, force, and acceleration.
Graphically a vector is represented by an arrow OP (Fig 1) defining the direction, the
magnitude of the vector being indicated by the length of the arrow. The tail end, O, of
the arrow is called the origin or initial point of the vector, and the head, P, is called
the terminal point or terminus.
ur
Analytically a vector is represented by a letter with an arrow over it, as A in Fig 1
ur
and its magnitude by A or just A. In printed works, bold faced type, such as A, is
ur
used to indicate the vector A while A or A indicates its magnitude. The vector OP
uuur
uuur
is also indicated as OP or OP; in such case the magnitude is denoted by OP, OP , or
OP .
12.1.2 A SCALAR is a quantity having magnitude but no direction. e.g. mass,
length, time, temperature, and any real number. Scalars are indicated by letters in
ordinary type as in elementary algebra. Operations with scalars follow the same rules
as in elementary algebra. Some scalars can take positive or negative values e.g.
electric charge, while others only make sense of they are restricted to be positive e.g.
length.
12.2 Vector Algebra
The operations of addition, subtraction, and multiplication familiar in the algebra of
numbers or scalars are, with suitable definition, capable of extension to an algebra of
vectors. The following definitions are fundamental:
1.
2.
Two vectors A and B are equal if they have the same magnitude and
direction regardless of the position of their initial points. Thus A = B in
Fig 2.
A vector having direction opposite to that of vector A but having the
same magnitude is denoted by -A as in Fig 3.
P
ur
A or A
A
B
A
-A
O
Fig 1
Fig 2
Fig 3
62
Chapter 1 Page 2
63
Chapter 1 Page 3
64
Chapter 2 Page 16
65
Chapter 2 Page 17
66
E
Statistics
13.1
Frequency Distributions
Summary (Refer to the Course Notes for Engineering Mathematics A , page 113
onwards)
Raw data may be re-arranged using a tally chart into a frequency distribution which
shows the frequency of occurrence of each class.
13.2 Histogram
The histogram is a diagram which represents the frequency distribution using
rectangles whose areas represent the frequencies.(If the class widths are all the same
then the heights represent the frequencies as well)
NOTE: The total area of the histogram is the total number of items i.e. the
‘Population’.
45
40
35
30
25
20
15
10
5
0
96
13.3
97
98
99
100 101 102 103 104
Frequency Polygon
Alternatively a frequency polygon may be produced by plotting the frequency of a
class against the class mid-point.
Note: the area under the polygon still equals the area of the histogram, i.e. the
population.
67
45
40
35
30
25
20
15
10
5
0
96
13.4
97
98
99
100 101 102 103 104
Frequency Curve
If the class intervals are made smaller and smaller then the changes in frequency
between classes would also become smaller until the frequency polygon has the
appearance of a smooth curve- the frequency curve.
Note: as before the area under the curve is equal to the population.
13.5
Arithmetic Mean
The mean value of the items in a set is defined to be the sum of all the values of the
items in the set divided by (the number of items in the set

Symbolically
x
x
N
For a grouped frequency distribution it would be

x
13.6
fx
f
Standard Deviation
The ‘deviation’ of a value, x, from the mean,  is defined to be x - 
The Mean Deviation is not very useful because it is always zero.
( Mean deviation =
f ( x   ) fx f
f



 0 )
f
f
f
f
However, the Mean Square Deviation is always positive and taking the Square Root
of this square deviation gives a useful measure of the dispersion of a distribution. It is
called the Standard Deviation and is represented by 
68
Mean Square Deviation =
f ( x   ) 2 fx 2 2 fx f 2
fx 2 2fx  2 f






f
f
f
f
f
f
f
Standard Deviation = Root Mean Square Deviation =  =
14
fx 2
 2
f
The Normal Distribution
Many frequency distributions approximate to the so-called ‘Normal Distribution’.
This distribution is defined mathematically by the function, f(x):

1
f ( x) 
e
 2
( x   )2
2 2
Thee graph of the normal distribution is a bell-shaped curve symmetrical about the
population mean  . The total area under the curve is always equal to exactly one.
Figure shows the normal curve for the case   50 and   10
0.05
0.04
0.03
0.02
0.01
0
0
20
40
60
80
100
Figure 1 A Normal Distribution with   50 and   10
Notice This function is zero when x =   and a maximum when x =  .
No actual distribution will follow this exactly (because of the ‘   ’ values) but
many follow it close enough to be described as ‘Normal’ and to make use of features
of the Normal Distribution.
Notice that in figure 1 that most of the visible area is contained between about x = 20
and x = 80. In general a useful rule of thumb is that most of the area lies between
x    3 and x    3 When sketching s normal distribution one usually takes
the approximate width of the function as lying between these two values.
69
For example, consider a distribution of resistors having a mean of 100  and a
standard deviation of 5  which are distributed normally. The distribution of a large
number of these would have a frequency curve positioned on the resistance, x, axis as
shown in Figure 2.
0.1
0.08
0.06
0.04
0.02
0
80
90
100
110
120
Figure 2 A Normal Distribution of Resistances with   100 and   5 
Another Normal distribution of resistors with mean 500  and a standard deviation of
10  is shown in Figure 3.
0.05
0.04
0.03
0.02
0.01
0
460
480
500
520
540
Figure 3 A Normal Distribution of Resistances with   500 and   10 
14.3 The Standard Normal Distribution
For convenience in dealing with different distributions a Standardised or Normalised
Normal Distribution is defined whose characteristics are that the area under the
curve is unity,1 ,the mean is zero and u represents the difference between a
variable value, x and its mean value,  expressed in standard deviations.
2
f (u ) 
and



1  u2
e
2
where
2
1  u2
e du  1
2
70
u=
x

The Areas under this standard normal curve have been tabulated and a table of areas
is given on page 72 of these notes, headed “Table 3” (The “3” is not significant, it is
there only because the publisher chose to make it the the third table in the book of
Statistical Tables from which it was taken).
14.4
Interpreting areas under normal curves
Let us now look at how we can we interpret Normal curves.
Take Figure 3 for example:
(1)


14.5
Looking at Figure 3 you see that half of the total area lies to the right of the
population mean   500 . This means that if we had a lot of these resistors
about half of them would be expected have resistance greater than 500 
Look at Figure 3 again and this time try to estimate roughly how much area
lies to the right of x = 520 . This area is useful because it tells us the
fraction of resistors that would be expected to be bigger than 520 .
Obviously different students will get different answers for their estimate of his
area. It’s not a very big area – probably less than about 0.05? In fact there is
a way we can find the area and all get the same answer with no guesswork.
This method uses Tables and is explained in the next section.
Finding normal areas using Tables
Look at the table of Normal areas found on page 72 of these notes. The
numbers in the table are the areas under the standard normal distribution that
has   0 and   1 . But this is apparently no use to us because our problem
in Figure 3 has   500 and   10 . However we can overcome this difficulty
x
by simply changing our variable from x to u where u 
.as follows

the u value corresponding to x = 520 is u 
x


520  500
2
10
(The value u = 2 is really telling us that our number x = 520 is 2 standard
deviations above the mean).
The Table on page 72 tells us that the area we want is A = 0.02275. We
interpret this to mean the proportion of resistors expected to have a resistance
greater than 520 ohms is only 0.02275 or approximately just over 2 resistors in
a thousand.
71
Insert Table 3
72
Question: “5000 resistors are distributed normally with a mean of 100  and a
standard deviation of 2 . Estimate the number of resistors with a resistance
greater than 103 .”
Answer: The link between an actual distribution and the standardised normal
distribution is the equation
x
u=

In this case it is necessary to find the u value corresponding to 103 
i.e.
u=
103  100
 15
.
2
Then using Table 3 find that when u = 1.5 the area in the tail is 0.0668.
Since the total area is 1 the area in the tail also represents the ‘fraction of the whole
distribution’ which has a u value greater than 1.5
This fraction applies to any Normal distribution so the fraction of resistors with u
greater than 1.5 (which also means resistance greater than 103  in this case) is also
0.0668 of the whole.(i.e. 0.0668 of 5000)
Therefore the number of resistors greater than 103 is 0.0668 x 5000 = 334.
Figure 5(a) Actual Distribution
Figure 5(b) Standardised Normal
5000 x
Total
Area =
5000
100
0.0668
Area
in Tail
= 334
= 0.0668
103

0
1.5
u
Answer :
For 103 
u=
103  100
 15
.
2
From Table 3 the fraction in the tail = 0.0668.
Therefore the number likely to be more than 103  is 0.0668 x 5000 = 334
Note: Table 3 gives the fraction ‘in the tail’
73
Question: Estimate the number of resistors between 98  and 103 .
Answer:
This time it is necessary to find two u values. ‘u’ will be positive for
resistors greater than the mean(103 ) and negative for resistors less
than the mean(98 ).
However because the Normal Distribution is symmetrical about the
mean, Table 3 is used for both but if u is positive then the fraction is in
the tail above the mean whereas if u is negative then the fraction is
in the tail below the mean.
For 98 
u=
98  100
 1
2
and fraction in the tail = 0.1587
For 103 
u=
103  100
 15
.
2
and fraction in the tail = 0.0668
0.1587
0.0668
-1
0
1.5
Therefore the fraction between 98  and 103  = 1 - (0.1587 + 0.0668)
= 0.7745
and the estimated number of resistors = 0.7745 x 5000 = 3873 (to nearest whole
number)
Question:
For the same distribution estimate the lowest resistance of the highest
1% of the resistors. (i.e. Use Table 3 ‘in reverse’ to find the u value
corresponding to a fraction in the tail of 0.01 and then convert to a
resistance for this distribution)
Answer:
1%
= 0.01
u
74
Table 3 shows that the fraction ‘0.01’ lies somewhere between the u values 2.32 and
2.33
x
Therefore we estimate the u value as approximately 2.325 and use u =
to find

x
x  100
2.325 =
so that x = 104.65 
2
Thus only 1% of the resistors will be expected exceed 104.65 
15
Probability
15.1
Theoretical Probability
When a coin is tossed it will come down either Heads or Tails. There are only these
two possible outcomes. It is not known in advance which way up the coin will land
and the outcome is said to be uncertain.
Probability is used to express our uncertainty and by definition a probability is a
number between 0 and 1.
If it is impossible for an event to occur the probability of it occurring is said to be 0
whereas
if it is certain that an event will occur the probability of it occurring is said to be 1.
The probability of obtaining a Head in one toss of a coin is one possibility out of two
equally likely possibilities and the probability is defined as
p(Head) =
possible ways of falling heads 1
 = 0.5
total possible ways of falling 2
Similarly the probability of a coin falling Tails =
1
= 0.5
2
Total Probability (which covers all possible events) = 1, meaning that the coin will
fall Heads or Tails
p(Head) + p(Tail) = 0.5 + 0.5 = 1
15.2
Empirical Probability
Often probabilities are estimated by experiment. In this case they are called empirical
probabilities and are found using
probability of an event =
total number of occurrences of the event
total number of trials
75
example: In a factory 500 samples of a particular product are tested and 20
are found to be defective. Estimate the probability that 1 sample
chosen at random would be defective
Answer:
p(defective) = 20/500 = 0.04
15.3 Representing Probabilities by Areas.
The area of a square with unit sides will have an area of unity and may be used to
represent the total probability so that in the case of tossing a coin half of the area
represents ‘heads’ and the other half ‘tails’; and for the die 1/6 th of the area
represents each face .
Heads
Tails
Area = 0.5 Area = 0.5
1
2
3
4
5
6
Referring back to a previous question
“5000 resistors are distributed normally with a mean of 100  and a standard
deviation of 2 . Estimate the number of resistors with a resistance greater than
103 .”
In this case it was necessary to find the u value corresponding to 103 
i.e.
u=
103  100
 15
.
2
Then using Table 3 it was found that when u = 1.5 the area in the tail is 0.0668.
Since the total area is 1, the area in the tail also represents the ‘fraction of the whole
distribution’ which has a u value greater than 1.5
It can now be seen that this area also represents the probability of the u value being
greater than 1.5
This fraction applies to any Normal distribution so the probability that a resistor
chosen at random from this population will have a resistance greater than 103  is
also 0.0668
76
15.4
Error Function
White noise in a digital communication system is a naturally occurring phenomenon
produced by the superposition of many randomly occurring events. Consequently it is
not possible to specify the instantaneous amplitude of the waveform at any given
instant. However these amplitudes follow a normal distribution so it is possible to
determine the probability that the amplitude will exceed a given value.
-x
0
+x
The mean value of white noise voltages is zero and use is made of the standard
deviation which is the root mean square deviation from the mean, , which, since the
mean is zero is the rms noise voltage.(See page 41)
The error function is described in similar terms to the normal distribution but with
the use of K/2 (instead of u) and area across the mean (instead of area in the tail).
The probability that the instantaneous noise amplitude is less than some value K is
probability(-K  v  K) = erf ( K / 2 ) 
1

 K/ 2
 e  y dy 
2
K/ 2
2

K/ 2
e
 y2
dy
0
This integral has been evaluated for values of K and is known as the error function.
e.g. If K=1 then the probability that the noise voltage will be within + or - one
standard deviation of the mean (i.e. that the noise voltage will be less than its rms
value) is found from Normal Error Function tables for erf(1/2) = erf(0.707) = 0.683
(An example of Error Function tables giving values of
erf ( x) 
1

x
 e  y dy 
2
x
2

x
e
 y2
dy
0
for values of x is shown on pages 417 and 418 of Telecommunications Engineering
by J.Dunlop and D.G.Smith: see over-page for copy)
77
The Complementary Error Function is defined as
erfc( x )  1  erf ( x )  1 
1

x
e
x
 y2
dy  1 
2

x
e
 y2
dy
0
Thus the probability that a noise voltage will exceed its rms value is
erfc(1/2) = 1 - 0.683 = 0.317
78
APPENDIX B
This covers more than is needed for your course but is a useful reference.
Hyperbolic functions - sinh, cosh, tanh, coth, sech, csch
DEFINITION OF HYPERBOLIC FUNCTIONS
Hyperbolic sine of x = sinh x = (ex - e-x)/2
Hyperbolic cosine of x = cosh x = (ex + e-x)/2
Hyperbolic tangent of x = tanh x = (ex - e-x)/(ex + e-x)
Hyperbolic cotangent of x = coth x = (ex + e-x)/(ex - e-x)
Hyperbolic secant of x = sech x = 2/(ex + e-x)
Hyperbolic cosecant of x = csch x = 2/(ex - e-x)
RELATIONSHIPS AMONG HYPERBOLIC FUNCTIONS
tanh x = sinh x/cosh x
coth x = 1/tanh x = cosh x/sinh x
sech x = 1/cosh x
csch x = 1/sinh x
cosh2x - sinh2x = 1
sech2x + tanh2x = 1
coth2x - csch2x = 1
FUNCTIONS OF NEGATIVE ARGUMENTS
sinh(-x) = -sinh x
cosh(-x) = cosh x
tanh(-x) = -tanh x
csch(-x) = -csch x
sech(-x) = sech x
coth(-x) = -coth x
79
ADDITION FORMULAS
sinh (x ± y) = sinh x cosh y ± cosh x sinh y
cosh (x ± y) = cosh x cosh y ± sinh x sinh y
tanh(x ± y) = (tanh x ± tanh y)/(1 ± tanh x.tanh y)
coth(x ± y) = (coth x coth y ± l)/(coth y ± coth x)
DOUBLE ANGLE FORMULAS
sinh 2x = 2 sinh x cosh x
cosh 2x = cosh2x + sinh2x = 2 cosh2x — 1 = 1 + 2 sinh2x
tanh 2x = (2tanh x)/(1 + tanh2x)
HALF ANGLE FORMULAS
sinh x/2 = ±
[+ if x > 0, - if x < 0]
cosh x/2 =
tanh x/2 = ±
[+ if x > 0, - if x < 0]
= <(sinh x)/(cosh x — 1) = (cosh x + 1)/sinh x
MULTIPLE ANGLE FORMULAS
sinh 3x = 3 sinh x + 4 sinh3 x
cosh 3x = 4 cosh3 x — 3 cosh x
tanh 3x = (3 tanh x + tanh3 x)/(1 + 3 tanh2x)
sinh 4x = 8 sinh3 x cosh x + 4 sinh x cosh x
cosh 4x = 8 cosh4 x — 8 cosh2 x + 1
tanh 4x = (4 tanh x + 4 tanh3 x)/(1 + 6 tanh2 x + tanh4 x)
POWERS OF HYPERBOLIC FUNCTIONS
sinh2 x = ½cosh 2x — ½
cosh2 x = ½cosh 2x + ½
80
sinh3 x = ¼sinh 3x — ¾sinh x
cosh3 x = ¼\\ cosh 3x + ¾cosh x
sinh4 x = 3/8 - ½cosh 2x + 1/8cosh 4x
cosh4 x = 3/8 + ½cosh 2x + 1/8cosh 4x
SUM, DIFFERENCE AND PRODUCT OF HYPERBOLIC FUNCTIONS
sinh x + sinh y = 2 sinh ½(x + y) cosh ½(x - y)
sinh x - sinh y = 2 cosh ½(x + y) sinh ½(x - y)
cosh x + cosh y = 2 cosh ½(x + y) cosh ½(x - y)
cosh x - cosh y = 2 sinh ½(x + y) sinh ½(x — y)
sinh x sinh y = ½(cosh (x + y) - cosh (x - y))
cosh x cosh y = ½(cosh (x + y) + cosh (x — y))
sinh x cosh y = ½(sinh (x + y) + sinh (x - y))
EXPRESSION OF HYPERBOLIC FUNCTIONS IN TERMS OF OTHERSIn
the following we assume x > 0. If x < 0 use the appropriate sign as indicated by
formulas in the section "Functions of Negative Arguments"
sinh x = u cosh x — u tanh x = u coth x = u sech x = u esch x = u
sinh x u
cosh x
u
u
tanh x u/
coth x
/u
/u u/
sech x 1/
1/u
csch x 1/u
1/
u/
l/
1/
u/
u
1/u
1/u
u
/u
81
/u 1/u
1/u
/u
1/
1/
/u u
u/
u/
u
GRAPHS OF HYPERBOLIC FUNCTIONS
y = sinh x
y = cosh x
y = tanh x
y = coth x
y = sech x
82
y = csch x
INVERSE HYPERBOLIC FUNCTIONS
If x = sinh y, then y = sinh-1 a is called the inverse hyperbolic sine of x. Similarly we
define the other inverse hyperbolic functions. The inverse hyperbolic functions are
multiple-valued and as in the case of inverse trigonometric functions we restrict
ourselves to principal values for which they can be considered as single-valued.
The following list shows the principal values [unless otherwise indicated] of the
inverse hyperbolic functions expressed in terms of logarithmic functions which are
taken as real valued.
sinh-1 x = ln (x +
cosh-1 x = ln (x +
-∞ < x < ∞
)
x ≥ 1 [cosh-1 x > 0 is principal value]
)
tanh-1x = ½ln((1 + x)/(1 - x))
-1 < x < 1
coth-1 x = ½ln((x + 1)/(x - 1))
x > 1 or x < -1
sech-1 x = ln ( 1/x +
0 < x ≤ 1 [sech-1 a; > 0 is principal value]
csch-1 x = ln(1/x +
)
)
x≠0
RELATIONS BETWEEN INVERSE HYPERBOLIC FUNCTIONS
csch-1 x = sinh-1 (1/x)
sech-1 x = cosh-1 (1/x)
coth-1 x = tanh-1 (1/x)
sinh-1(-x) = -sinh-1x
tanh-1(-x) = -tanh-1x
coth-1 (-x) = -coth-1x
83
csch-1 (-x) = -csch-1x
GRAPHS OF INVERSE HYPERBOLIC FUNCTIONS
y = sinh-1x
y = cosh-1x
y = tanh-1x
84
y = coth-1x
y = sech-1x
y = csch-1x
RELATIONSHIP BETWEEN HYPERBOLIC AND TRIGONOMETRIC
FUNCTIONS
sin(ix) = i sinh x
cos(ix) = cosh x
tan(ix) = i tanh x
csc(ix) = -i csch x
sec(ix) = sech x
cot(ix) = -i coth x
sinh(ix) = i sin x
cosh(ix) = cos x
tanh(ix) = i tan x
csch(ix) = -i csc x
sech(ix) = sec x
coth(ix) = -i cot x
85
PERIODICITY OF HYPERBOLIC FUNCTIONS
In the following k is any integer.
sinh (x + 2kπi) = sinh x
csch (x + 2kπi) = csch x
cosh (x + 2kπi) = cosh x
tanh (x + kπi) = tanh x
sech (x + 2kπi) = sech x
coth (x + kπi) = coth x
RELATIONSHIP BETWEEN INVERSE HYPERBOLIC AND INVERSE
TRIGONOMETRIC FUNCTIONS
sin-1 (ix) = isinh-1x
sinh-1(ix) = i sin-1x
cos-1 x = ±i cosh-1 x
cosh-1x = ±i cos-1x
tan-1(ix) = i tanh-1x
tanh-1(ix) = i tan-1x
cot-1(ix) = -i coth-1x
coth-1 (ix) = -i cot-1x
sec-1 x = ±i sech-1x
sech-1 x = ±i sec-1x
csc-1(ix) = -i csch-1x
csch-1(ix) = -i csc-1x
86
APPENDIX C
Amplitude changes in the modulating wave (signal) are used to vary the
instantaneous frequency of the carrier wave from its unmodulated value.
It is possible to write a mathematical expression for the frequency modulated signal :
The constant is the modulation index for FM. It is defined as follows:
The Greek letter δ represents the frequency deviation and fm represents the
modulating frequency that causes the deviation. As in the case of AM, this time
domain representation of the FM signal can be converted to an equivalent frequencydomain expression that includes the carrier and sidebands. Because the mathematics
required for this conversion are quite complex, we will only consider the result:
The Jn(x) functions are known as Bessel Functions of the First Kind. Graphs of Jn(x)
look like slowly decreasing sine and cosine functions. The Jn(x) functions are a
closely related family of functions in the same way that sin(nx) and cos(nx) for a
family of similar functions.
The zeroth order Bessel function, J0(m) determines the amplitude of the carrier. The
nth Bessel function Jn(m) determines the amplitude of the nth pair of sidebands. There
are two important concepts contained in the expression shown above:
The amplitude of the carrier depends on m. the modulation index. This is quite
different from AM, where the amplitude of the carrier was independent of the value of
m
There are an infinite number of sidebands. Thus the theoretical bandwidth of FM is
infinite.
87
An infinite bandwidth signal would be very difficult to transmit. Fortunately, the
higher order sidebands in FM have extremely low amplitude and may be ignored. For
example: if the modulation index is 5, only the first 7 sidebands are significant in
value.
There is a rule of thumb, known as Carson’s Rule, that predicts the bandwidth
occupied by the significant sidebands of an FM signal, based on the maximum
modulation frequency and its corresponding modulation index:
Parameter fm is the frequency of the modulating signal, δmax is the maximum
deviation, and m is the corresponding modulation index. If a range of frequencies is
used to modulate the carrier, the maximum modulating frequency and its
corresponding modulation index are used.
Commercial FM broadcasting uses a maximum deviation of 75 KHz and a maximum
modulating frequency of 15 KHz. Substituting these values into Carson’s Rule gives:
B = 2*(75+15) = 180 KHz.
The carrier amplitude of an FM signal is determined by the value of J0(m). There are
certain values of m for which the carrier amplitude is zero. These permit a technician
to measure FM modulator linearity. This is done as follows:
A sine wave of known frequency is applied to the FM modulator. The amplitude of
the modulating signal is slowly increased while the FM carrier is observed on a
spectrum analyzer. As the amplitude of the modulating signal is increased, the carrier
will decrease to zero. The first zero carrier condition occurs when m = 2.4. At this
point, the deviation δ is 2.4 * modulating frequency. The amplitude of the input signal
is recorded. Then the amplitude of the input is increased more until the second zero
carrier condition is reached. At this point, m = 5.5 and δ is 5.5 * modulating
frequency. The amplitude of the input signal is recorded again. If the modulator is
linear then:
(Amplitude at m = 5.5)/(Amplitude at m = 2.4) = 2.917
88