Roberts(2003) Fundamentals of Signals and Systems

EE2D1 – Fundamentals of Signals and Systems (Maths) – Course Overview Clive Roberts

Course Overview Introduction During this course we will be studying the mathematics required for engineering transforms and analysis in communications and control. We will study:

• Fourier Analysis; • Laplace Transforms; • z-Transforms; • Statistics and Probability.

Key Texts I list two key texts:

• E Kreyszig, Advanced Engineering Mathematics, Wiley, 1999, ISBN 0-471-33328-X

• G James, Advanced Modern Engineering Mathematics, Prentice Hall, 1999, ISBN 0-201-59621-0

However there are a large number of texts available in the library in these areas. It is better to pick a text that you find easy to follow and understand. The notes supplied each week will be comprehensive, however a text will provide further information and extra examples. The notes and other useful information will also be available at www.eee.bham.ac.uk/robertc Tutorials Tutorials (staffed by PGTAs) will take place on Thursday at 11am during weeks 3, 5, 7, 9. You will be expected to hand in work to be marked prior to these tutorials by Tuesday at 11am prior to the tutorial. Support Maths Optional support maths (staffed by PGTAs) will be available on Wednesday at 2pm in NG15 during weeks 3, 6, 8, 10. This will provide an opportunity for those that are having some difficulty to go through the material again in a non- lecture environment and ask questions having been through the material. Reading / Self-Study Week There will be no lectures for this course during week 9 (only a tutorial). This is to provide you with some time to review the course material, catch up with any outstanding work, revise for the class test and commence the coursework exercise. Class Test The class test will be held during the tutorial session of week 11. This will account for 10% of the course’s final mark. Coursework The coursework, given in week 8, will account for 20% of the course’s final mark. Examination During the summer examination period you will sit a 1½ hour paper.

EE2D1 – Fundamentals of Signals and Systems (Maths) – Course Overview Clive Roberts

Timetable Week 1 Tuesday at 2pm – Course Overview Lecture – LT1 Week 2 Monday at 11am – Introduction to Fourier Analysis Lecture – LT1 Wednesday at 11am – Fourier Series Examples and Applications Lecture – LT1 Week 3 Monday at 11am – Introduction to Fourier Transforms Lecture – LT1 Wednesday at 11am – Fourier Transforms Examples and Applications Lecture – LT1 Wednesday at 2pm – Fourier Support Maths – NG15 Thursday at 11am – Fourier Tutorials – Seminar Rooms Week 4 Monday at 11am – Further Fourier Analysis Lecture – LT1 Wednesday at 11am – Introduction to Laplace Transforms Lecture – LT1 Week 5 Monday at 11am – Laplace Transforms Examples and Applications Lecture – LT1 Wednesday at 11am – Laplace Transform Application to DEs Lecture – LT1 Thursday at 11am – Laplace 1 Tutorials – Seminar Rooms Week 6 Monday at 11am – Laplace Transforms Examples and Applications Lecture – LT1 Wednesday at 11am – Further Laplace Transforms Lecture – LT1 Wednesday at 2pm – Laplace Support Maths – NG15 Week 7 Monday at 11am – Intro. to Discrete Systems and z-Transforms Lecture – LT1 Wednesday at 11am – z-Transforms Examples and Applications Lecture – LT1 Thursday at 11am – Laplace 2 Tutorials – Seminar Rooms Week 8 Monday at 11am – MatLab Lecture – LT1 Wednesday at 11am – Coursework Lecture – LT1 Wednesday at 2pm – z-Transform Support Maths – NG15 Week 9 (Reading / Self-study Week) Thursday at 11am – z-Transorm Tutorials – Seminar Rooms Week 10 Monday at 11am – Further z-Transforms Lecture – LT1 Wednesday at 11am – Introduction to Probability and Statistics Lecture – LT1 Wednesday at 2pm – Support Maths – NG15 Week 11 Monday at 11am – Statistics Examples and Applications Lecture – LT1 Wednesday at 11am – Probability Examples and Applications Lecture – LT1 Thursday at 11am – Class Test – Seminar Rooms

EE2D1 – Signals and Systems – Lecture Notes 1 Clive Roberts

Fourier Analysis

Introduction Fourier Analysis is a family of mathematical techniques, all based on decomposing signals into sinusoids. Fourier Analysis is named after Jean Baptiste Joseph Fourier – a French mathematician and physicist. Fourier was interested in heat propagation, and presented a paper in 1807 on the use of sinusoids to represent temperature distributions. The paper contained a novel claim that any continuous periodic signal could be represented as the sum of properly chosen sinusoidal waves. One of the reviewers of the paper, Lagrange, voted against the publication of the paper insisting that the technique could not be used to represent signals with corners (discontinuous slopes such as square waves). It was not until after Lagrange’s death that the work was officially published – some time during the French Revolution, so Fourier probably had better things to think about!! Well, who was right? A good question... Lagrange was right in his assertion that a summation of sinusoids can not exactly form a signal with corner; however, you can get exceptionally close. So close, in fact, that the difference between the actual and reconstructed waveform has zero energy – close enough for us!! This phenomenon is nowadays referred to as Gibb’s Effect – more about that later. Fourier Analysis can be broken into four categories:

1. Periodic – Continuous The examples here include: sine waves, square waves and any other waveform that happens to repeat itself in a regular pattern continuously ( ∞± ). The transform for this type of signal is referred to as Fourier Series.

2. Aperiodic – Continuous This includes, for example, decaying exponentials and the Gaussian curve. These signals extend to both positive and negative infinity without repeating themselves in a periodic pattern. This type of transform is simply known as the Fourier Transform.

3. Periodic – Discrete These are discrete signals that repeat themselves in a periodic fashion from negative to positive infinity. This class of transform is known as the Discrete Fourier Transform or DFT.

4. Aperiodic – Discrete These signals are only defined at discrete points between positive and negative infinity and do not repeat themselves in a periodic manner. This type of transform is called the Discrete Time Fourier Transform.

You may wonder why sinusoids are used rather than, say, square waves or saw tooth waves. There are an infinite number of ways to decompose a signal, so it is possible; however, our aim is to end up with something that is easier to deal with than the original signal. The component sine and cosine waves are simpler than the original signal because they have a property known as sinusoidal fidelity - if the input to a linear system is sinusoidal then the output will also be a sinusoidal at exactly the same frequency as the input (i.e. only a change in amplitude and phase may be seen). In


electronics a circuit is often considered by its frequency response, which produce graphs of how a circuit’s gain and phase vary with frequency.

Figure 1 - The four categories of Fourier Anaylsis

Other important properties of sine and cosine waves which we should be familiar with are:

• Sines and cosines do not have any jumps of kinks in them; • Sines and cosines can be differentiated or integrated as many times as you like

and they don’t develop any discontinuities; • When we differentiate or integrate sines and cosines they stay the same shape.

Periodic Waveforms (revision) In order to undertake a study of Fourier series we therefore need a good understanding of periodic functions, especially sine and cosine functions. Some of the important definitions and properties are listed below:

• The function ( )

+=+=

ωφ

ωφω tAtAtf sinsin)( is a sine wave of amplitude

A, angular frequency ω , frequency π

ω2

, period ωπ2

=T and phase angle φ .

The time displacement is define to be ωφ

. As shown in Figure 2. Similar

remarks can be made about the function ( )φω +tAcos and together the sine and cosine functions form a class of functions known as sinusoids or harmonics.

• It is particularly important that you understand how to integrate these

functions:


o ∫ ∫ +=+−= cn

tntnc

ntn

dttnω

ωω

ωω

ωsin

coscos

.sin

for n = ±1, ±2, ….

• Sometimes a function occurs as the sum of a number of different sine and cosine components such as:

o ttttf 111 4sin7.02sin8.0sin2)( ωωω ++=

Note in particular that the angular frequencies of all the components are integer multiples of angular frequency 1ω . The component with the lowest frequency (or largest period) is t1sin2 ωπ . The quantity 1ω is called the fundamental angular frequency and this component is called the fundamental or first harmonic. The component with the angular frequency 12ω is called the second harmonic and so on. A consequence of this is that the resulting function, f(t), is periodic and has the same frequency as the fundamental. Some harmonics may be missing, for example, above the third harmonic is missing. In power systems application a common value for 1ω maybe 100π as this corresponds to a frequency of 50 Hz.

Fourier Series The Fourier series approximates to a function f(t) by using a trigonometric polynomial of degree N as follows:

o ∑=

++≈N

nnn ntbnta

atf

1

0 )]sin()cos([2

)(

Assuming that f(t) is continuous on the interval ππ ≤≤− t , the coefficients na and

nb can be computed by the formulas:

o ∫−

=π

ππdttfa ).(

10

for the constant term and:

o ∫∫−−

==π

π

π

π ππdtnttfbdtnttfa nn ).sin()(

1).cos()(

1

Example 1 Consider the periodic function:

o

<<<<−

=π

πttt

tf0,

,0,0)(

with period π2 as shown in Figure 5. Using the equation for 0a we get:

o ∫∫ ===−

ππ

π

πππ 0

0 2.

1).(

1dttdttfa


which gives a constant term of 42

0 π=

a.

The coefficients of the cosine terms are computed using the equation for an.

o ∫−

=π

ππdtnttfan ).cos()(

1

o ∫=π

π 0

).cos(1

dtnttan

o π

π 02

)cos(1

)sin(1

+= nt

nnt

nt

an = [ ]1)cos(1

2−π

πn

n

Similarly the sine terms are computed as:

o ∫−

=π

ππdtnttfbn ).sin()(

1

o ∫=π

π 0

).sin(1

dtnttbn

o [ ])cos(1

)sin(1

)cos(1

02

ππ

π

nn

ntn

ntnt

bn −=

+−=

The series approximation can be rewritten using the identity relationship

nn )1()cos( −=π and by knowing that an = 0 when n is an even integer. The result is:

o ∑=

++≈N

nnn ntbnta

atf

1

0 )]sin()cos([2

)(

which expands as:

o .......)3cos(92

)cos(2

4)( −−−≈ tttf

πππ

........)3sin(31

)2sin(21

)sin( ++−+ ttt

Figure 2 shows the plots for the Fourier approximation for Example 1. One to try before the next lecture….. Consider the periodic function:

o ππ <<−+= tttf ,1)( with period π2 . Sketch f(t) and calculate a concise expression for the Fourier Series.

Answer =

−−≈ ∑

=

N

n

n ntn

tf1

sin)1(2

1)(


-4 -3 -2 -1 0 1 2 3 4-1

0

1

2

3

4

t

f(t)

N=5f(t)

-4 -3 -2 -1 0 1 2 3 4-1

0

1

2

3

4

t

f(t)

N=20f(t)

Figure 2 - Fourier series approximations for Example 1


Fourier Series (cont.) Now we have tried a couple of examples, there are a few things we should be aware of.

Firstly, the term2

0a represents the average value (or dc component) of the waveform.

Secondly, the period of integration that we have use up until now has been π2 , but

we can use the Fourier Series over any complete period, i.e. 2Tt −= to

2Tt = . If we

think carefully about the period of integration, it may make the process simpler. The Fourier Series equations then become:

o ∑=

+

+≈

N

nnn T

tnbT

tnaa

tf1

0 2sin2cos2

)( ππ

Assuming that f(t) is continuous on the interval 22

TtT ≤≤− , the coefficients na

and nb can be computed by the formulas:

o ∫−

=2

2

0 ).(2

T

T

dttfT

a

for the constant term and:

∫∫−−

=

=

2

2

2

2

.2sin)(2.2cos)(2T

Tn

T

Tn dt

Ttntf

Tbdt

Ttntf

Ta ππ

Thirdly, the Fourier Series of even and odd functions can be computed with significantly less effort than that needed for functions without such symmetry. The properties that define an even or odd function are:

• An even function has a graph that is symmetric with respect to the vertical axis (t=0) and therefore satisfies the equation )()( tftf −= . For the even function, )(tf e , a range of integration that is symmetrical about the vertical axis, where t = 0, gives:

o ∫∫ =−

ππ

π 0

).(2).( dttfdttf ee

Based on these results, if f(t) is an even function:

∑=

+≈N

nn nta

atf

1

0 )]cos([2

)(

• An odd function is symmetric with respect to the origin and satisfies the

equation )()( tftf −=− . The integration over symmetrical range about the vertical axis for an odd function )(tfo is zero:


o 0).( =∫−

dttfo

π

π

Based on these results, if f(t) is an odd function:

o ∑=

+≈N

nn ntb

atf

1

0 )]sin([2

)(

• A rotated function is symmetric about the zero axis, and the wave shape of alternate half-cycles is identical but reversed in sign and satisfies

)2()( Ttftf +−−= . There is no constant term when rotational symmetry

exists, as the dc component is zero. If the function is also odd, only odd-harmonic sine terms will appear in the Fourier Series. An even function with rotational symmetry will have a Fourier Series consisting of only odd harmonic cosine terms.

Example 2 (an odd example) Consider the periodic function:

o ππ <<−+= tttf ,1)( with period π2 . This is an odd function as )()( tftf −=− ; therefore there will be no an terms.

The dc component of the waveform

20a

is:

o ( ) 12

)1(12

0 =+−++

=ππa

Now we can work out the sine components:

o ∫−

=π

ππdtnttfbn ).sin()(1

o ∫−

+=π

ππdtnttbn ).sin()1(1

o [ ])cos(2)sin(1)cos(112 π

π

π

π

nn

ntn

ntn

tbn −=

+

+−=

−

The series approximation can be rewritten using the identity relationship nn )1()cos( −=π and by knowing that an = 0 when n is an even integer. The result is:

o ∑=

+≈N

nn ntb

atf

1

0 )]sin([2

)(

which we can write concisely as:

o ∑=

−−+≈

N

n

ntn

tf1

2

)sin()1(21)(

or expand as:

o .......3sin322sinsin21)( ttttf +−+≈


Example 3 (an even example) Consider the periodic function:

o

<<−<<−

−<<−

=

ππ

ππ

ππ

t

t

t

tf

2,022,42,0

)(

with period π2 . This is an even function as )()( tftf −= therefore there will be no bn terms.

By inspection, the dc component of the waveform

20a

is:

o 22

0 =a

Now we can work out the cosine components:

o ∫−

=2

2

).cos()(1π

ππdtnttfan

o ∫−

=2

2

).cos(41π

ππdtntan

o

=

=

− 2sin8)sin(14 2

2

πππ

π

π

nn

ntn

an

The result is:

o ∑=

+≈N

nn nta

atf

1

0 )]cos([2

)(

which we can write expanded as:

o .......5cos583cos

38cos82)( ttttf

πππ+−+≈

Half-range Series Sometimes an engineering function is not periodic but is only defined over a finite interval. In cases like this Fourier analysis can still be used. In these cases we attempt to make the waveform periodic in some manner. For example, we may reflect the function in the vertical axis and then repeat in periodically so that the result is a periodic even function. We have performed what is called a periodic extension of the

given function. Within the period of interest of t = 0 to 2Tt = , nothing has changed

but a periodic function has been constructed. We can find the Fourier Series of this periodic function and within the period of interest this will converge to the required function. What happens outside the period of interest is not important. In addition, the periodic function is now even, so the Fourier Series will contain no sine terms. An alternative method would also be to reflect the waveform in both the vertical and t axes to gain an odd periodic function.


Whichever method we choose, the resulting Fourier Series only gives a representation

of the original function in the interval 2

0 Tt << and as such is termed as a half-range

Fourier Series. Similarly we refer to half-range sine series for odd series containing only sine terms and half-range cosine series for even series containing only cosine terms.


Tutorial Sheet 1

1. Find the Fourier Series representation of the function:

o

<<<<−−

=π

πt

ttf

0,40,4

)( period π2


o 0,)( 2 <<−+= ttttf ππ period π2


o

<<<<−+

=10,001),1(2

)(ttt

tf period 2


o

<<<<−

=π

πt

tttf

0,00,

)(2

period π2


o

<<

<<−−=

20,

02,)( TtA

tTAtf period T

6. Find the half-range cosine series representing the function:

o π<<= tttf 0sin)(

7. Graph an appropriate periodic extension of: o 10)( <<= tetf t

and find its half-range cosine series.

8. Find the half-range sine series representing the function: o 202)( <<−= tttf

9. Find the average power developed across a 2Ω resistor by a current signal with a period π2 given by:

o

<<<<−−

=π

πt

ttti

0,00,

)( period π2


Applications of the Fourier Series Parseval’s Theorem If the function f(t) is periodic with period T and has Fourier coefficients of an and bn the Parseval’s theorem states:

o ( )∑∫=−

++==N

nnn ba

adttftf

1

222

022

2.))((1)(

π

ππ

This is frequently used in power calculations. Example 4 Find the average power developed across a 1Ω resistor by a voltage signal with a period π2 given by:

o 2

3cos32sincos)( ttttv +−=

Here, v(t) is already expressed as a Fourier Series with a1 = 1, a3 = 2

1 and b2 = 31− .

All other Fourier coefficients are 0. The instantaneous power is Rv 2

and hence the

average power over one period is given by: :

o dttftv .))((1)( 22 ∫−

=π

ππ

o Vtv 36.121

311)(

2222 =

+

−+=

o WRtv

Pav 68.02

)( 2

==

Frequency Spectrum In general, the Fourier Series of a periodic function with period T seconds contains the fundamental sinusoid and numerous harmonics, some of which may be zero. The plot of the magnitude of the frequency components is called the amplitude, or

frequency spectrum. The frequency components are spaced T

f 10 = hertz apart. On a

graph the spectrum is a series of points (or lines) and is called a discrete spectrum. A discrete Fourier spectrum is characteristic of all periodic functions.


Figure 1 - Discrete Fourier Spectrums of common waveforms

Complex Notation An alternative notation for Fourier Series involving complex number is often used in practical engineering applications. For this we use the Euler relationship:

o ϑϑϑ sincos je j ±=±

from which we can obtain expressions for ϑcos and ϑsin :

o jeeee jjjj

2sin

2cos

ϑϑϑϑ

ϑϑ−− −

=+

=

which enables us to rewrite our general approximation of the Fourier Series as:

o ∑=

+

+≈

N

nnn T

tnbT

tnaa

tf1

0 2sin2cos2

)( ππ

o ∑=

−+

++≈

−−N

nn

Ttnj

n jeebeea

atf

Ttnj

Ttnj

Ttnj

1

2

0

222)(

222 ππππ


o ∑=

−−

+

+−

+≈N

n

Ttnj

nnTtnj

nn ejba

ejbaa

tf1

220

222)(

ππ

which we can write as

o ∑−=

≈N

Nn

Ttnj

nectfπ2

)(

where

o ,....2,122

=+

=−

= − njba

cjba

c nn

nnn

and 2

00

a=c . It can be shown that the Fourier coefficients, c , are the given by: n

o dtetfT

cT

T

Ttnj

n .).(1 2

2

2

∫−

−=

π

Example 5 Find the complex Fourier series representation of the function with period T defined by:

o

<<−=

otherwise

TtTtf

,044,1)(

We find

o dteT

cT

T

Ttnj

n .1 4

4

2

∫−

−=

π

o 4

4

2

21

T

T

Ttnj

n

Ttnj

eT

c

−

−

−=

π

π

o

−

−=

−22

21 ππ

π

jnjn

n eenj

c

o

−=

−

jee

nc

jnjn

n 21 22

ππ

π

o Ttnj

n enn

cππ

π

2

2sin1 −

=

therefore:

o TtnjN

Nnen

ntf

πππ

2

2sin1)(

−

−=∑=

Frequency Response of a Linear System If a sine wave of amplitude Ai is applied to a linear system then the amplitude Ao of the output is given by:


o io AjGA )( ω= and the phase shift, φ , is given by:

o )( ωφ jG∠= Note that Ao and φ are dependent upon ω . It is important to note that )( ωjG is a frequency dependent function. It is now possible to analyse the effect of applying a generalised periodic waveform to a linear system. The first stage is to calculate the Fourier components of the input waveform. The amplitude and phase shift of each of the output components is then calculated using )( ωjG . Finally, the output components are added to obtain the output waveform. Example 6 Consider the low-pass filter in the diagram below. Using Kirchoff’s voltage law and Ohm’s law we obtain:

o oi viRv +=

C

R

Vin Vout

For the capacitor:

o Cj

ivo ω=

Eliminating i gives:

o )1( RCjvvRCjvv oooi ωω +=+=

o RCj

jGAA

vv

i

o

i

o

ωω

+===

11)(

It is convenient to convert )( ωjG to a polar form.

o RCRC

jGωω

ω12 tan)(1

01)(−∠+

∠=

o RCRC

jG ωω

ω 1

2tan

)(11)( −∠

+=


Therefore:

o 2)(1

1)(RC

jGω

ω+

=

o RCjG ωω 1tan)( −−=∠ Note that the circuit is a low-pass filter; it allows low frequencies to pass easily and rejects high frequencies. The cut-off point of the filter, that is the at which significant frequency attenuation begins to occur, can be varied by changing the values of R and C. The quantity RC is usually known as the time constant for the system, τ . When

3.0=τ the equations of gain and phase shift reduce to:

o 209.01

1)(ω

ω+

=jG

o ωω 3.0tan)( 1−−=∠ jG If we examine the response of the system to a given input:

o period is

<<−−<<

=0,1

0,1)(

tt

tfπ

ππ2

The waveform is a rotated function so will not contain any dc component, 02

0 =

a ,

and will only contain odd harmonic sine components. Therefore calculating the Fourier Series reduces to:

o ∑=

≈N

nn ntbtf

1sin)(

and

o ∫−

=π

ππdtnttfbn .sin)(1

therefore:

o

−+−= ∫∫

−

π

ππ 0

0

.sin.sin1 dtntdtntbn

o

−+

=

−

π

ππ 0

0 coscos1n

ntn

ntbn

o ( )( ) ([ ]0coscoscos0cos1−−−+−−= nn

nbn ππ

π)

o ( ) )cos1(2cos221 nn

nn

bn ππ

ππ

−=−=

The values of the first few (odd) coefficients are:

o π

ππ

4)cos1(21 =−=b

o π

ππ 3

4)3cos1(32

3 =−=b


o π

ππ 5

4)5cos1(52

5 =−=b

The next stage is to evaluate the gain and phase changes of the Fourier components. n = 1

o 11 =ω

o 96.0109.01

1)( 1 =×+

=ωjG

o °−=−=∠ − 7.163.0tan)( 11ωjG

n = 3

o 33 =ω

o 74.0909.01

1)( 3 =×+

=ωjG

o °−=−=∠ − 0.429.0tan)( 13ωjG

n = 5

o 55 =ω

o 55.02509.01

1)( 5 =×+

=ωjG

o °−=−=∠ − 3.565.1tan)( 13ωjG

The output of the system can therefore be calculated as:

o ( ) )3.565sin(5

2.2)0.423sin(396.27.16sin84.3

0 °−+°−+°−= ππ

ππ

ππ

v

Figure 2 - Input and Output of the low-pass filter


Fourier Transforms Introduction From the work to date we have seen that almost any periodic signal can be represented as a linear combinations of sine and cosine waves of various frequencies and amplitudes. All frequencies are integer multiples of the fundamental frequency. However, many waveforms are not periodic, examples include pulse signals and noise signals. The Fourier techniques can still be useful by the applications of the Fourier Transform. The Fourier Transform is used in communications engineering and signal processing. For example, it can be used to analyse the processes of modulation, which involves superimposing an audio signal on to a carrier signal, and demodulation, which involves removing the carrier signal to leave the audio signal. Under certain conditions it can be shown that a non-periodic function f(t), can be expressed not as the sum of sine and cosine waves bus as an integral:

o ( )∫∞

+=0

sin)(cos)()( ωωωωω dtBtAtf

where:

o ∫∫∞

∞−

∞

∞−

== dtttfBdtttfA .sin)(1

)(.cos)(1

)( ωπ

ωωπ

ω

Provided that: 1. f(t) and f’(t) are piecewise continuous in every finite interval;

2. ∫∞

∞−

dttf )( exists.

At a point of discontinuity of f(t) the integral representation converges to the average value of the right hand and left hand limits. As with Fourier Series, and equivalent complex representation exists which is more commonly used:

o ∫∞

∞−

= ωωπ

ω deFtf tj .).(21

)(

where:

o ∫∞

∞−

−= dtetfF tj .).()( ωω

These equations form what is referred to as a Fourier Transform Pair. The Fourier Transform of f(t) is )(ωF which is sometimes written as )(tfℑ . Similarly f(t) is the Inverse Fourier Transform of )(ωF , usually denoted by )(1 ωF−ℑ . The Fourier Transform of f(t) is defined to be:

o dtetfFtf tj∫∞

∞−

−==ℑ ωω ).()()(


There is no universal convention concerning the definition of these integrals and a number of variants are still correct. Therefore, it is important to be aware of possible differences when consulting text books. Example 1 Find the Fourier Transform of the function tetutf −= )()( , where u(t) is the unit step

function i.e.

<<∞−∞<<

0001

tt

.

o dtetfF tj∫∞

∞−

−= ωω )()(

o dteeF tjt∫∞

−−=0

)( ωω

o dteF tj∫∞

+−=0

)1()( ωω

o ∞+−

+−

=0

)1(

)1()(

ωω

ω

je

Ftj

o ω

ωj

F+

=1

1)( as 0)1( →+− tje ω as ∞→t

Some Properties of the Fourier Transform

Figure 1 - Homogeneity of the Fourier Transform


Figure 2 - Additivity of the Fourier Transform

The time and frequency domains are alternative ways of representing signals. The Fourier Transform is the mathematical relationship between these two representations. If a signal is modified in one domain, it will also be changed in the other domain, although usually not in the same way. Mathematical operations, such as addition, scaling and shifting, also have a matching operation in the opposite domain. These relationships are called Properties of the Fourier Transform, how a mathematical change in one domain results in a mathematical change in the other domain. The Fourier Transform is linear, that is, it possesses the properties of homogeneity and additivity. Homogeneity means that a change in amplitude in one domain produces an identical change in amplitude in the other domain, i.e. when the amplitude of a time domain waveform is changed, the amplitude of the sine and cosine waves making up that waveform must also change by an equal amount. This is shown in Figure 1, and represented mathematically as:

o )( fkfk ℑ=ℑ


Additivity of the Fourier Transform means that the addition in one domain corresponds to addition in the other domain. An example of this is shown in Figure 2. Since the two time domain signals add to produce a new time domain signal, the two corresponding spectra add to produce a new spectrum. This is represented mathematically as:

o gfgf ℑ+ℑ=+ℑ Example 2 We saw in Example 1 that:

o ωj

etu t

+=ℑ −

11

)(

Also:

o dteetuetu tjtt .).()( 22 ω−−∞

∞−

− ∫=ℑ

o dteetu tjt .)(0

)2(2 ∫∞

+−− =ℑ ω

o ∞+−

−

+−

=ℑ0

)2(2

)2()(

ω

ω

je

etutj

t

o ωj

etu t

+=ℑ −

21

)( 2

Therefore:

o ωω jj

etuetu tt

++

+=+ℑ −−

21

11

)()( 2

o )2)(1(

12)()( 2

ωωωω

jjjj

etuetu tt

+++++

=+ℑ −−

o )2)(1(

23)()( 2

ωωω

jjj

etuetu tt

+++

=+ℑ −−

Figure 3 shows how the phase is affected when the time domain waveform is shifted to the left or right. This is referred to as the First Shift Theorem. The magnitude has not been included, as it is not changed by the shift. In Figures 3a to 3d the waveform is gradually shifted from having a peak at 128 samples to have it at sample 0. The time domain waveform in Figure 3 is symmetrical around a vertical axis, i.e. the left and right sides are mirror images of each other. Signals with this type of symmetry are referred to as having linear phase, because the phase of their frequency spectrum is a straight line. Signals that don’t have the left-right symmetry are called nonlinear phase, and have phase that are something other than a straight line. When the time domain waveform is shifted to the right, the phase remains a straight line, but experiences a decrease in slope. When the time domain is shifted to the left, there is an increase in the slope. This is the main property you need to remember from this section; a shift in the time domain corresponds to changing the slope of the phase.


This is represented mathematically by the First Shift Theorem:

o )()( aFtfe jat −=ℑ ω where a is a constant


Example 3 Show that the Fourier transform of:

o <<−

=otherwise

ttf

0223

)(

is given by ω

ωω

2sin6)( =F . Then use the shift theorem to find the Fourier

Transform of )(tfe jt− .

o ∫− −

−−

−

==2

2

2

2

33)(ω

ωω

ω

je

dteFtj

tj

o

−

−=

−

ωω

ωω

jee

Fjj 22

3)(

o

−=

−

ωω

ωω

jee

Fjj

26)(

22

from Euler’s jee jj

2sin

ϑϑ

ϑ−−

=

o ωω

ω 2sin6

)( =F

We have ωω

ω 2sin6

)()( ==ℑ Ftf . Using the first shift theorem with a = -1 we

obtain:

o )1(2sin1

6)1()( +

+=+=ℑ − ω

ωωFtfe jt


Fourier Transforms (continued) First and Second Shift Theorems We can represent the First Shift Theorem (frequency shift) mathematically as:

o )()( aFtfe jat −=ℑ ω where a is a constant We can represent the Second Shift Theorem (time shift) mathematically as:

o )()( ωFeatf jat−=−ℑ where a is a constant Spectra In the Fourier analysis of periodic waveforms we stated that although a waveform physically exists in the time domain it can be regarded as comprising of components with a variety of temporal frequencies. The amplitude and the phase of these components are obtained through Fourier coefficients an and bn. This is known as a frequency domain description. Plots of amplitude versus frequency and phase versus frequency are together known as the spectrum of a waveform. Periodic function have discrete or line spectra as we have previously seen. Only a discrete set of frequencies is required to synthesize a periodic waveform. However, when analyzing non-periodic waveforms via the Fourier Transform we find that a continuous range of frequencies is required. Instead of discrete spectra we have a continuous spectra. The modulus of the Fourier Transform, )(ωF , gives the spectrum amplitude, while it argument ))(arg( ωF describes the spectrum’s phase. Amplitude Modulation Amplitude modulation is a technique that allows audio signals to be transmitted as electromagnetic radio waves. The maximum frequency of audio signals is typically 10 kHz. If these signals were to be transmitted directly then it would be necessary to use a very large antenna. This can be seen by calculating the wavelength of an electromagnetic wave of frequency 10 kHz using the formula:

o λfc = where c is the velocity of an electromagnetic wave is a vacuum (3 × 108 ms-1), f is the frequency of the wave and λ is its wavelength. λ therefore is equal to 30,000 m. It can be shown that an antenna must have dimensions of at least one-quarter of the wavelength of the signal being transmitted if it is to be reasonably efficient. Clearly a very large antenna would be needed to transmit a 10 kHz signal directly. The solution is to have a carrier signal of a much higher frequency than the audio signal which is usually termed the modulation signal. This allows the antenna to be a reasonable size as a higher frequency signal has a lower wavelength. The arrangement for mixing the two signals is shown in Figure 1. If an amplitude modulated signal is given by:

o ttxt cωφ cos)()( = where cω is the angular frequency of the carrier signal and x(t) is the modulation signal. Using Euler’s Identities )(tφ can be expressed as:


X)(tx

tcωcosCarrier Signal

Audio Signal Modulated Signalttxt cωφ cos)()( =

Antenna

Figure 1 - Block diagram of amplitude modulation

o 2

)()(tjtj cc ee

txtωω

φ−+

=

Taking the Fourier Transform and using the first shift theorem gives (this represents a time shift):

o

+

ℑ=Φ=ℑ−

2)(

)()()(tjtj cc ee

txtωω

ωφ

o

ℑ+

ℑ=Φ=ℑ−

2)(

2)(

)()(txetxe

ttjtj cc ωω

ωφ

o ))()((21

)()( cc XXt ωωωωωφ ++−=Φ=ℑ

where )()( txX ℑ=ω , the frequency spectrum of the modulation signal. Figure 2a shows an audio signal with a DC bias such that the signal always has a positive value. Figure 2b shows that its frequency spectrum is composed of frequencies from 300 Hz to 3 kHz, the range needed for voice communication, plus a spike for the DC component. All other frequencies have been removed by analog filtering. Figures 2c and 2d show the carrier signal, a pure sinusoid of much higher frequency than the audio signal. In the time domain, amplitude modulation consists of multiplying the audio signal by the carrier wave. As shown in 2e, this results in an oscillatory waveform that has an instantaneous amplitude proportional to the original audio signal. The envelope of the carrier wave is equal to the modulating signal. This signal can be routed to an antenna, converted into a radio wave, and then detected by a receiving antenna. This results in a signal identical to 2e being generated in the radio receiver's electronics. A detector or demodulator circuit is then used to convert the waveform in 2e back into the waveform in 2a. Since the time domain signals are multiplied, the corresponding frequency spectra are convolved. That is, 2f is found by convolving 2b and 2d. Since the spectrum of the carrier is a shifted delta function, the spectrum of the modulated signal is equal to the audio spectrum shifted to the frequency of the carrier. This results in a modulated


Figure 2 - The Amplitude Spectrum of the he Amplitude Modulated Signal with πω 3=c

Figure 3 - Amplitude Modulation

spectrum composed of three components: a carrier signa;, an upper sideband, and a lower sideband. These correspond to the three parts of the original audio signal: the DC component, the positive frequencies between 0.3 and 3 kHz, and the negative frequencies between -0.3 and -3 kHz, respectively. Even though the negative frequencies in the original audio signal are somewhat elusive and abstract, the resulting frequencies in the lower sideband are as real as you could want them to be.


For example, consider the frequency spectrum for television transmission. A standard TV signal has a frequency spectrum from DC to 6MHz. By using these frequency-shifting techniques, 82 of these 6 MHz wide channels are stacked end-to-end. For instance, channel 3 is from 60 to 66MHz, channel 4 is from 66 to 72 MHz, channel 83 is from 884 to 890 MHz, etc. The television receiver moves the desired channel back to the DC to 6MHz band for display on the screen. This scheme is called frequency Domain multiplexing. Convolution and Correlation Convolution is an important technique in signal and image processing. It provides a means of calculating the response (output) of a system to an arbitrary input signal if the impulse response is known. The impulse response is the response of the system to an impulse function. Convolving the input signal and the impulse response results in the response to the arbitrary system. Correlation is a second important technique. It can be used to determine the time delay between a transmitted signal and a received signal as they might occur in radar or sonar detection equipment. Convolution and the Convolution Theorem It f(t) and g(t) are two real piecewise continuous functions, their convolution, which is denoted by f * g, is defined as:

o ∫∞

∞−

−=∗ λλλ dtgfgf )()(

f * g is itself a function of t, so we sometimes will write (f * g)(t). Convolution is an integral with respect to λ (a dummy variable). The convolution theorem states that convolution in the time domain corresponds to multiplication in the frequency domain:

o If )()( ωFtf =ℑ and )()( ωGtg =ℑ then… o )()( ωω GFgf =∗ℑ

This theorem gives us a technique for calculating the convolution of two function using the Fourier Transform, since:

o )()(1 ωω GFgf −ℑ=∗ So it is possible to find the convolution of f(t) and g(t) by:

1. finding g the corresponding Fourier Transforms, )(ωF and )(ωG ;

2. multiplying these together to form )()( ωω GF ; 3. finding the inverse Fourier Transform which gives us f * g.

Correlation and Correlation Theorem If f(t) and g(t) are two piecewise continuous function their correlation, which is denoted as f ê g, is defined as:

o f ê g = ∫∞

∞−

− λλλ dtgf )()(

This formula is the same as that for convolution except that the function g is not folded; that is we have )( tg −λ here rather than )( λ−tg .


Tutorial Sheet 2 1. Find the Fourier Transform of the function:

o

<≤≤−

−<

=t

t

t

tfπ

ππ

π

01

0

)(

2. Find the Fourier Transform of the function:

o

<≤<

≤≤−−<≤−

−<

=

tt

tt

t

tf

πππππ

πππ

2022

11

221

20

)(


o

<≤<

≤≤−

−<≤−−<

=

tt

t

tt

tf

πππ

ππ

πππ

022

122122

10

)(


o

<

≤≤−

−<

=

t

tt

t

tf

2022

20

)(π

ππ

π


o

<

≤≤−+

−<

=

t

tt

t

tf

20222

20

)(π

πππ

π


o

≤>

=−

00

)(tete

tft

t


Laplace Transforms The Laplace transform is a well-established mathematical technique for solving differential equations. It is named in honor of the great French mathematician, Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. As illustrated in Figure 1, the Laplace transform changes a signal in the time domain into a signal in the s-domain, also called the s-plane. The time domain signal is continuous, extends to both positive and negative infinity, and may be either periodic or aperiodic. The Laplace transform allows the time domain to be complex.

Figure 1 - The Laplace Transform

As shown in Figure 1, the s-domain is a complex plane (i.e. there are real numbers along the horizontal axis and imaginary numbers along the vertical axis). The distance along the real axis is expressed by the variable, σ . Likewise, the imaginary axis uses the variable, ω , the natural frequency. This coordinate system allows the location of any point to be specified by providing values for σ and ω . Using complex notation, each location is represented by the complex variable, s, where: ωσ js += . Just as with the Fourier transform, signals in the s-domain are represented by capital letters.


For example, a time domain signal, )(tx , is transformed into an s-domain signal, )(sX , or alternatively, ),( ωσX . The s-plane is continuous, and extends to infinity in

all four directions. Just as the Fourier transform analyses signals in terms of sinusoids, the Laplace transform analyses signals in terms of sinusoids and exponentials. From a mathematical standpoint, this makes the Fourier transform a subset of the more elaborate Laplace transform. Figure 1 shows a graphical description of how the s-domain is related to the time domain. To find the values along a vertical line in the s-plane (the values at a particular σ ), the time domain signal is first multiplied by the exponential curve: te σ− . The left half of the s-plane multiplies the time domain with exponentials that increase with time (σ < 0), while in the right half the exponentials decrease with time (σ > 0). Next take the complex Fourier transform of the exponentially weighted signal. The resulting spectrum is placed along a vertical line in the s-plane, with the top half of the s-plane containing the positive frequencies and the bottom half containing the negative frequencies. Take special note that the values on the y-axis of the s-plane ( 0=σ ) are exactly equal to the Fourier transform of the time domain signal. As discussed previously the Fourier Transform can be given as:

o ∫∞

∞−

−= dtetfF tj .).()( ωω

This can be expanded into the Laplace transform by first multiplying the time domain signal by the exponential term:

o [ ] dteetfFsF tjt ..).(),()( ωσωσ −∞

∞−

−∫==

The location in the complex plane can be represented by the complex variable, s, where ωσ js += . This allows the equation to be reduced to:

o ∫∞

∞−

−= dtetxsF st .).()(

Figure 2 shows three pairs of points in the s-plane: A and A’, B and B’, and C and C’. Since each of these pairs has specific values for σ and ω± , there are two waveforms associated with each pair: tet σω −).cos( and tet σω −).sin( . For instance, points A and A’ are at a location of 5.1=σ and 40±=ω , and therefore correspond to the waveforms: tet 5.1)40cos( − and tet 5.1)40sin( − . Figure 3 shows an example of a time domain waveform, its frequency spectrum and its s-domain representation. The example time domain signal is a rectangular pulse of width two and height one. As we have previously seen earlier in the course the complex Fourier transform of this signal is a sinc function in the real part, and zero in the imaginary part. The s-domain is an undulating two-dimensional signal.


Figure 2 - Waveforms associated with the s-domain

Figure 3 -Time, frequency and s-domains


An analogy of how the Laplace transform is used in signal processing is…. Imagine you are traveling by train at night between two cities. Your map indicates that the path is very straight, but the night is so dark you cannot see any of the surrounding countryside. With nothing better to do, you notice an altimeter on the wall of the carriage and decide to keep track of the elevation changes along the route. Being bored after a few hours, you strike up a conversation with the conductor: “Interesting terrain,” you say. “It seems we are generally increasing in elevation, but there are a few interesting irregularities that I have observed”. Ignoring the conductor's obvious disinterest (and the scared look in his eyes!!), you continue: “Near the start of our journey, we passed through some sort of abrupt rise, followed by an equally abrupt descent. Later we encountered a shallow depression”. Thinking you might be dangerous or demented, the conductor decides to respond: “Yes, I guess that is true. Our destination is located at the base of a large mountain range, accounting for the general increase in elevation. However, along the way we pass on the outskirts of a large mountain and through the center of a valley”. Now, think about how you understand the relationship between elevation and distance along the train route, compared to that of the conductor. Since you have directly measured the elevation along the way, you can rightly claim that you know everything about the relationship. In comparison, the conductor knows this same complete information, but in a simpler and more intuitive form: the location of the hills and valleys that cause the dips and humps along the path. While your description of the signal might consist of thousands of individual measurements, the conductor's description of the signal will contain only a few parameters. To show how this is analogous to signal processing, imagine that we are trying to understand the characteristics of some electric circuit. To aid in our investigation, we carefully measure the impulse response and/or the frequency response. However, this does not mean that you know the information in the simplest way. In particular, you understand the frequency response as a set of values that change with frequency. Just as in our train analogy, the frequency response can be more easily understood in terms of the terrain surrounding the frequency response. That is, by the characteristics of the s-plane. With the train analogy in mind, look back at Figure 3. How does the shape of this s-domain aid in understanding the frequency response? It doesn't! The s-plane in this example provides no insight into why the frequency domain behaves as it does. This is because the Laplace transform is designed to analyze a specific class of time domain signals: impulse responses that consist of sinusoids and exponentials. If the Laplace transform is taken of some other waveform the resulting s-domain is meaningless. Systems that belong to this class are extremely common in science and engineering. This is because sinusoids and exponentials are solutions to differential equations One of the things we know about the impulse response is that it will be composed of sinusoids and exponentials, because these are the solutions to the differential equations that govern the system. Try as we might, we will never find this type of system having an impulse response that is, for example, a square pulse or triangular waveform. Also, the impulse response will be infinite in length. That is, it has nonzero values that extend from 0=t to ∞=t . This is because sine and cosine waves have a constant amplitude, and exponentials decay toward zero without ever actually reaching it. If the system we are investigating is stable, the amplitude of the impulse response will become smaller as time increases, reaching a value of zero at ∞=t .


There is also the possibility that the system is unstable, for example, an amplifier that spontaneously oscillates due to an excessive amount of feedback. In this case, the impulse response will increase in amplitude as time increases, becoming infinitely large. Even the smallest disturbance to this system will produce an unbounded output. The center column in Figure 5 shows the impulse response of an RLC notch filter. It contains an impulse at t = 0, followed by an exponentially decaying sinusoid. As illustrated in (a) through (e), we will probe this impulse response with various exponentially decaying sinusoids. Each of these probing waveforms is characterized by two parameters: ω , that determines the sinusoidal frequency, and σ , that determines the decay rate. In other words, each probing waveform corresponds to a different location in the s-plane, as shown by the s-plane diagram in Figure 4. The impulse response is probed by multiplying it with these waveforms, and then integrating the result from −∞=t to ∞=t . This action is shown in the right column. Our goal is to find combinations of σ and ω that exactly cancel the impulse response being investigated. This cancellation can occur in two forms: the area under the curve can be either zero, or just barely infinite. All other results are uninteresting and can be ignored. Locations in the s-plane that produce a zero cancellation are called zeros of the system. Likewise, locations that produce the “just barely infinite” type of cancellation are called poles. Poles and zeros are analogous to the mountains and valleys in our train analogy, representing the terrain “around” the frequency response.

Figure 4 - Pole-Zero Example


Figure 5 - Probing the Impulse Response


Laplace Transforms (cont.) The Laplace Transform has the same properties to the Fourier Transform, namely: Additivity

o )()()( gfgf lll +=+ To find the Laplace Transform of a sum of function, we sum the Laplace Transforms of the individual functions. Homogeneity:

o )()( fkkf ll = If we multiply a function by a constant, k, then the corresponding transform is also multiplied by k. Example 1 Find the Laplace Transform of the following:

o tet 25 2 −

o tt etet lll 2525 22 −=−

o 1

21025

32

−−=−

sset tl

First Shift Theorem:

o )()( asFtfe at +=−l We obtain )( asF + by replacing every s in F(s) by s + a. The variable s has been shifted by an amount a. Example 2 Use the first shift theorem to calculate the Laplace Transform of ttetf t 5sin)( 3−= .

o 22 )25(

105sin

+=

ss

ttl from the tables

o )3()(3 +=− sFtfe tl the first shift theorem

o ( )( )22

3

253

)3(10)(

++

+=−

s

stfe tl

o ( )22

3

346

)3(10)(

++

+=−

ss

stfe tl

Second Shift Theorem:

o )()()( sFedtfdtu sd−=−−l


The function, )()( dtfdtu −− , is obtained by moving )()( tftu to the right (delaying) by an amount d. Example 3

Given 9

2)(

+=

ss

tfl , find )2()2( −− tftul

Use the second shift theorem with d = 2.

o 9

2)2()2((

2

+=−−

−

sse

tftus

l

Laplace Transforms of Derivatives and Integrals Later we will use the Laplace Transform to solve differential equations. In order to do this we need to be able to find the Laplace Transform of derivatives of functions. Let f(t) be a function of t, and f’ and f’’ the first and second derivatives of f. The Laplace Transform of f(t) is F(s). Then:

o )0()(' fssFf −=l o )0(')0()('' 2 fsfsFsf −−=l

where )0(f and )0('f are the initial values of f and f’. The general case for the Laplace Transform to an nth derivative is thus:

o )0(...)0(')0()( )1(21)( −−− −−−−= nnnnn ffsfssFsfl Another useful and important result is:

o )(1

).(0

sFs

dttft

=

∫l

Example 4 Given x(0) = 2 and x’(0) = -1, calculate the expressions for the Laplace Transform of

xxx +− '3''2 . o 2)()0()(' −=−= ssXxssXxl o 12)()0(')0()('' 22 +−=−−= ssXsxsxsXsxl

Therefore: o )()2)((3)12)((2'3''2 2 sXssXssXsxxx +−−+−=+−l o 84)()132('3''2 2 +−+−=+− ssXssxxxl

where the Laplace Transform of x(t) is X(s). Example 5 The voltage, v(t), across a capacitor of capacitance C is given by:

o ∫=t

dttiC

tv0

).(1

)(


Taking the Laplace Transform gives:

o )(1

)( sICs

sV =

where the Laplace Transform of v(t) is V(s) and the Laplace Transform of i(t) is I(s). Inverse Laplace Transforms Before we attempt to solve differential equations using the Laplace Transform we must consider the inverse Laplace Transform. We now must consider the problem of finding a function f(t), having been given the Laplace Transform, F(s). The Laplace Transform tables and the linearity properties of the Laplace Transform help us to do this. Example 6

Find the inverse Laplace Transform of 3

16s

o 23

1 816

ts

=

−l

from the Laplace Transforms Tables and knowledge of homogeneity The inverse Laplace Transform of a fraction is often best found by expressing it as its partial fractions, and finding the inverse transform of these. Example 7

Find the inverse Laplace Transform of ss

s−−

2

14.

o 1)1(

14142 −

+=−−

=−−

sB

sA

sss

sss

To solve: o BssAs +−=− )1(14

substituting for s = 1 gives: o 3=B and 1=A

therefore:

o tess

311

31 11 +=

−+

−− ll

We can also solve the inverse Laplace Transform using complex numbers. We use this case when all the factors in the denominator are linear (i.e. no quadratic factors). Example 8

Find the inverse Laplace Transform of 136

32 ++

+ss

s.

First we factorise the denominator:

o 2

)13(43662

42 −±−=

−±−=

aacbb

s


o js 232

166±−=

−±−=

The denominator can then be factorised as (s – a)(s – b) where ja 23 +−= and jb 23 −−= .

o bs

Bas

Absas

sss

s−

+−

=−−

+=

+++

))((3

1363

2

To solve: o )()(3 asBbsAs −+−=+

substituting for s = -3 + 2j gives: o )4()23(2 jAbjAj =−+−=

o 21

=A and 21

=B

therefore:

o

−+

−= −−−

bsassF

1211

21

)( 111 lll

o 2

)(1btat ee

sF+

=−l

o )(21

)( )23()23(1 tjtj eesF −−+−− +=l

o )(21

)( 2231 jtjtt eeesF −−− +=l

o )2sin2cos2sin2(cos21

)( 31 tjttjtesF t −++= −−l

o tesF t 2cos)( 31 −− =l


Solving Differential Equations using the Laplace Transforms In engineering, differential equations are commonly used to model dynamic systems. These are systems that change with time. Examples include electronic circuits with time-dependent currents and voltages, a chemical production line in which pressures, tank levels, flow rates, etc. vary with time and a semiconductor device in which hole and electron densities change with time. Up until now we have considered how to find the Laplace Transform of a function of time and how to find the inverse Laplace Transform. We now are going to apply this to finding the particular solution to differential equations. The initial conditions are automatically satisfied when solving an equation using the Laplace Transform. They are contained in the transform of the derivative terms. The Laplace Transform of the equation is found. This transforms the differential equation into an algebraic equation. The transform of the dependent variable is found and then the inverse transform is calculated to give the required particular solution. Example 1 Solve the differential equation:

o 3)0(0 ==+ xxdydx

using the Laplace Transform. The Laplace Transform of each term is found:

o )0()(),()( xsXsdydx

sXx −=

= ll

o 3)()0()()'( −=−==

ssXxsXsx

dydx

ll

Therefore o 0)(3)( =+− sXssX o 3)()( =+ sXssX o 3)()1( =+ sXs

o 1

3)(

+=

ssX

Taking the Laplace Transform using the Laplace Transform tables gives: o tetx −= 3)(


Tutorial Sheet 3 1. Solve the following equation using Laplace Transforms:

o 22''' −=−+ yyy 1)0(',2)0( == yy

2. Solve the following equation using Laplace Transforms: o texxx −=−+ 2'2'' 0)0(')0( == xx

3. Solve the following equation using Laplace Transforms: o 466'5'' −=+− txxx 2)0(',1)0( == xx

4. Consider the circuit in Figure 1. Before the switch is closed, the capacitor has

a voltage V across it. If the switch is closed at t = 0 a current will flow in the circuit and the voltage, v, across the capacitor decays with time. Derive a differential equation for the circuit, and hence (using the Laplace Transform)

prove that RCt

evv−

= )0( for 0≥t .

5. Suppose an electronic thermometer is used to measure the temperature of an oven. The sensing element does not respond instantly to changes in the oven temperature because it takes a long time for the element to heat up or cool down. Provided the electronic circuitry does not introduce further time delays then the differential equation that models the thermometer is given by:

o 0vvdt

dvm

m =+τ

where vm = measured temperature, vo = oven temperature, τ = time constant of the sensor. For convenience the temperature is measured relative to the ambient room temperature, which forms a base line for temperature measurement. If the sensing element of an electronic thermometer has a time constant of 2 seconds. If the temperature of the oven increases linearly at the rate of 3 oCs-1 starting from an ambient room temperature of 20 oC at t = 0, calculate the response of the thermometer to the changing oven temperature.

Figure 1


Laplace Transforms in Control Engineering Transfer Functions It is possible to obtain a mathematical model of an engineering system that consists of one or more differential equations. We have seen how the solution to differential equations can be found using the Laplace Transform - this leads to the concept of a transfer function. Consider the differential equation assuming it models an engineering system:

o 0)0()()()(

0 ===+ xxtftxdt

tdx

The f(t) represents the input to the system and x(t) represents the output, or response of the system to the input f(t). Taking the Laplace Transform of the differential equation gives:

o )()()( 0 sFsXxssX =+− o )()()1( sFsXx =+

therefore:

o s

sGsFsX

+==

11

)()()(

As we know from our studies in Systems Engineering, )()(

sFsX

is called the transfer

function. It is the ratio of the Laplace Transform of the output to the Laplace Transform of the input. It is usually denoted as G(s). The transfer function provides an algebraic relationship between the input and the output of an engineering system, thus it allow the analysis of dynamic systems based on differential equations. The example above assumes zero initial conditions. This is necessary to form the transfer function, without this assumption the relationship between the input and the output would have been more complicated. The relationship would vary depending on how much energy was stored in the system at

0=t . By assuming zero initial condition, the transfer function depends purely on the system characteristics. If the effect of the combination of a systems input and initial conditions is required, it is necessary to carry out a full solution of the differential equations as previously considered. In practice however, there are many cases where the simplified treatment provided by the transfer function is perfectly acceptable. Transport Lag Transport Lag is a term used to describe the time delay that may be present in certain engineering systems. An example would be a conveyor feeding a furnace with coal supplied by a hopper. Varying the opening at the base of the hopper will vary the amount of fuel supplied to the furnace, but there is a time delay before this changed quantity of fuel reaches the furnace. The time delay depends on the speed and length of the conveyor. Mathematically the function describing variation in the quantity of fuel entering the furnace is a time-shifted version of the function describing the variation in the quantity of fuel placed on the conveyor. Therefore, let:

o =)()( tqtu quantity of fuel placed on the conveyor, where u(t) is the unit step.

o =−− )()( dtqdtu quantity of fuel entering the furnace


o d = time delay introduced by the conveyor o l = length of conveyor (m) o v = speed of the conveyor (ms-1)

The input to the conveyor is )()( tqtu . The output from the conveyor is

)()( qtqdtu −− . If the conveyor is moving at a constant speed then dl

v = and so

vl

d = . The transfer function that models the conveyor belt can be obtained by using

the second shift theorem. o )()()()()( sQetqedtqdtusQ sdsd

d−− ==−−= ll

Transport lags can cause difficulty when trying to control a system because of the delay between taking a control action and its effect being felt. In this example, increasing the quantity of fuel on the conveyor does not lead to an immediate increase in fuel entering the furnace. The difficulty of controlling the furnace temperature increases as the transport lag introduced by conveyor increases.

Conveyor Velocity, v

Conveyor Length, l

Coal

Furnace

Figure 1 – Coal is fed into the furnace via the conveyor belt

Poles, zeros and the s-plane Most transfer functions for engineering systems can be written as rational functions (i.e. as the ratio of two polynomials in s, with a constant factor, K).

o )()(

)(sQsP

KsG =

P(s) is of order m, and Q(s) is of order n; for a practical system m < n. Hence G(s) may be written as:

o ))...()(())...()((

)(21

21

n

m

pspspszszszsK

sG−−−−−−

=

The values of s that make G(s) zero are known as the system zeros and correspond to the roots of P(s) = 0, that is s = z1, z2,…., zm. The values of s that make G(s) infinite are known as the system poles and correspond to the roots of Q(s) = 0, that is s = p1, p2, …., pn. Poles may be real or complex. Complex poles always occur in complex conjugate pairs whenever the polynomial Q(s) has real coefficients. Engineers find it useful to plot these poles and zeros on an s-plane diagram. A complex plane plot is used with a real axis label of σ and an imaginary axis label of


ωj . Poles are marked as crosses and zeros are marked as small circles. Figure 2 shows an s-plane plot for the transfer function:

o )31)(31)(1(

)2)(3(3)(

jsjssss

sG−++++

+−=

The benefit of this approach is that it allows the character of a linear system to be determined by examining the s-plane plot. The transient response of the system can be easily visualised by the number and position of the system poles and zeros.

Pole-Zero Map

Real Axis

Imag

Axi

s

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3-3

-2

-1

0

1

2

3

Figure 2 – Poles and zeros plotted for the transfer function:

)31)(31)(1()2)(3(3

)(jsjss

sssG

−+++++−

=

If 1

1)(

pssX

−= where the pole p1 is given by a + bj, then

o )sin(cos)( 1 btjbteeeetx atjbtattp +=== The real part of the pole, a, gives rise to an exponential term and the imaginary part, b, gives rise to an oscillatory term. If a < 0 the response, x(t), will decrease to zero as

∞→t . Given a system with transfer function G(s), input signal R(s) and output signal C(s), then:

o )()()(

sGsRsC

=

o )()()( sRsGsC = and so:

o )(.))...()(())...()((

)(21

21 sRpspspszszszsK

sCn

m

−−−−−−

=

The poles and zeros of the system are independent of the input that is applied. All that R(s) contributes to the expression for C(s) is extra poles and zeros.

When s

sR1

)( = (i.e. a unit step input):


o spspsps

zszszsKsC

n

m

))...()(())...()((

)(21

21

−−−−−−

=

o sB

psA

psA

psA

sCn

n 1

2

2

1

1 ....)( +−

++−

+−

=

Taking the inverse Laplace Transform gives us: o 121 ....)( 21 BeAeAeAtc tp

ntptp n ++++=

If the system is stable then p1, p2, …., pn will have negative real parts and their contribution to c(t) will vanish as ∞→t . The response caused by the system poles is often called a transient response because it decreases with time for a stable system. The component of the transient response due to a particular pole is independent of the system input and is determined by the nature of the system poles. It is now possible to derive a series of rules relating the transient response of the system to the position of the system poles in the s plane. Rule 1 The poles may be either real or complex, but for a particular pole it is necessary for the real part to be negative if the transient caused by the pole is to decay with time. Otherwise the transient response will increase with time and the system will be unstable. Poles of a linear system must all lie in the left half of the s-plane for stability. Rule 2 The further a plot is to the left of the imaginary axis, the faster its transient decays. This is because its transient contains a larger negative exponential term. For example, e-5t decays faster than e-2t. The poles near to the imaginary axis are termed the dominant poles as their transients take longer to decay. It is quite common for engineers to ignore the effects of poles that are more than five or six time further away from the imaginary axis than the dominant poles. Rule 3 For a real system, poles with imaginary components occur as complex conjugate pairs. The transient resulting from this pair of pole has the form of a sinusoidal term multiplying an exponential term. For a pair of stable poles, (i.e. with a negative real part – see Rule 1), the transient can be sketched by drawing a sinusoid confined with a decaying exponential envelope. A complex system may have many poles and zeros but by plotting them on the s-plane the engineer begins to get a feel for the character of the system. The form of the transients relating to particular poles or pairs of poles can be obtained using the above rules. The magnitude of the transients, that is the values of A1, A2, …., An depends on the system zeros. It can be shown that having a zero near a pole reduces the magnitude of the transient relating to that pole. Engineers often deliberately introduce zeros into a system to reduce the effect of unwanted poles. If a zero coincides with the pole, it cancels it and the transient corresponding to that pole is eliminated.


Difference Equations and the z-Transform Introduction The Laplace Transform has been shown to be useful for the solution of differential equations, construction of transfer functions and control theory. Laplace methods are used when the variables are considered to be continuous. The z transform plays a similar role for discrete systems. We will consider the z transform and its uses in solving linear constant coefficient difference equations. Difference Equations Difference equations are the discrete equivalent of differential equations. The terminology is similar and the methods of solution have several aspects in common. Difference equations arise whenever an independent variable can have only discrete values. They are a growing are of importance in engineering due to their usefulness when dealing with discrete-time systems such as microprocessors. Consider a simple difference equation:

o 10][]1[ =−+ nxnx The dependent variable is x, while the independent variable is n. A solution to the difference equation is obtained by knowing the dependent variable for each value of interest of the independent variable. Therefore, the solution takes the form of a sequence. There are often many different sequences that satisfy a difference equation. The general solution embraces all of these and all possible solutions can be obtained from it. Example 1 Show nAnx 2][ = , where A is a constant, is a solution of:

o 0][2]1[ =−+ nxnx Therefore:

o nAnx 2][ = o nn AAnx 222]1[ 1 ==+ + o 02222][2]1[ =−=−+ nn AAnxnx

If in addition we are given a condition, perhaps 3]0[ =x , the constant A can be

equated. If nAnx 2][ = , then AAx == 02]0[ , and therefore 3=A . The solution is thus )2(3][ nnx = . This is the specific solution and satisfies both the difference equation and the given condition. The order is the difference between the highest and lowest arguments of the dependent variable. The equation:

o nnxnxnx =−+−+ ][7]1[]2[3 is second order because the difference between 2+n and n is 2.

o ]2[7]1[]1[ −=−−+ nxnxnx


is third order because the difference between 1+n and 2−n is 3. As with differential equations, in general the higher the order the more difficult the difference equation is to solve. In some difference equations the dependent variable occurs only once. These are considered to be of zero order. We refer to these as non-recursive difference equations because calculations of the value of the dependent variable does not require knowledge of the previous values. In addition, difference equations of order 1≥ are referred to as recursive difference equations because their solution requires knowledge of previous values of the dependent variable. The difference equation:

o 1][ 2 ++= nnnx has zero order and so is a non-recursive difference equation. Sometimes an equation or expression can be written in different ways. We need to be able to rewrite equations so that comparisons can be made. When general solutions to equations are found usually the equation is initially written in a standard form, therefore there may be a need to rewrite equations. Example 2 Rewrite the equation so that the highest argument of the dependent variable is 1+n .

o 7]2[2]2[]3[ ==+−+ xnnxnx The highest argument in the equation is 3+n ; this is reduced by 2 to 1+n . To do this n is replace by 2−n .

o 7]2[)2(2][]1[ =−=−+ xnnxnx Block Diagram Representation of Difference Equations Many engineering systems can be modelled by means of difference equations. It is possible to represent a difference equation using a block diagram. Difference equations operate on discrete time data and therefore a continuous signal is sampled before use. In the most common form of sampling, a sample is taken at regular intervals, T. We will refer to the sequence that has been obtained by sampling a continuous waveform at intervals T as x[n]. Several components are used in the block diagram. The delay block (Figure 1) has the effect of delaying the sequence by one sample interval (i.e. a delay of T).

T x[n-1]x[n]

Figure 1 - A delay block

Another block diagram element represents the addition of a constant to a sequence. A sequence can be scaled by a constant also. Both of these elements are shown in Figure 2.


a

+

x[n]

x[n]+a

k

x

x[n]

kx[n]

Figure 2 - Adding a constant to a sequence and scaling a sequence by a constant

Example 3 A simple example of a discrete time filter is described by the difference equation:

o ][]1[][ nxnayny =−− where ][nx is the input sequence, ][ny is the output sequence and a is a constant. If a is positive then the filter behaves as a low-pass filter which rejects high frequencies but allows low frequencies to pass. If a is negative then the filter behaves as a high-pass filter. A block diagram for the filter is shown in Figure 3. This is a recursive filter because calculations of ][ny requires knowledge of previous values of the output sequence. The block diagram contains a feedback path – this is a feature of recursive difference equations.

∑

x T

y[n]x[n]

-a

y[n-1]-ay[n-1]

Figure 3 - A discrete time filter

Example 4 - Numerical Solutions of Difference Equations Given:

o 1]0[][]1[ ==−+ xnnxnx determine x[1], x[2] and x[3].

o 1]1[

0]0[]1[1

=∴=−

=

xxx

n

o 2]2[

1]1[]2[2

=∴=−

=

xxx

n


o 4]3[

2]2[]3[3

=∴=−

=

xxx

n

Example 5 From Example 3 the solution for a low-pass discrete-time filter is:

o ][]1[][ nxnayny =−− If a is positive then the filter is a low-pass filter. Choosing a = 0.5 we will examine the response of the filter to a unit step input applied at n = 0, i.e.

o <

=otherwise

nnx

100

][

Assuming 0][ =ny for .1−≤n o ]0[]1[5.0]0[ xyy +−= o 1)0(5.0]0[ +=y o 1]0[ =y

Similarly: o ]1[]1[5.0]1[ xyy +−= o 1)1(5.0]0[ +=y o 5.1]0[ =y

and so on.


z-Transforms The sequence Nkkf ∈],[ may have been created by sampling a continuous signal. We define its z-transform to be:

o [ ] ∑∞

=

−==0

][)(k

kzkfkfZzF

We can see that the z-transform creates an infinite series, thus:

o [ ] [ ] ....]3[]2[]1[

032

++++=z

fz

fz

ffkfZ

In most engineering applications it is not necessary to work with the infinite series as it is often possible to express this in a closed form. The closed form is generally valid for values of z within a region known as the radius of absolute convergence as will become apparent during some of the examples given. Example 1

Find the z-transform of the sequence defined by

≠=

=0001

][kk

kf

o [ ] ∑∞

=

−=0

][k

kzkfkfZ

o [ ] [ ] ....]3[]2[]1[

032

++++=z

fz

fz

ffkfZ

o [ ] ....000

132

++++=zzz

kfZ

o 1)( =zF This is sometimes called the Kronecker delta sequence, ][kδ . Example 2 Find the z-transform of the sequence defined by Nkkkf ∈= ,][ .

o [ ] ∑∞

=

−=0k

kkzkfZ

o [ ] ....321

032

++++=zzz

kfZ

o [ ]

+++= ....

321

12zzz

kfZ

If we use the binomial theorem to express 21

1−

−

zas an infinite series, we find that:

o 2)1(

)(−

=z

zzF

Note that the process of taking the z-transform converts the sequence f[k] into the continuous function F(z).


Sampling a continuous signal Most of the signals encountered in the real world are continuous in time. This means they have a signal level for every value of time over a particular time interval of interest. An example would be the temperature of a room. This type of signal can be modelled using a continuous mathematical function in which for each value of t there is a continuous signal level, f(t). Many engineering system have signals which are only of interest at particular points (which are often equally spaced and separated by time T). These are referred to as discrete signals. They are modelled by a mathematical function that is only defined at certain points in time. If we have a continuous signal, f(t), defined for 0≥t which we sample (measure) at intervals of time T we obtain a sequence of sample values of f(t), i.e. f[0], f[1], f[2], … It can be shown that sampling a continuous signal does not lose the essence of the signal provided that the sampling rate is sufficiently high. It is often convenient to represent a discrete signal as a series of weighted impulses. The strength of each impulse is the level of the signal at the corresponding point in time; using:

o ∑∞

=

−=0

)(][)(*k

kTtkftf δ

the * indicates that f(t) has been sampled. This is a useful mathematical way of representing a discrete signal as the properties of the impulse function lend themselves to a value that exists for a short interval of time. Note, no sampling method has zero sampling time, but provided the sampling time is much smaller that the sampling interval, then this is a valid mathematical model of a discrete system. We can apply the z-transform directly to a continuous function, f(t), if we regard the function as having been sampled at discrete intervals of time. The Relationship Between the z Transform and the Laplace Transform We have defined the z-transform independently of any other transforms. However, there is a close relationship between the z-transform and the Laplace transform, the z-transform is regarded as the discrete equivalent of the Laplace transform. If the continuous signal f(t) is sampled at intervals of time, T, we obtain a sequence of sampled values Nkkf ∈],[ . As we already know this sequence can be regarded as a train of impulses:

o ∑∞

=

−=0

)(][)(*k

kTtkftf δ

Taking the Laplace transform, we have:

o ∫ ∑∞ ∞

=

− −=0 0

)(][)(*k

st dtkTtkfetf δl

o ∑ ∫∞

=

∞− −=

0 0

)(][)(*k

st dtkTtekftf δl

assuming that it is permissible to interchange the order of summation and integration.

o ∑∞

=

−=0

][)(*k

skTekftfl

Now making the variable z = esT


o ∑∞

=

−=0

][)(*k

kzkftfl

which is the definition of the z-transform. Example 3

Consider the function tetutf −= )()( which has the Laplace transform 1

1)(

+=

ssF .

Suppose f(t) is sampled at intervals T to give the sequence kTekf −=][ , for k = 1, 2, 3…

a) Find the z-transform of f[k] b) Make the change of variable sTez = to obtain )(* sF c) Show that provided the sample interval T is sufficiently small, TF*(s)

approximates the Laplace transform F(s)


Coursework Assignment During the course we have studied the use of Fourier Series, Fourier Transform, Laplace Transform and z-Transform. As we have discussed, these tools are very useful in many engineering applications. You should write a 2500-4000 word report summarising qualitatively the principle of each of the techniques, typical application areas and variants of the techniques we have considered. In addition, for each technique you should detail a real life engineering application you have found through research (library, web, research papers) explaining why the chosen technique is used and how the work has been implemented. Your report should be submitted to reception by Monday 27th January 2003. Late submissions will be penalised at 2% per day.


z-Transforms (cont.) Because of the relationship between the Laplace and z transforms many of the properties are similar. Linearity If f[k] and g[k] are two sequences, then:

o ][][][][ kgZkfZkgkfZ +=+ First Shift Theorem If f[k] is a sequence and F(z) is its transform, then:

o ])1[....]1[]0[()(][ 1 −+++−=+ − izffzfzzFzikfZ iii In particular, if i = 1:

o ]0[)(]1[ zfzzFkfZ −=+ If i = 2:

o ]1[]0[)(]2[ 22 zffzzFzkfZ −−=+ Second Shift Theorem The function f(t)u(t) is defined by:

o

<≥

=000)(

)()(tttf

tutf

where u(t) is the unit step function. The function )()( iTtuiTtf −− , where i is a positive integer, represents as shift to the right (a time delay) of i sample intervals. The second shift theorem states:

o )(][][ zFzikuikfZ i−=−− Inversion of the z-transform Just as it is necessary to invert Laplace transforms we need to be able to invert z-transforms, and as before we make use of the tables, partial fractions and shift theorems. Example Find the sequence whose z-transform is:

o )4)(5(

2)(

2

+−−

=zz

zzzF

The first stage is to divide the expression for F(z) by z:

o )4)(5(

12)(+−

−=

zzz

zzF

We now split the RHS into partial fractions using the standard techniques, giving:

o 4

15

1)(+

+−

=zzz

zF


Multiplying back through by z gives:

o 45

)(+

+−

=z

zz

zzF

Which we now invert using the tables giving us kkkf )4(5][ −+= The z-transform and Difference Equations As we have previously discussed, the z-transform is used to solve difference equations. Example Solve the difference equation 0][3]1[ =−+ kyky , 4]0[ =y . Taking the z transform of both sides of the equation gives us:

o 00][3]1[ ==−+ ZkykyZ Since Z0 = 0. Using the properties of linearity we find:

o 00][3]1[ ==−+ ZkyZkyZ

Using the first shift theorem on the first of the terms on the LHS we obtain: o 0][34][ =−− kyZzkyzZ

Writing )(][ zYkyZ = , this becomes:

o zzYz 4)()3( =− so that:

o 3

4)(

−=

zz

zY

The function on the RHS is the z-transform of the required solution. Inverting this using the tables we find kky )3(4][ = .

Tutorial Sheet 4 1. Use z-transforms to solve 6][3]1[ −=−+ kxkx , 1]0[ =x

Answer - )3(23][ kkx −= 2. Use z-transforms to solve kkxkx 2][]1[2 =−+ , 2]0[ =x

Answer - 3

)5.0(5

32 kk

+

3. Use z-transforms to solve 12][]1[ +=++ kkxkx , 0]0[ =x

Answer - k 4. Use z-transforms to solve 0][16]1[8]2[ =++−+ kxkxkx , 10]0[ =x , 20]1[ =x

Answer - )4(5)4(10 kk k−


Probability and Statistics Introduction to Probability Consider a machine which manufactures electronic components. These must meet a certain specification. The quality control department regularly samples the components. Suppose, on average, 92 out of 100 components meet the specification. Imagine that a component is selected at random and let A be the outcome that a component meets the specification; let B be the outcome that a component does not meet the specification. Therefore we say the probability of A occurring is

92.010092 = and the probability of B occurring is 08.0100

8 = . The probability is

therefore considered to be a measure of the likelihood of the occurrence of a particular outcome. We should note that the probability of all possible outcomes is 1. The process of selecting a component is called a trial. The possible outcomes are called events. In the above example there are two possible events, A and B. If we attempt to define probability in a more formal way… Let E be an event. We wish to obtain the probability that E will occur, i.e. P(E), when a trial takes place. In order to do this we repeat the trial a large number of times, n. We count the number of times that an event E occurs (denoted by m). We can conclude that:

o nm

EP =)(

The larger the number of trials that take place, the more confident we are of our estimate of the probability of E occurring. For example, consider the trial of tossing a coin. If we wish to calculate the probability of a head occurring then measuring the result of 1000 tosses of the coin is likely to give a more accurate estimate that measuring the results from 10 tosses. Example 1 Machine A and B make components, which are then placed on a conveyor belt. Of those made by machine A, 93% are acceptable. Of those made by machine B, 95% are acceptable. Machine A makes 60% of the components and machine B makes the rest. Find the probability that a component selected at random from the conveyor belt is:

a) made by machine A b) made by machine A and acceptable c) made by machine B and acceptable d) made by machine B and unacceptable


Mutually Exclusive Events If we have a machine that manufactures car components. Suppose each components falls into one of the following categorie s:

o top quality; o OK; o not very good; o rubbish.

After many samples have been taken and tested, it is found that under certain circumstances the probability that a component falls into a categories is as follows:

Category Probability top quality 0.18

OK 0.65 not very good 0.12

rubbish 0.05 The four categories cover all possibilities and so the probability must sum to 1. Example 2 Using the data in the table calculate the probability that a component selected at random is either standard or top quality. In this example the events A and B could not possibly occur together. A component is either top quality or OK but cannot be both. We therefore define A and B as being mutually exclusive. If Ei and Ej are mutually exclusive we denote this by:

o ∅=∩ ji EE where ∅ represents an empty set, i.e. we are saying that ji EE ∩ is an impossible event and can never occur. Complementary Events Consider an electronic circuit. Either the circuit work or it doesn’t. Let A be the event that the circuit works correctly, and B be the event that the circuit does not work correctly. Either A or B must happen when the circuit is switched on, and one excludes the other. The events A and B are said to be complementary. Two events are considered to be complementary if they are mutually exclusive and in a single trial where either A or B must happen. Therefore:

o 1)()( =+ BPAP and

o 0)( =∩ BAP Instead of using the events A and B we tend to refer to the events as A and A . If the probability that a component last 3 years or less is the event D and this has the probability P(D) = 0.08. We can say:

o D: the component last 3 years or less;


o D : the component last more than 3 years. D and D are complementary events so:

o 1)()( =+ DPDP

o 92.008.01)(1)( =−=−= DPDP Multiplication Law If two machines, M and N, both manufacture components. Of the components made by machine M , 92% are of an acceptable standard and 8% are rejected. For machine N, only 80% are of an acceptable standard and 20% are rejected. If all the components are manufactured by machine M the P(E) = 0.92, and if all the components are manufacture by machine N the P(E) = 0.8 (where E is the event that a component is of an acceptable standard). If half the components are manufactured by machine M and half by machine N then P(E) = 0.86. Why?? This leads to the idea of conditional probability. If we define events:

o A: the component is manufactured by machine M; o B: the component is manufactured by machine N.

Then the probability that a component is of an acceptable standard, given it is manufactured by machine M, is written as P(E|A). This is interpreted as the conditional probability of E given A. Similarly P(E|B) is the probability of E happening, given B has already happened. Consider events A and B for which are mutually exclusive ( ∅=∩ BA ). Suppose we know that event A has occurred and we need to find the probability that B occurs, i.e. P(B|A). Knowing that event A has occurred we can restrict our attention to set A. Event B will occur if any outcome is in BA ∩ . Therefore:

o )(

)()|(

APBAP

ABP∩

=

The multiplication law therefore states: o )().|()( APABPBAP =∩

Example 3 A manufacturer studies the reliability of a certain component so that suitable guarantees can be given: 83% of components remain reliable for at least 5 years; 92% remain reliable for at least 3 years. What is the probability that a component which has remained reliable for 3 years will remain reliable for 5 years?


For independent events E1 and E2: o )()|( 121 EPEEP = o )()|( 212 EPEEP = o )()()( 2121 EPEPEEP =∩

The concept of independence maybe applied to more than two events. Three or more events are independent if every pair is independent. A compound event may comprise several independent events. The multiplication law is extended in an obvious way:

o )()()()( 321321 EPEPEPEEEP =∩∩ o )()()()()( 43214321 EPEPEPEPEEEEP =∩∩∩

Example 4 The probability that a component is fault is 0.04. Two components are picked at random. Calculate the probability that:

a) both components are faulty; b) both components are not faulty; c) one of the components is faulty; d) one of the components is not faulty; e) at least one of the components is faulty; f) at least one of the components is not faulty.


Example Class Questions Question 1 The life spans of 5000 electrical components are measured to assess their reliability. The life span (L) is recorded and the results are shown in the table below. Find the probability that a randomly selected component will last for:

a) more than 3 years? b) between 3 and 5 years? c) less than 4 years?

Lifespan of components Number

5>L 500 54 ≤< L 2250 43 ≤< L 1850

3≤L 400 Question 2 Machines M and N manufacture a component. The probability that the component is of an acceptable standard is 0.95 when manufactured by machine M and 0.84 when manufactured by machine N. Machine M supplies 65% of components; machine N supplies 35%. A component is picked at random.

a) What is the probability that the component is of an acceptable standard?

b) What is the probability that a component is of an acceptable standard and is made by machine M?

c) What is the probability that the component is of an acceptable standard given it is made by machine M?

d) What is the probability that the component was made by machine M? e) What is the probability that the component was made by machine M

given it is of an acceptable standard? f) The component is not of an acceptable standard. What is the

probability that it was made by machine N? Question 3 Machine 1 manufactures an electronic chip, A, of which 90% are acceptable. Machine 2 manufactures an electronic chip, B, of which 83% are acceptable. Two chips are picked at random (one of each kind). Find the probability that they are both acceptable. Question 4 Machines 1, 2 and 3 manufacture resistors A, B and C respectively. The probabilities of their respective acceptabilities are 0.9, 0.93 and 0.8. One of each resistor is selected at random.

a) Find the probability that they are all acceptable; b) Find the probability that at least one resistor is acceptable.

Roberts(2003) Fundamentals of Signals and Systems

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