Introduction to Fourier Analysis

Electrical and Electronic Principles

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

The following presentation is a part of the level 4 module -- Electrical and Electronic Principles. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st

year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Introduction Complex Periodic Waveform “Make up” of the Fourier Series Value of Ao Value of An Value of Br Observation of Waveform Modifying Waveform Inserting Value of N Numerical Integration Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:Green D C, Higher Electrical Principles, Longman 1998 Hughes E , Electrical & Electronic, Pearson Education 2002Hambly A , Electronics 2nd Edition, Pearson Education 2000Storey N, A Systems Approach, Addison-Wesley, 1998

Semiconductor Theory

It was discovered by Joseph Fourier, that any periodic wave (any wave that repeats itself after a given time, called the period) can be generated by an infinite sum of sine wave which are integer multiples of the fundamental (the frequency which corresponds to the period of the repetition. Lets look at a periodic wave. Here is an example plot of a signal that repeats every second.

0

1

2

3

4

5

6

7

8

9

10

0 0.5 1 1.5 2

This wave is not sinusoidal and therefore it is difficult to imagine that we can add sine waves together to produce this waveform.

Fourier did show that such a waveform can be generated using sine and cosine wave of different sizes or sine waves with different phase angles.        Fourier showed how it was possible to determine the sizes of the sine and cosine waves using integration techniques.

Introduction to Fourier Analysis

It has been shown that any complex periodic waveform can be written as a infinite sum of sinusoidal waves of different amplitude and phase shift which are all integer frequency multiples of the fundamental waveform which is the frequency of the original complex periodic waveform.

V = V1 Sin (t 1) + V2 Sin (2t 2) + V3 Sin (3t 3) + V4 Sin (4t 4) + …

To remove the requirement for the angle we can say that each component has a sin part and a cos part:

1

V1

A1

B1

e.g.V1 Sin (t 1) = A1 Cos (t) + B1 Sin (t)

The values of A and B can be positive, negative or zero.

In addition we must allow for a D.C. component to the wave and this is quantified using A0. We can therefore write the complex periodic waveform as:

10 ))()((n nn tnSinBtnCosAAv

Introduction to Fourier Analysis

The constants can be determined using the following integrals:

Tt

t

dttfT

A )(1

0

Tt

tn dttntfT

A )cos()( 2

Tt

tn dttntfT

B )sin()( 2

The values of the constants are called the Fourier Coefficients and the whole is called the Fourier Series.

Introduction to Fourier Analysis

By observation we can make certain decisions as to the “make up” of the Fourier Series. See below.

)()( tftf these will have only A constants (all B = 0) the series will have only Cos functions. There is no A0 component – balanced about 0.

Even Functions

Odd Functions

)()( tftf these will have only B constants (all A = 0) the series will have only Sin functions. There is no A0 component – balanced about 0.Introduction to Fourier Analysis

Half Wave Symmetrical

)2

()(T

tftf these will have no even harmonics. There is no A0 component – balanced about 0.

Example

5v

-1v

T

T/2

Introduction to Fourier Analysis

Value of A0

We have two parts

Tt

t

dttfT

A )(1

0

T

T

T

dtdtT

A2

2

00 15

1

/

/

T

T

Ttt

TA

2

200 15

1

20

2

510

TT

T

TA

22

41

22

510

T

T

TT

TA

Introduction to Fourier Analysis

Value of An

Again we have two parts

T

T

T

n dttndttnT

A2

2

0

52

/

/

)cos()cos(

T

T

T

n tnn

tnnT

A2

2

0

152)sin()sin(

2

110

2

52 Tn

nTn

n

Tn

nTAn

sinsinsin

Tt

tn dttntfT

A )cos()( 2

We can replace by using:T

f 2

2 Introduction to Fourier Analysis

T

Tn

n

T

T

Tn

n

T

T

Tn

n

T

TAn 2

2

2

2

22

2

2

52

sinsinsin

nn

nn

nn

An sinsinsin1

215

216

nn

nn

An sinsin

Now we put the values of n in:

n=1

n=2

0216

1

sinsinA

042

12

32

sinsinA

All A values are zero!

Introduction to Fourier Analysis

Value of Bn

Once again we have two parts

T

T

T

n dttndttnT

B2

2

0

52

/

/

)sin()sin(

T

T

T

n tnn

tnnT

B2

2

0

152)cos()cos(

2

115

2

52 Tn

nTn

nn

Tn

nTBn

coscoscos

Tt

tn dttntfT

B )sin()( 2

We can again replace by using:T

f 2

2 Introduction to Fourier Analysis

T

Tn

n

T

T

Tn

n

T

n

T

T

Tn

n

T

TBn 2

2

2

2

22

5

2

2

2

52

coscoscos

nn

nnn

nn

Bn coscoscos1

2155

n

nn

nn

Bn5

216

coscos

Now we put the values of n in:

n=1

n=2

1251652

161

coscosB

02

5

2

13

2

54

2

12

2

62

coscosB

Introduction to Fourier Analysis

n=3

n=4

n=5

3

12

3

5

3

12

3

56

3

13

3

63

coscosB

04

5

4

1

4

6

4

58

4

14

4

64

coscosB

5

12

5

5

5

1

5

6

5

510

5

15

5

65

coscosB

From the results so far we can see that all even harmonics are 0 and that we have a common value of 12/ with this being divided by the harmonic, i.e. 1, 3, 5, 7, etc.

etctttv ...)sin()sin()sin(

5

5

13

3

1122

Introduction to Fourier Analysis

-2

-1

0

1

2

3

4

5

6

Fundamental only )sin( tv 12

2

-2

-1

0

1

2

3

4

5

6

Fundamental + 3rd Harmonic

)sin()sin( ttv

3

3

1122

-2

-1

0

1

2

3

4

5

6

)sin()sin()sin( tttv

5

5

13

3

1122

-2

-1

0

1

2

3

4

5

6

This includes the 7th, 9th, and 11th harmonics

By observation we should have been able to deduce that the waveform was ODD and only calculated the B (Sin only) values and also that it was half wave symmetrical which meant that we would have known that B2, B4, B6 etc would be 0.

Example

5v

T

3T/4

Is the wave EVEN, ODD or Half Wave Symmetrical?

NOIntroduction to Fourier Analysis

Can we modify it to make it fit into one of the categories?

5v

T

3T/8 5T/8

EVEN with two areas to integrate

EVEN with one area to integrate

5v

T

T/8 7T/8

Value of A0

Tt

t

dttfT

A )(1

0

87

80 5

1 /

/

T

T

dtT

A

8780 5

1 //TTtT

A

7538

30

8

301

8

5

8

3510 .

T

T

TT

TA

Introduction to Fourier Analysis

Value of An

87

8

52 /

/

)cos(T

Tn dttnT

A

87

8

52T

Tn tn

nTA

/)sin(

8

5

8

752 Tn

n

Tn

nTAn

sinsin

Tt

tn dttntfT

A )cos()( 2

Tf

22

T

Tn

n

T

T

Tn

n

T

TAn 8

2

2

5

8

72

2

52

sinsin

44

75

nn

nAn sinsin

Now we put the values of n in:

n=1

n=2

n=3

n=4

n=5

2

10

2

1

2

15

44

751

sinsinA

2

1011

2

5

4

2

4

72

2

52

sinsinA

32

10

2

1

2

1

3

5

4

3

4

73

3

53

sinsinA

0004

5

4

4

4

74

4

54

sinsinA

52

10

2

1

2

1

5

5

4

5

4

75

5

55

sinsinA

Introduction to Fourier Analysis

0

1

2

3

4

5

6

-1

0

1

2

3

4

5

6

Introduction to Fourier Analysis

-1

0

1

2

3

4

5

6

-1

0

1

2

3

4

5

6

-1

0

1

2

3

4

5

6

Example

10v

T/2

TBy observation the function is EVEN and Half Wave Symmetrical.

It is also obvious that the average value is 5

A0 = 5v

Introduction to Fourier Analysis

Numerical Integration can be used if the waveform is complicated and the results need not be too accurate.

Angle (Rad) Function FnxCos(angle) FnxCos(3xangle) Fnxcos(5xangle)

0 0 0 0 0

0.31415927 1 0.951056516 0.587785252 6.12574E-17

0.62831853 2 1.618033989 -0.618033989 -2

0.9424778 3 1.763355757 -2.853169549 -5.51317E-16

1.25663706 4 1.236067977 -3.236067977 4

1.57079633 5 3.06287E-16 -9.18861E-16 1.53144E-15

1.88495559 6 -1.854101966 4.854101966 -6

2.19911486 7 -4.114496766 6.657395614 -3.00161E-15

2.51327412 8 -6.472135955 2.472135955 8

2.82743339 9 -8.559508647 -5.290067271 4.96185E-15

3.14159265 10 -10 -10 -10

3.45575192 9 -8.559508647 -5.290067271 9.92273E-15

3.76991118 8 -6.472135955 2.472135955 8

4.08407045 7 -4.114496766 6.657395614 -6.86007E-15

4.39822972 6 -1.854101966 4.854101966 -6

4.71238898 5 -9.18861E-16 2.75658E-15 -1.34761E-14

5.02654825 4 1.236067977 -3.236067977 4

5.34070751 3 1.763355757 -2.853169549 -2.20494E-15

5.65486678 2 1.618033989 -0.618033989 -2

5.96902604 1 0.951056516 0.587785252 -2.94025E-15

6.28318531 0

Sum -40.86345819 -4.851839996 -2

Average -2.043172909 -0.242592 -0.1

Value -4.086345819 -0.485184 -0.2

A0 = 5

A1 = -4.086

A2 = -0.485

A3 = -0.2

This produces

-1

1

3

5

7

9

11

Introduction to Fourier Analysis

This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.

© 2009 University of Wales Newport

This work is licensed under a Creative Commons Attribution 2.0 License.

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