Interpretation of the Physical Significance of the Fourier

Interpretation of the Physical Significance of the Fourier Analysis of Signals using Complex Exponentials

By

A. Idi Aminu

Department of Basic Sciences, Adamawa State College of Agriculture, Ganye.

E-mail: [email protected]

Abstract

Fourier analysis amount to finding the Fourier coefficient of the signal in question and then substituting these

coefficients into the general complex Fourier series

jnx

no eCC expression, from which the properties of

the signal been analyzed can be deduced. Through this analysis, it was found that the saw-tooth wave form

consists of a continuous sine series:

.....3

3

12

2

1

2)( tSintSintSin

ppty

and in the same manner the

triangular wave form was analyzed and found to consist of discrete sine series:

.....5

5

13

3

122

)( xSinxSinSinxpp

ty

. Through this work I have used the complex exponential jnxe

method in finding the Fourier coefficients

dxexfC jnx

n

)(2

1 and e then used trigonometric approach to

authenticate my results. At the end of this work I have also used the Fourier analysis to analyze how well do the

half and the full wave rectifier diode valve converts an A.C to a D.C. And for the half-wave rectifier diode the

expression was: .2

1.....4

15

12

3

12)( tSinItCostCos

IItI o

oo

While that of a full-wave rectifier

the expression was: ......45

12

3

142)(

tCostCostI

Hopefully by the end of this work, the reader

can be able to analyze a signal or an electronic gadgets such as the diode-valve and the like, by the use of

Fourier analysis and predict with precision their behavior by just looking at the final expression, also called the

design equation.

Keywords:

Fourier analysis, Fourier coefficients, complex exponentials, discrete sine series, discrete cosine series, full and

half wave rectifier, diode valve, square wave, triangular wave, and the saw-tooth wave

DISCUSSIONS

Any periodic function or signal can be expressed

as a sum of sinusoids e.g

.2

)( SinnxbCosnxaa

xf nno To explain this

expression in a layman’s language. Suppose you

have a full loaf of bread and then divide this loaf

of bread into two equal halves. Each half of the

bread can be thought of as the average value of the

signal, D.C, ao/2, value (zero-frequency). Now

keep one half of the bread as it is, but divide the

other half into unequal pieces. Now if you

continue to add the divided pieces of the bread to

the other half of the bread, you will eventually get

back full loaf of bread. And if you want to modify

the shape of the bread, you will simply refuse to

add the some divided pieces of the bread to the

other half. Obviously now, you can modify the

GSJ: Volume 9, Issue 5, May 2021 ISSN 2320-9186 822

GSJ© 2021 www.globalscientificjournal.com

GSJ: Volume 9, Issue 5, May 2021, Online: ISSN 2320-9186 www.globalscientificjournal.com

2

T

y

P

Y

2t

shape of this bread into an infinite number of

different shapes. Now you can think of the sine

and cosine as the divided pieces of the bread. The

same procedure is applied during audio

compression, take or record a voice as a signal e.g

an audio speech, decompose these into the average

and the sine terms. You can now drop or neglect

some of these sine terms and them recombined the

remaining terms to give back the original signal.

The argument is that, the human ear cannot be able

to notice any difference between the original and

the modified (sound). All these procedures are a

function of the Fourier coefficients therefore all

the reason why the concepts of finding the Fourier

coefficients are important.

FOURIER ANALYSIS OF SIGNALS:

1. THE TRIANGULAR WAVE

y(t) t t

(Fig. 1)

The triangular wave is defined by:

Y = 0 at t = 0

Y = p at t = T/2

Y = 0 at t = T

And this repeats every period T, as shown in fig.

(1) from which we can deduce;

2

02 TtT

pty

And

TtTT

tTpy

2

)(2

Recall that: the Fourier series in complex form is:

n

jnx

no eCCxf )(

Where Co and Cn are the coefficients to be

calculated.

Using: dxexfC jnx

n

)(2

1

So that,

dteT

tTp

T

ptC tj

T

n

T

T

.0

0

2

2

)(221

On integrating we get,

T

o

T

T

ttT

t

T

pC

2

2

2)(

2

2 2

0

2

2

Substituting the limits we get,

82

2

2

2

8

2 22

2

22

2

2

2

TT

T

pTT

T

pT

T

pCo

4

2

42

4

ppppp

pCo

2

pCo

For Cn using

dtetTT

pdtet

T

pC tjn

T

tjn

n

T

T

2

2

)(22

2

0

2

1nC

2nC

So that 21 nnn CCC

From,

dtetT

pC

T

tjn

n

2

1

0

2

2 (using integration by parts)

That is VUVUI 1



3

2

1

0

2222

12T

tjntjn

n enj

ejn

t

T

pC

Substituting the limit we get,

2222

10

2

222

njee

jn

T

T

p TT jnjn

22222

1)1(

2

)1(2

nnjn

T

T

p n Where j2 = -1

If n = odd then,

222

)2(

2

)1(21 njn

T

T

pCn

For 2nC we get

dtetTT

pC

T

tjn

n

T

2

2)(

22

After integrating, the first term and second term by

part

2222

21

22

2 T

pe

nje

jn

te

jn

TC

T

tjntjn

T

tjn

n

TT


2222222

1

2

12 22

2 vnj

e

jn

eT

nj

e

jn

eT

jn

eT

jn

eT

T

pC

tjnjnTjnTjnjnTjn

n

TT

For n = even

2222222

)1(1)1()1(222 njnjjn

T

jn

T

jn

T

T

pC

nn

n

For n = odd (1, 3, 5)

222

2

2

3222 njn

T

jn

T

T

pCn

Where j2 = -1

222

2

2

22 njn

T

T

pCn

So that 21 nnn CCC

22222

2

2

)2()2(

2

)1(2

T

p

jn

T

nnjnC

n

n

222222222

2

4

8840

2

n

p

n

p

nT

p

nT

pCn

22

2

n

pCn

Now substituting these coefficients into the

Fourier series expression we get,

tjn

no

tjn

no eCCeCCtf )(

But 2

pCo and

22

2

n

pCn

therefore,

2

5

2

3

22

5

2

3

22 531...

531...

2

2)(

n

e

n

e

n

e

n

e

n

e

n

epptf

tjtjtjtjtjtj

....523212

4

2)(

2

55

2

33

22

tjtjtjtjtjtj eeeeeepptf

But tee tjtj

cos2

(Euler’s formula)

.....5cos

5

13cos

3

1cos

4

2)(

222ttt

pptf

Interpretation of the physical significance of the

triangular wave expression

(i). It consist of all cosine terms, apart from P/2

(average value)

(ii). The convergence is fast

2

1

nmeaning by just

adding a few terms you will get back the ‘original’

signal or function.

(iii). It is a discrete cosine series (ideal for image

processing)

PROOF: USING TRIGONOMETRIC

APPROACH FOR THE TRIANGULAR WAVE:



4

P

0

From

11

)(n

n

n

no tSinbtCosnaatf

2

20

)(2121T

T

T

o dtT

tTp

Tdt

T

pt

Ta

Integrating w.r.t t, we get

T

T

o

T

T

T

t

T

pt

T

pt

T

pa

2

2

2

2

22

2

2 2

22

0

2

2

Substituting the limits we get

82

2

2

2

8

2 22

2

22

2

2

2

TT

T

pTT

T

pT

T

pao

24

2

4444

ppppppp

p

2

pCo

2

pao Q.E.D

For the coefficient an using:

tdtCosnT

tTatdtCosn

T

at

Ta

T

T

T

n

2

20

)(222

Integration of the first term by parts and

substituting the limits we get

1

2

42221

tnCos

nT

pan

Part (1)

Integrating the second term by parts and

substituting the limit we get

TT

n

TT

tCosnnT

ptSinn

nT

pa

22

2 222

414

But 0tSinn for any value of n

2

41

2

40

222222

tnCostCosn

nT

ptnCos

nT

p

But 2t

12222

nCosCosnn

pan

For n = even, 0, 2, 4 we get

an = 0

12222

CosCosn

pan

nn Cn

p

n

pa

2222

4112

Q.E.D

(2). SQUARE WAVEFORM (High Frequencies)

(Fig 2)

The above square wave can be defined by the

function f(x) such that

0)( xf from 0 x

and

pxf )( from x0

To analyze the properties of the square wave all

we need is the Fourier coefficients given by;

dxexfC jnx

n

)(2

1

So that,

2222

10

0

0

ppx

pdxepC xj

n

2

pCo

For Cn using;



5

P

y(t) y

X

00

1

22

1

jnxjnx

n ejn

pdxepC


11

22

0

jnjn

n ejn

p

jn

e

jn

epC

01)1(2

12

njn

njn

pe

jn

pC

for n =

even

But if n = odd 1, 3, 5 we get

jn

p

jn

p

jn

pC n

n

2

21)1(

2

jn

pCn

Writing the Fourier series we get

jnx

no eCCxf )(

But 2

pCo and

jnx

pCn

jn

eppxf

jnx

2)(

...531

...5312

)(5353

j

e

j

e

j

e

j

e

j

e

j

eppxf

jxjxjxjxjxjx

....52322

2

2)(

5533

j

ee

j

ee

j

eeppxf

jxjxjxjxjxjx

But Sinxj

ee jxjx

2 (Euler’s formula)

....5

5

13

3

12

2)( xSinxSinSinx

ppxf

(i). All the cosine have vanished showing that it

may be used for audio compression.

(ii). The convergence is slow

n

1 meaning a

square wave have a lot of high frequencies. This

implies that a lot of terms should be added

together before obtaining the ‘original’ signal.


APPROACH FOR THE SQUARE WAVE

From

11

0)(n

n

n

n tSinnbtCosnaaty

2222

1

0

00

ppt

ppdta

2

0

pa Q.E.D

00 a

0

21

1

n

pCosn

n

ppSinntdtbn

for n = odd and 0

for n = even

pbn

2 n = 1, 3, 5 Q.E.D

(3) THE SAW TOOTH WAVEFORM:

Y

t T 2T 3T

For linear displacement w.r.t time then; y = P at t

= 0 to y = 0 at t = T and more. We can then safely

write;

T

tT

p

y or

T

tp

T

tTpy 1



6

Recall that: by using the complex analysis we

have,

jnx

neCxf )( and in order to analyze this

signal we need its Fourier coefficients by using;

dxexfC jnx

n

)(2

1

If

T

jn

T

o eT

t

T

pe

T

t

T

pC

0

00

0

11

2222

2

0

2 pT

T

p

T

TT

T

ptt

T

pC

T

o

2

pCo

For Cn, using;

dxexfC jnx

n

)(2

1

dtteT

pdte

T

pdte

T

tp

TC

T

tjn

T

tjn

T

tjn

n

0

2

00

11

Let

TtjnT

tjn

njn

e

T

pdte

T

pC

00

1


01111

1

Tjn

pe

jnT

p

jnjn

e

T

pC tjn

tjn

n

Let

dtetT

pC

T

tjn

n

0

22

Integrating by parts we get,

Ttjntjn

nj

e

jn

et

T

p

0

222

1

Substituting the limit

we get

22222

10

2 njnj

e

jn

eT

T

pC

TjnTjn

n

0

1112222222 jn

T

T

p

njnjjn

T

T

pCn

jn

pCn

22

jn

pCCC nnn

20

21

jn

pCn

2

Writing the Fourier series we get,

1

)(n

jnx

neCty but 2

pCo and

jn

pCn

2

nj

eppeCCty

tjn

n

tjn

no

22)(

1

...321

...321

...22

)(3232

j

e

j

e

j

e

j

e

j

e

j

eppty

tjtjtjtjtjtj

...232222

2

2)(

3322

j

ee

j

ee

j

eeppty

tjtjtjtjtjtj

But tSinj

ee tjtj

2 (Euler’s formula)

...3

3

12

2

1

2)( tSintSintSin

ppty

INTERPRETATION OF THE PHYSICAL

SIGNIFICANCE OF THE ABOVE SAW-

TOOTH ANALYSIS EXPRESSION

i. It contains Sine terms only

ii. It is a continuous Sine series (Ideal for audio

compression)

iii. It shows a slow convergence

n

1which

means it may require a lot of terms to add up

to the ‘original’ signal.

iv. It also indicates that while amplitude

decreases the frequency increases, which



7

+

- Input Voltage

Input

diode

Output R

implies that the convergence to the original

signal is assured.

PROOF: USING THE TRIGONOMETRIC

APPROACH FOR THE SAW-TOOTH WAVE

Using;

1 12

)(n n

nno tSinbtCosna

aty

To analyze the properties of this signal we need

their Fourier coefficients, ao, an and bn

dtT

t

T

pdtty

Ta

TT

o

00

1)(1

By integrating w.r.t t we get,

2222

2

0

2 pT

T

p

T

TT

T

p

T

tt

T

pa

T

o

oo Cp

a 2

Q.E.D

T

n tdtCosnT

t

T

pa

0

12

by integrating we get,

TTT

n tdtSinnnT

tSinnn

t

TtSinn

nT

pa

000

11

1112

0tSinn for any value of n

T T

n tdtSinnTn

ptdtSinn

nTT

pa

0 0

21100

2

tCosnTn

p

1

222

But 2T

212

22Cosn

Tn

pan 0 na

For coefficient bn

T TT

n tdtSinnT

ttdtSinn

T

ptdtSinn

T

t

T

pb

0 00

21

2

By integrating w.r.t to t we get,

TCosn

nntCosn

nT

pbn

1112

n

p

Tn

p

Tn

pb

T

n

2

22

nn Cn

pb

Q.E.D

FOURIER ANALYSIS OF ELECTRONIC

COMPONENTS (DIODE-VALVE)

HALF-WAVE RECTIFIER:

After passing through the diode (i.e a half wave

rectifier) the sinusoidal current, tSinItI m )(

will be rectified. The rectified current function can

be written as;

tSinIm 20 Tt

I(t) =

0 02 tT

As shown below;

I(t) + +

T -T/2 0 T/2 T 3T/2

t

With time period T. its Fourier expansion in

complex form is:

tjn

no eCCtI )(

Hence the Fourier analysis of this signal involves

finding the Fourier coefficients and then



8

substituting this coefficient into the general

Fourier expansion like this:

2

2

2

2

0

2

1T

T

T

TtCosI

etSinIC mtj

mo


0)0(2

2 CosCosCosCoswI

C Tmo

mmm III 2

21)1(

2

m

o

IC

For the coefficient Cn we have

0

0

00

2

10

2

1

2

1dtetSinIdtetdtSinIC

tjn

m

tjn

mn

0

0

2dtetSin

IC

tjnmn

But by this identity;

)()()(22

22 cbxbCoscbxaSin

ba

ecbxSine

xx

Where a = 2, b = 1 and c = 0

Similarly,

tCostjnSinba

eIdtetSin

IC

tjn

mtjnmn

122 22

0

So that substituting the limits we get,

200

1)(1)( 22

0

22

m

jnjn

n

ICosjSin

jn

eCosjnSin

jn

eC

210

1

11(0

1 22

m

n

n

I

nn

eC

But jSinnCosne n but 0Sinn and

nCosn 1

21

)1(

21

)1(

1

)1(222

m

n

m

n

n

I

n

I

nnC

01

1)1(2

nC

n

n for n = odd 1, 3, 5

But

22 1

2

1

1)1(

nnC

n

n

for n = even 0, 2, 4

)1(1

2

2 22 n

I

n

IC mm

n

)1( 2n

IC m

n

But for n = 1 then,

0)11(1

mm IIC

Undefined

Therefore, since this is unacceptable for n = 1 we

get

From T

2 but

2

TT

2

2T T

22

T

2T

Using and after substituting the limits ,0 we

get

m

j

ICosjSinj

eCosjnSin

j

eC

00112

122

0

221

mIC

10

11

1

11

)10)(1(

2

122221



9

22

1

2

1

2

11

mm

IIC

21

mIC

Therefore,

mI

C 0 , 21 n

IC m

n

and 2

1mI

C

Therefore, the general Fourier expression is;

n

tjn

n CeCCtI 10)(

121)( C

n

eIItI

n

tjn

mm

tSinIeeeeII

tI m

tjtjtjtj

mm

2...

)41(2)21(2...

)41(2)21(2...

2)(

2

4

2

2

2

4

2

2

tSinIeeeeII

tI m

tjtjtjtj

mm

2...

15232

2)(

4422

21 n

IC m

n

but Coswtee tjtj

2

22

(Euler’s

formula)

tSinItCostCosII

tI mm

02

1...4

15

12

3

12)(


SIGNIFICANCE OF THE ABOVE HALF-

WAVE RECTIFIER ANALYSIS EXPRESSION:

- The expression indicates discrete (even) cosine

series.

- The convergence is fast .1

12

n Implies by

just adding few terms we can get the original

signal.

- The input sinusoidal current is SinwtIm but the

output sinusoidal current is SinwtIm21 . This

means only half of this sinusoidal current was

converted into D.C. But this properly is useful

especially where a smooth d.c. is not required.

For example this circuit arrangement can be

used in the construction of battery charger,

which charges batteries which uses electrolyte.

PROOF: USING THE TRIGONOMETRIC

APPROACH FOR THE HALF-WAVE

RECTIFIER ANALYSIS:

From

1 1

0 )()()(n n

nn tnSinbtnCosaatI

For the coefficient 0a we have,

02

00

0

22

CosCos

T

ItCos

T

ItdtSin

T

Ia Tmmm

TT

But 2T

mmmm IIICos

Ia 2

211

21

20

00 CI

a m

(Q.E.D)

For the coefficient an we have,

2

0

2T

tdtCosntSinT

Ia m

n

Using trig. Identities;

)()(2

1vuSinvuSinSinuCosv

2

0

)()(2

T

tntSintntSinT

Ia m

n

Integrating w.r.t t, and substituting the limit, we

get

)1(

0

)1(2

)1(

)1(

)1(

2

1

2

2

n

Cos

n

tnCos

n

tnCos

T

Ia m

n

But 2T



10

+ve

-ve 0

0

D2

D1

O

Output DC signal

n

nCos

n

nCosIa m

n1

1

1

1

2

But from trigonometric identities we get,

SinASinBCosACosBBACos )(

21

1

1

1 mn

I

n

CosnCos

n

SinnSinCosnCosa

But 0SinSinn for any value of n

222 1

1)1(

)1(

1

)1(

2cos2

2 n

I

n

CosnI

n

nIa

n

mmmn

01

112

n

Ia m

n

for n = odd 1, 3, 5

But

nm

n Cn

Ia

)1(

22

(Q.E.D)

For b1 using trigonometric we get

2

0

1

2T

tdtSinntSinT

Ib m But n = 1

2

0

2

1

2T

tdtSinT

Ib m But tCostSin 21

2

12

2

0

1 )21(2

2T

dttCosT

Ib m

Integrating and substituting limits we get,

But 2T

0

2

22

2221

SinIT

T

Ib mm

But 0Sinn for any value of n

02

1 mIb

112

CI

b m (Q.E.D)

FULL-WAVE RECTIER: here I want to use the

Fourier analysis to see if the Full-Wave rectifier as

it is been called converts A.C into a complete D.C.

The full wave rectifier can thought of as

inverting the positive peaks to a negative troughs

and vice-versa as shown below:

This is what the circuit diagram shows, but let us

prove it if this true using theory and equation.

To analyze the properties of the full wave rectifier

all we need is the Fourier coefficients from this

expression;

tjn

n eCCtI 0)(

For a full wave rectifier its defined by this

expression;

tSintI )( t0

tSin 0 t

For the coefficient C0, we have,

Since the given function is an even function then,

dtetSindtetIC tjtjn

0

0

0

0

1)(

1

A.C



11

dttSinC

0

0

1 Integrating and substituting

the limits we get,

111

)1()1(11

00

tCosC

20 C

For the coefficient Cn we have, and since the

function is an even function then,

dtetSinC tjn

n

0

1

Using the identity;

cbxbCoscbxaSinba

ecbxSine

axx

22

2

Where a = 2, b = 1 and c = 0

Similarly by using this identity we get,

bCosbxaSinbxba

etSine

tjntjn

22

Where a = -jn, b = 1 and c = 0

022

0

0.101

1CosjnSin

jn

etdtSineC

tjntjn

n


0101

1

1

122

CosjnSinn

CosjnSinn

eC

jn

n

But jSinnCosne jn

10

1

1

1

11122 nn

C

n

n

22 1

21

1

111

nn

n

if n = even, 0, 2, 4

12

2

nCn

Hence the Fourier series of this analysis becomes;

n

tjn

n eCCtI 0)(

But

20 C and

12

2

nCn

...1412

...1412

...22

)(2

4

2

2

2

4

2

2 tjtjtjtj eeeetI

...15232

222)(

4422 tjtjtjtj eeeetI

But tCosee tjtj

2

...4

15

12

3

142)( tCostCostI


SIGNIFICANCE OF THE FOURIER ANALYSIS

OF THE FULL WAVE RECTIFIER

EXPRESSION:

- The original frequency ( ) has been

eliminated and now the lowest frequency of

oscillation is 2 . Indicating frequency of

oscillation is always increasing by (2n) where

n is the position of the term in the series.

- The convergence is fairly fast

1

12n

. It

means the conversion of the A.C to D.C is

instantaneously.

- The expression contains cosine terms only,

showing that a full wave rectifier does a good

job of approximating an A.C to a D.C.



12


APPROACH FOR THE FULL-WAVE

RECTIFIER.

From

11

0)(n

n

n

n tSinnbtCosnaatI

Since the function is even function then,

)1()1(11

)(1

0

0

0

tCosttdSina

00

2Ca

(Q.E.D)

For the coefficient an

0

)(2

ttdCosntSinan

Using trig. Identities;

)()(2

1vuSinvuSinSinuCosv

0

)()(2

12tntSintntSinan

Integrating w.r.t t and substituting the limit we get,

n

nCos

n

nCosan

1

)(1

)1(

)(1

2

2

Using: Cos(A+B) = CosA CosB – SinA SinB

2

2

1

1

)1(

1

n

CosnCos

n

SinnSinCosnCosan

But 0SinnSin for any value of n

)1(

1)1(2

)1(

12

)1(

222222 nn

Cosn

n

Cosna

n

n

22 1

22

1

1)1(2

nna

n

n

if n = even, 0,

2, 4

nn Cn

a

)1(

42

(Q.E.D)

CONCLUSION

The most important parameter in the Fourier series

is the Fourier coefficients. In this work, because of

the flexibility of the Fourier series I have used two

‘flavours’ in finding the Fourier coefficients viz:

The complex exponential method and

trigonometric approach to find the Fourier

coefficients and then use these Fourier coefficients

to generate the Fourier series to analyze and

interpret the physical significance of the Fourier

series generated. Here I have covered both the

triangular, square, saw-tooth waves forms as well

as the half and full wave rectifiers diodes and went

ahead to interpret their physical significance. For

example the discrete Cosine series obtained from

the analysis of the triangular wave form is good

tool for image processing. The continuous Sine

series from the analysis of the saw-tooth wave

form is also a good tool for audio compression.

While the half wave rectifier analysis shows that it

can be used for the construction of battery

chargers which uses electrolyte and the full wave

rectifier analysis is ideal for the construction of

battery chargers which do not use electrolyte. As

this work is just an ‘opening’ for the analysis and

interpretation of signals using Fourier analysis, I

recommend that readers should also strive to look

for other uses of this analysis. Above all thank you

very much, Jean-Baptiste Fourier, the founder of

the Fourier series.

REFERENCES:

Birkhauser (2006). Journal of Fourier analysis and

application ISSN 10695869. New York: McGraw

Hill

Boashash B. ed (2003). Time frequency signal

analysis and processing. A comprehensive

reference, oxford: Elsevier science 0-08-044335-4

Campbell, George, Foster, Ronald (1989). Fourier

integrals for practical applications.



13

Chatfield, Chris (2004). The analysis of time

series: An Introduction Test.

Douandi Keotxea, Javier (2001), Fourier analysis,

American mathematical society, ISBN 0-8218-

2172-5

Dym. H; McKean H. (2009), Fourier series and

integrals, academic press, ISBN 978-0-12-226451-

1

Folland, G. B. (1992) Fourier analysis and its

application, pacific Groove CA: Brooks.

Folland, Gerald (2006) Harmonic analysis in

phase space, Princeton University press

Gupta, B. D. (2010) mathematical physics (Fourth

Edition).

Grafakos, Loukas, (2004) Classical and modern

Fourier analysis, Prentice-Hall ISBN 0-13-

035399-X.

Hans G. Feichtinger (2015). The journal of Fourier

analysis and applications 10(4): 516-610. CRC

Press

Jason D. (2004) Excellent, Elementary and

Concise introduction to Fourier analysis. (June 12)

Morriso N. (1994). Introduction to Fourier

analysis: New York, Wiley online library

Kammler, David (2000), A first course in Fourier

analysis Princeton Hall, ISBN 0-13-578782-3

Rudin, Walter (2007) Real and complex analysis

(3rd

edition) Singapore: McGraw Hill. ISBN 0-07-

100276-6

Stein, Elias, Weiss, Guido (2001) Introduction to

Fourier analysis on Euclidean spaces, Princeton

university press, ISBN 978-0-691-08078-9

Sogge, C.D (2003) Fourier integrals in classical

analysis. New York: Cambridge university press.

Stoica, Petre; Mosses, Randolph (2005) Spectral

analysis of signals. Nj: Prentice Hall.

Spiegel, M.R. (2004). Theory and problems of

Fourier analysis of signals. Nj: Prentice Hall.



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Interpretation of the Physical Significance of the Fourier