Chapter 3: Fourier Series

Da-Chuan Cheng, PhD 1112/04/20 Da-Chuan Cheng, PhD 1

Chapter 3: Fourier Series

• Introduction– In 1808, Fourier wrote the first version of his

celebrated memoir on the theory of heat “Theorie Analytique de la Chaleur”. He made a detailed study of trigonometric series, which he used to solve a variety of heat conduction problems.

– Nearly two centuries after Fourier’s work, the series that bears his name is still important, practically and theoretically, and still a topic of current research.

Da-Chuan Cheng, PhD 2Da-Chuan Cheng, PhD 2

Computation of Fourier series : Real form

1 10 )sin()cos()(

:seriesFourier in the , and ts,coefficienFourier thecompute we interval On the

k kkk

kk

kxbkxaaxf

baπ,x-π

Important properties of Fourier series:

otherwise0

0 if2

1 if1

)cos()cos(1

kn

kn

dxkxnx(1)

otherwise0

1 if1)sin()sin(

1 kndxkxnx(2)

.,integer allfor 0)sin()cos(

1kndxkxnx

integers. are ,kn

(3)

,2,1k



,

)2sin(,

)sin(,

2

1,

)cos(,

)2cos( ,

xxxx

An equivalent way of starting this theorem is that the collection

is an orthonormal set of functions in ]),([2 L

Da-Chuan Cheng, PhD 4

Matlab program


% Confirm page 3.L=37; % L can be any positive integer.t=[-pi:(2*pi)/L:pi]; % range is from -pi to (pi- dt)t=t(1:end-1);y=cos(t)/sqrt(L/2); % equation in page 3. y=cos(t).disp(['The length of vector y=cos(t) is : ']);sum(y.^2) % confirm the length=1.figure(1); plot(t,y); hold on;

長度不是 π﹐而是 L/2

The length of vector y=cos(t) is :

ans =

1.0000



z1=sin(t)/sqrt(L/2); % equation in page 3. y=sin(t).disp(['The length of vector y=sin(t) is : ']);sum(z1.^2) % confirm the length=1.figure(1); plot(t,z1); hold on;

The length of vector y=sin(t) is :

ans =

1.0000


y1=cos(2*t)/sqrt(L/2); % y=cos(2t).disp(['The length of vector y1=cos(2t) is : ']);sum(y1.^2) % confirm the length=1.figure(1); plot(t,y1,'y');

The length of vector y1=cos(2t) is :

ans =

1


% Confirm cos(t)sin(t)=0z = cos(t).*sin(t);sum(z)

ans =

4.9960e-016

% confirm cos(t)*sin(2t)=0z = cos(t).*sin(2*t);sum(z)

ans =

-1.1102e-015




Proof of the properties on page 2:

The derivations of the first two equalities use the following identities:

.sinsincoscos))cos((sinsincoscos))cos((kxnxkxnxxknkxnxkxnxxkn

) if(0)sin()sin(

2

1

)))cos(())(cos((2

1coscos

knkn

xkn

kn

xkn

dxxknxknkxdxnx



dxnxnxdxkn )2cos1)(2/1(cos then,1 If 2

proof.) the(Complete.21)/1( then ,0 If

dxkn

Equation (2) and (3) can be proved in a similar way.

Fourier coefficients computation:

1 1

0 )sin()cos()(k k

kk kxbkxaaxfAssume

nxdxkxbkxaanxdxxfa

k kkn cos)sin()cos(1

cos)(1

: find To10

Note that k starts from 1, why? (Hint: see page 2, Eq.(1))



nn

kk

kk

kk

kk

k kk

aa

dxnxkxb

dxnxkxanxn

a

dxnxkxb

dxnxkxadxnxa

dxnxkxbkxaa

00

)cos()sin(1

)cos()cos(1

|)sin(11

)cos()sin(1

)cos()cos(1

)cos(1

)cos()sin()cos(1

1

10

1

10

10



1)cos()(1

nadxnxxf n

1)sin()(1

Similarly,

nbdxnxxf n

000 2

1)(

2

1: find To adxadxxfa

實際上就是訊號 f(x)的平均值



Theorem:

1 1

0 )sin()cos()(k k

kk kxbkxaaxfIf

then,

dxxfa )(

2

10

1)cos()(1

ndxnxxfan

1)sin()(1

ndxnxxfbn

The Fourier coefficients for a given function are unique.


Matlab implementation


% Page 13.L=64; % L can be any positive integer.t=[-pi:(2*pi)/L:pi]; % range is from -pi to (pi- dt)t=t(1:end-1); f=rand(1,L).*sin(rand(1,L)*2); figure; plot(t,f); a0=mean(f) a1=sum(f.*cos(t))/L/2b1=sum(f.*sin(t))/L/2


Change the representation form

)]cos(,),[cos(),cos(; kkkxx see p.13

)]cos(,),[cos(),cos(;

kka

xkaxa

)]cos(,),[cos(),2

2cos(;

kk

a

xkaxa

or



Theorem:

1 10 )/sin()/cos()(

k kkk axkbaxkaaxfIf

then, a

adxxf

aa )(

2

10

1)2

2cos()(

1

ndx

a

nxxf

aa

a

an

1)2

2sin()(

1

ndx

a

nxxf

ab

a

an



Even and odd functions

).()( if odd is );()( ifeven is

function; a be :Let :Definition

xfxffxfxffRRf

The following properties follow from the definition.

Even X Even = EvenEven X Odd = OddOdd X Odd = Even

If F is an even function, then

aa

adxxFdxxF

0)(2)(

If F is an odd function, then 0)( a

adxxF



Theorem:

a

k

a

kk

dxaxkxfa

a

dxxfa

a

axkaaxf

0

00

10

)/cos()(2

)(1

with)/cos()(

cosines. haveonly will][ interval on the seriesFourier its then function,even an is )( If

-a,axf

sines. haveonly will][ interval on the seriesFourier its then function, oddan is )( If

-a,axf

a

k

kk

dxaxkxfa

b

axkbxf

0

1

)/sin()(2

with)/sin()(



Gibbs phenomenon

Approximation of square wave in 5 steps

The height of the blip is approximately the same no matter how many terms are considered in the partial sum.


Computation of Fourier series (1) : Complex form

Complex form of Fourier Series

Often, it is more convenient to express Fourier series in complex form using the complex exponentials due to the simple computational properties of these functions.

Definition:

1 where)sin()cos(

:is lexponentiacmplex thenumber t, realany For

ititeit

This definition is motivated by substituting x=it into the Taylor series for xe

)sin()cos()!5!3

()!4!2

1(

!4

)(

!3

)(

!2

)()(1

with !4!3!2

1

5342

432

432

tittt

titt

itititite

itxxxx

xe

it

x



Lemma:

itit

stiisit

stiisit

itit

it

itti

ieedt

deeeeee

eee

ee

)(

/

1||

)(

)(

)2(

Theorem:

]).,([Lin lorthonorma is ,2,1,0,1,2, ,2

functions ofset The 2

ne int

Proof:

mnmn

mni

edtedteeee

tmnitmniimtimt

if2 if0

)(,

)()(intint



Theorem:

dtetfa

teatf

n

nn

int

int

)(2

1then

, interval on the )( If

Combine with the previous theorem in Chap2 p.7, we get the following theorem:

Proof: To find na we simply substitute f(t) into:

dtntmtntmtintmtintmta

dtntintmtimtadteea

nn

nn

tin

n

imtn

))sin()sin()sin()cos()cos()sin()cos()(cos(2

1

))sin()))(cos(sin()(cos(2

1

2

1

Use the properties in p.2.


.,integer allfor 0)sin()cos(

1mndxmxnx

dtntmtntmtintmtintmtann

))sin()sin()sin()cos()cos()sin()cos()(cos(2

1

Property (3)

0 0

(1) n≠m

otherwise0

0 if2

1 if1

)cos()cos(1

mn

mn

dxmxnx

otherwise0

1 if1)sin()sin(

1 mndxmxnx

=0


dtntmtntmtann

))sin()sin()cos()(cos(2

1

(2) n=m=0 000 22

1)0cos(

2

1aadta

(3) n=m 1≧ nnn adtadtntnta

2

1))(sin)((cos

2

1 22



So the complex Fourier series of f is

k

tki

n

tinn e

k

iea )12(

)12(

2

Example:

0 if1

0 if1)(

t

ttf

The n-th complex Fourier coefficients is:

even isn if0

odd isn if2]1)[cos(

))]cos(1(1)[cos(2

]||[)(

1

2

1

2

1

2

1)(

2

1 00

0

0

n

i

n

ninn

n

i

eein

dtedtedtetfa tintintintinntin



t=[-pi:2*pi/1024:pi];t=t(1:end-1);fs=zeros(1,length(t));N=100;K=[-N:N];for k=K; fs=fs+2/((2*k+1)*pi)*sin((2*k+1)*t);end; figure(1); subplot(121); plot(t,fs); fc=zeros(1,length(t));for k=K, fc=fc-2/((2*k+1)*pi)*cos((2*k+1)*t);end;figure(1); subplot(122); plot(t,fc);

Matlab implementation

k

tkiek

i )12(

)12(

2

0 if1

0 if1)(

t

ttf



Result: k=-10:10

real part imaginary part



Result: k=-50:50




Result: k=-100:100




Result: k=-10000:10000




Theorem: The set of functions

]).,([Lfor basis lorthonormaan is ,2,1,0,1,2, ,2

2 aana

e a

t/in

a

a

atinn

n

atinn dtetf

aαetf // )(

2

1 then ,)( If

Example:

01 if1

10 if1)(

t

ttf

n

ni

ninninn

i

eein

dtedtedtetfa

ndst

tintintintintnin

]1)[cos(

]))sin()(cos()01()01())sin()(cos([2

]||[)(

1

2

1

2

1

2

1)(

2

1

term2 term1

01

10

0

1

1

0

1

1



Relation between the real and complex Fourier series

If f is a real valued function, the real form of its Fourier series can be derived from its complex form and vice versa. For simplicity, we discuss this derivation on the interval -π t π, but this discussion also holds for other ≦ ≦intervals as well.

.)(2

1 where)(

termpositive

10

termnegative

1

dtetfeetf tin

nn

tinn

n

tinn

If f is real valued, then because nn

ntintin

n dtetfdtetf

)(2

1)(

2

1

n≠0



termnegative

1

termpositive

10)(

n

tinn

n

tinn eetf

10 Re2)(

}Re{2

n

tinnetf

zzz



00 )(2

1adttf

)(2

1

))sin())(cos((2

1

1for )(2

1

nn

tinn

iba

dtntinttf

ndtetf

similar to p. 13

(see p. 13)



10

10

10

)sin()cos(

))sin())(cos((Re

Re2)( Recall

nnn

nnn

n

tinn

ntbnta

ntintiba

etf

Exactly the same to page 13.


Homework

• 請自行找出一個長 32的信號﹐寫 Matlab程式將其 Fourier series的係數 (a0, an, bn)找出。

• Issue date: 5/5

• Due date: 19/5

Documents

Chapter 3: Fourier Series