Ch. 11 Fourier Analysisocw.snu.ac.kr/sites/default/files/NOTE/Ch11 Pt II.pdf · 2019. 3. 15. · Seoul National Univ. 2 11.6 Orthogonal Series. Generalized Fourier Series Standard

Seoul NationalUniv.

1

Ch. 11 Fourier AnalysisPart II

서울대학교

조선해양공학과

서유택

2018.10

※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다.

Seoul NationalUniv.

2

11.6 Orthogonal Series. Generalized Fourier Series

Standard Notation for Orthogonality (직교성) and Orthonormality (정규직교성)

For orthonormal functions y0, y1, y2, … with respect to weight function r(x) (> 0) on

interval a ≤ x ≤ b

Orthogonal Series, Orthogonal Expansion or Generalized Fourier Series

To find Fourier constants (a0, a1, …) of f(x): Multiplying both sides by ryn

0

, 1

b

m n m n mn

a

m ny y r x y x y x dx

m n

2,

b

a

y y y r x y x dx

0 0 1 1

0

m m

m

f x a y x a y x a y x

0 0 0

, ( , )

b b b

n n m m n m m n m m n

m m ma a a

f y rf y dx r a y y dx a ry y dx a y y

Due to the Orthogonality 2 2

( , ) ,n n n n n n n na y y a y f y a y

* 직교성 (orthogonality): 서로 다른 것끼리는 공통점이 없다라는 뜻. 선형대수학에서 두 벡터 사이의 내적이 0이라는 것으로 정의하며,한 벡터가 다른 벡터의 성분을 조금도 가지고 있지 않다는 것을 말함

* 정규직교성 (orthonormarlity): 선형대수학에서 두 단위 벡터 (크기가 1) 사이의 내적이 0이라는 것

0

1

( ) cos sinn n

n

f x a a nx b nx

Seoul NationalUniv.

3


Orthogonal Series, Orthogonal Expansion or Generalized Fourier Series

0 0 1 1

0

m m

m

f x a y x a y x a y x

2 2

, 1 0 1 2

b

m

m m

am m

f ya r x f x y x dx m , , ,

y y

2

, n n nf y a y

,

b

n n

a

f y rf y dx

Seoul NationalUniv.

4

[Review] Ch. 5 Series Solutions of ODEs. Special Functions

In the previous chapters, linear ODEs with constant coefficients (상수계수)

can be solved by algebraic methods (대수적인 방법), and that their

solutions are elementary functions (초등함수) known from calculus (미적분).

For ODEs with variable coefficients (변수계수) the situation is more

complicated, and their solutions may be nonelementary functions.

In this chapter, the three main topics are Legendre polynomials, Bessel

functions, and hypergeometric functions (초기하함수).

Legendre’s ODE and Legendre polynomials are obtained by the power series

method (거듭제곱급수 또는 멱급수 해법).

Bessel’s ODE and Bessel functions are obtained by the Frobenius method,

an extension of the power series method.

21 '' 2 ' 1 0x y xy n n y

2 2 2'' ' 0x y xy x y

* Elementary functions (초등함수): 다항 함수, 로그 함수, 지수 함수, 삼각 함수와 이들 함수의 합성 함수들을 총칭* Hypergeometric functions (초기하함수): 거듭제곱 급수로 나타내지는 일련의 특수 함수들을 총칭

Seoul NationalUniv.

5

Power Series (거듭제곱):

Coefficients: a0, a1 , a2, …

Center: x0

Power Series in powers of x if x0 = 0:

Ex 1 Maclaurin series

2

0 0 1 0 2 0

0

m

m

m

a x x a a x x a x x

2

0 1 2

0

m

m

m

a x a a x a x

2

0

2 3

0

2 2 4

0

2 1 3 5

0

11 1, geometric series

1

1! 2! 3!

1cos 1

2 ! 2! 4!

1sin

2 1 ! 3! 5!

m

m

mx

m

m m

m

m m

m

x x x xx

x x xe x

m

x x xx

m

x x xx x

m

[Review] 5.1 Power Series Method (거듭제곱급수해법, 멱급수해법)

The power series method is the standard method for solving linear

ODEs with variable coefficients.

등비급수

Seoul NationalUniv.

6


Example 1 Fourier-Legendre Series: an eigenfunction expansion

2

0 0 1 1 2 2 0 1 2

0

3 1

2 2m m

m

f x a P x a P x a P x a P x a a x a x

1

1

2 1 0 1 2

2m m

ma f x P x dx m , , ,

1

2

1

2( ) = 0 1 2

2 1m mP P x dx m , , ,

m

1 1

1

1 1

2 1 3 3(sin ) sin 0.95493

2 2m m

ma x P x dx a x x dx

2

1

b

m m

am

a r x f x y x dxy

For instance ( ) sinf x x

1 3 5 7 9

11

sin 0.95493 ( ) 1.15824 ( ) 0.21929 ( ) 0.01664 ( ) 0.00068 ( )

0.00002 ( ) ...

x P x P x P x P x P x

P x

1)(0 xP

xxP )(1

)13(2

1)( 2

2 xxP

)35(2

1)( 3

3 xxxP

)33035(8

1)( 24

4 xxxP

2

0

2 2 ! 11 or from Ch.5.2

2 ! ! 2 ! 2 2

Mm n m

n nm

n m n nP x x M

m n m n m

r(x) = 1

2,

b

a

y y y r x y x dx

Seoul NationalUniv.

7

Bessel’s equation:

Apply the Frobenius method

Substitute the series with undetermined

coefficients and its derivatives.

Indicial equation:

2 2 2'' ' 0, where 0x y xy x y

0r r

0

m r

m

m

y a x

2 2

0 0 0 0

1 0m r m r m r m r

m m m m

m m m m

m r m r a x m r a x a x a x

2

0 0 0

2

1 1 1

2

2

1 0 0

1 1 0 1

1 0 2, 3, s s s s

r r a ra a s

r ra r a a s

s r s r a s r a a a s

[Reference] 5.4 Bessel’s Equation. Bessel Functions Jv(x)

1 2,r v r v

Seoul NationalUniv.

8

Coefficient Recursion (계수 점화) for r = r1 = v

For s = 2m,

1 1

2 3 5

2 1 0 0 ( 0)

2 0 0s s

a a v

s sa a a a

2 2 2 2 2 22

12 2 2 0 , 1, 2,

2m m m mm ma a a a m

m m

2 02

1

2 1a a


2

0 0 0

2

1 1 1

2

2

1 0 0

1 1 0 1

1 0 2, 3, s s s s

r r a ra a s

r ra r a a s

s r s r a s r a a a s

4 2 02 4

1 1

2 2 2 2 2! 1 2a a a

2 02

1 , 1, 2,

2 ! 1 2

m

m ma a m

m m

Seoul NationalUniv.

9


Bessel Functions Jv(x) for Integer v = n

Choose

Bessel function of the first kind of order n:

(n차 제 1종 Bessel 함수)

2

20

1

2 ! !

m m

n

n m nm

xJ x x

m n m

0

1

2 !na

n

2 2

1, 1, 2,

2 ! !

m

m m na m

m n m

0)1(2

2

yx

n

x

yy

2 2 2'' ' 0x y xy x y

0

m r

m

m

y a x

1 3 5( , ... 0)r n a a a

Seoul NationalUniv.

10


Example 2 Fourier-Bessel Series

Step 1. Bessel’s equation as a Strurm-Liouville equation

The Bessel function Jn(x) with fixed integer satisfies Bessel’s equation

We set

Dividing by x and using

1 2

1 2

' 0

' 0

k y x k y x at x a

l y x l y x at x b

Orthogonality on interval

' ' 0p x y q x r x y

at

2 2 2'' ' 0x y xy x y

Bessel’s equation

(From Theorem 1 in Sec. 11.5)

Strurm-Liouville equation

Seoul NationalUniv.

11


Theorem 1 Orthogonality of Bessel Functions

For each fixed nonnegative integer n the sequence of Bessel functions of the

first kind forms an orthogonal set (직교집합) on the

interval 0 ≤ x ≤ R with respect to the weight function r(x) = x, that is,

,1 ,2, , n n n nJ k x J k x

, ,

0

0 , fixed

R

n n m n n jxJ k x J k x dx j m n

has infinitely many zeros. Assume that

Step 2. Orthogonality

a0,1

a1,1

a0,2a1,2

a0,3

a0,4

a1,3

2

20

1

2 ! !

m m

n

n m nm

xJ x x

m n m

Seoul NationalUniv.

12


Step 3. Fourier-Bessel Series

because the square of the norm is

2 2

, 1 0 1 2

b

m

m m

am m

f ya r x f x y x dx m , , ,

y y

Seoul NationalUniv.

13

11.7 Fourier Integral

Equalizer

Seoul NationalUniv.

14


/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

A real function is represented by a complex series; a series in which the coefficients are complex numbers** ( )f x nc

*Mary L. Boas, Mathematical Methods in the Physical Science., Second Edition, John Wiley & Sons, 1966, p647-648**Zill D.G., & Cullen M.R., Advanced Engineering Mathematics, Third Edition, Jones and Bartlett, 2006, p670

-expand a periodic function

Function expansion

-in series of sine, cosine, and complex**

-infinite but discrete set of frequencies

Q1. is it possible to represent a function which is not periodic by something analogues to a Fourier series?

nc

2 3023

0, 1, 2,...n

......

Q2. can we somehow extend or modify Fourier series to cover the case of a continuous set of frequencies?

n

recall, an ‘integral’ is a ‘limit of a sum’

Fourier Series Fourier Integral

Fourier Series

x

( )f x

2 22

2f

T p p

2 2

2T p p

* ω (angular frequency): also referred to by the terms angular speed, radial frequency, circular frequency, orbital frequency, radian frequency

Seoul NationalUniv.

15


A real function is represented by a complex series ; a series in which the coefficients are complex numbers** ( )f x nc

*Mary L. Boas, Mathematical Methods in the Physical Science., Second Edition, John Wiley & Sons, 1966, p647-648**Zill D.G., & Cullen M.R., Advanced Engineering Mathematics, Third Edition, Jones and Bartlett, 2006, p670

Function expansion

Fourier Series

1 ˆ( ) ( )2

i xf x f e d

1ˆ ( ) ( )2

i xf f x e dx

Fourier Integralextend

-represent nonperiodic functions

-integral of sine, cosine, and complex**

-continuous set of frequencies

n


Fourier Series Fourier Integraln


Fourier Series Fourier Integral-expand a periodic function

-in series of sine, cosine, and complex**

-infinite but discrete set of frequencies

corresponding

ˆ( )f

interval-valued variable n

ncthe set of coefficients

continuous variable

become a function ˆ ( )f

nc

2 3023

2 2

2T p p

0, 1, 2,...n

......

x

( )f x

/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

2 22

2f

T p p

Seoul NationalUniv.

16

Ex. 1 Rectangular Wave

Consider the periodic rectangular wave fL(x) of period 2L > 2 given by


0 1

1 1 1

0 1

L

L x

f x x

x L

1 1 1

lim0 otherwise

LL

xf x f x

: Nonperiodic function

which we obtain from fL

Seoul NationalUniv.

17



Sol)

1 1 1

lim0 otherwise

LL

xf x f x

1 1 1

0

1 1 0

0 for all

sin1 1 1 2 2

, cos cos2

n

n

b n

n

n x n x La dx a dx dx

nL L L L L L L

L

fL is even

Amplitude Spectrum of fL: Sequence of Fourier coefficients

2L=4, L=2

1 1

sin / sin /1 2 1 2( ) cos( / ) cos( )

/ /n

n n

n L n Lf x n L

L L n L L L n L

1

sin / 220.636

2 / 2a

2

sin 2 / 220

2 2 / 2a

0

1

cos evenn

n

nf x a a x f

L

1

sin oddn

n

nf x b x f

L

n = 2 n = 4

Seoul NationalUniv.

18

[Reference] Wave Characteristics

Seoul NationalUniv.

19



Sol)

1 1 1

lim0 otherwise

LL

xf x f x

2L=8, L=4

1

sin / 420.450

4 / 4a

2

sin 2 / 420.318

4 2 / 4a

3

sin 3 / 420.150

4 3 / 4a

4

sin 4 / 420

4 4 / 4a

1 1

sin / sin /1 2 1 2( ) cos( / ) cos( )

/ /n

n n

n L n Lf x n L

L L n L L L n L

n = 4

Seoul NationalUniv.

20



Sol)

1 1 1

lim0 otherwise

LL

xf x f x

2L=16, L=8

1 5

2 6

3 7

4 8

sin / 8 sin 5 / 82 20.243 0.117

8 / 8 8 5 / 8

sin 2 / 8 sin 6 / 82 20.225 0.075

8 2 / 8 8 6 / 8

sin 3 / 8 sin 7 / 82 20.196 0.034

8 3 / 8 8 7 / 8

sin 4 / 8 sin 8 / 82 20.159 0.000

8 4 / 8 8 8 / 8

a a

a a

a a

a a

1 1

sin / sin /1 2 1 2( ) cos( / ) cos( )

/ /n

n n

n L n Lf x n L

L L n L L L n L

Seoul NationalUniv.

21


From Fourier Series to Fourier Integral

0

1

cos sin , L n n n n n

n

nf x a a w x b w x w

L

1

1 1 cos cos sin sin

2

L L L

L n L n n L n

nL L L

f v dv w x f v w vdv w x f v w vdvL L

1

1 1cos cos sin sin

2

L L L

L n L n n L n

nL L L

f v dv w x w f v w vdv w x w f v w vdvL

1

1 1n n

n n ww w w

L L L L

0

1

2

1cos , 1, 2,

1sin , 1, 2,

L

L

L

n

L

L

n

L

a f x dxL

n xa f x dx n

L L

n xb f x dx n

L L

0

0

( ) ( ) written ( )lim limb

a ba

f x dx f x dx f x dx

Let’s assume that the resulting nonperiodic function f(x) as L→ (e.g., 1/L

→0) is absolutely integrable on the x-axis; that is, the following (finite!)

limits exist.

Seoul NationalUniv.

22


From Fourier Series to Fourier Integral

0

1lim cos cos sin sin

LL

f x f x wx f v wvdv wx f v wvdv dw

1

1 1cos cos sin sin

2

L L L

L n L n n L n

nL L L

f v dv w x w f v w vdv w x w f v w vdvL

an(n)

w

0

Fourier Integral : cos sin

1 1 cos , sin

f x A w wx B w wx dw

A w f v wvdv B w f v wvdv

A w B w

L→∞에 따라서 an(n)은 더 촘촘해짐

1 0

n n

n

g w w g w dw

1 w

L

Seoul NationalUniv.

23


Theorem 1 Fourier Integral

If f (x) is piecewise continuous in every finite interval and has a right-hand

derivative and a left-hand derivative at every point and if the integral

exists, then f (x) can be represented by a Fourier integral

with A and B given by .

At a point where f (x) is discontinuous the value of the Fourier integral equals

the average of the left- and right-hand limits of f(x) at that point.

0

0

lim lim written

b

a ba

f x dx f x dx f x dx

0

cos sinf x A w wx B w wx dw

1 1

cos , sinA w f v wvdv B w f v wvdv

Seoul NationalUniv.

24


Ex. 2 Single Pulse, Sine Integral

Find the Fourier integral representation of the function

1 1

0 1

xf x

x

11

11

1

1

0

1 1 sin 2sincos cos

1 1sin sin 0

2 cos sin

wv wA w f v wvdv wvdv

w w

B w f v wvdv wvdv

wx wf x dw

w

0

/ 2 0 1cos sin

: / 4 1

0 1

xwx w

dw xw

x

Dirichlet's Discontinuous Factor

Sol)

0


디리클레

Seoul NationalUniv.

25


0

sin : Si

uw

u dww

Sine Integral

0 0 0

sin sin2 cos sin 1 1

a a aw wx w wxwx wf x dw dw dw

w w w

Sine integral Si(u) and

integrand

Sol-continued)

0

sinParticular interest when 0

2

wx dw

w

w wx t We set

1 1

0 0

1 sin 1 sin 1 1Si 1 Si 1

x a x at tdt dt a x a x

t t

(1 )dt t

xdw w

dt dw

t w 0 0 ( 1)w a t x a

w wx t ( 1)dt t

xdw w

dt dw

t w 0 0 ( 1)w a t x a

피적분 함수

0

2 cos sin

wx wf x dw

w

Seoul NationalUniv.

26


Sol-continued)

1 1

0 0

1 sin 1 sin( )

1 1Si 1 Si 1

x a x at t

f x dt dtt t

a x a x

0

sin

2

wdw

w

0

sinSi

uw

u dww

1/ Si(64)

1/ Si(32)

1x

0x 1/ {Si 8 Si 8 }

2x 1/ {Si 24 Si 8 } 1/ {Si 48 Si 16 } 1/ {Si 96 Si 32 }

1/ {Si 32 Si 32 }

1/ Si(16) 1/ Si(32)

Gibbs phenomenon: The shift of the oscillations toward the points of discontinuity -1 and 1

0.5x 1/ {Si 12 Si 4 } 1/ {Si 48 Si 16 }

1

1

1/ 2

0

1/ {Si 24 Si 8 }

8a 16a 32a y

Seoul NationalUniv.

27


Fourier Cosine Integral and Fourier Sine Integral

Fourier Cosine Integral: f (x) is an even function → B(w) = 0

Fourier Sine Integral: f (x) is an odd function → A(w) = 0

0 0

2cos , cosf x A w wxdw A w f v wvdv

0 0

2sin , sinf x B w wxdw B w f v wvdv

1cos

1sin

A w f v wvdv

B w f v wvdv

Seoul NationalUniv.

28


Ex. 3 Laplace Integrals

We shall derive the Fourier sine integrals of f (x) = e-kx, where x > 0 and k > 0. The result will be

used to evaluate the so-called Laplace integrals.

1. Fourier Cosine Integral

2. Fourier Sine Integral

2 2 2 2

0 0

2 2 2cos sin coskv kvk w k

A w e wvdv e wv wvk w k k w

2 2 2 2

0 0

2 2 2sin sin coskv kvw k w

B w e wvdv e wv wvk w w k w

f (x)

Sol)

2 2

0

2 cos kx k wxf x e dw

k w

2 2

0

2 sin kx w wxf x e dw

k w

2 2

0

cos( 0, 0)

2

kxwxdw e x k

k w k

2 2

0

sin ( 0, 0)

2

kxw wxdw e x k

k w

Laplace integral

Laplace integral

0

2cosA w f v wvdv

0

2sinB w f v wvdv

0


Seoul NationalUniv.

29

11.8 Fourier Cosine and Sine Transforms

Integral Transform

Transformation in the form of an integral that produces from given functions new functions

depending on different variable, ex) Laplace transform

Tools used in solving ODEs. PDEs.

Fourier Cosine Transform

Fourier cosine integral:

Set

Fourier Cosine Transform

(from to ):

Inverse Fourier Cosine Transform

(from to ):

0 0

2 cos , cosf x A w wxdw A w f v wvdv

ˆ2 / andcA w f w v x

0

2ˆ cosc cf f w f x wxdx

F

0

2 ˆ coscf x f w wxdw

0

2cosA w f v wvdv

0

2sinB w f v wvdv

f x ˆcf w

ˆcf w f x

0

stF s f e f t dt

L

Seoul NationalUniv.

30


Fourier Sine Transform

Fourier sine integral:

Set

Fourier Sine Transform

(from to ):

Inverse Fourier Sine Transform

(from to ):

0 0

2sin , sinf x B w wxdw B w f v wvdv

ˆ2 / sB w f w

0

2ˆ sins sf f w f x wxdx

F

0

2 ˆ sinsf x f w wxdw

f x ˆcf w

ˆcf w f x

0

2cosA w f v wvdv

0

2sinB w f v wvdv

Seoul NationalUniv.

31


Ex. 1 Fourier Cosine and Fourier Sine Transforms

Find the Fourier cosine and Fourier sine transforms of the function

0

0

k x af x

x a

0

0

2 2 sinˆ cos

2 2 1 cosˆ sin

a

c

a

s

awf w k wxdx k

w

awf w k wxdx k

w

f (x)

Sol)

Q : Note that for these transforms do not exist. (Why?) 0f x k const x

ˆ ˆ,c sf w f w 에서 정적분의 극한 값이 존재하지 않음 (a=)

0


F

0


F

Seoul NationalUniv.

32


Linearity

Let f (x) be continuous and absolutely integrable on the x-axis and piecewise continuous

on every finite interval

0 0 0

2 2 2[ ( )]cos cos cosc

c c

af bg af x bg x wxdx a f x wxdx b g x wxdx

a f b g

F

F F

Theorem 1 Cosine and Sine Transforms of Derivatives

Let f (x) be continuous and absolutely integrable on the x-axis, let f ʹ(x) be piecewise

continuous on every finite interval, and let f (x) → 0 as x →∞. Then

2

0 ,c s s cf x w f x f f x w f x

F F F F

2

2

20

20

c c

s s

f x w f x f

f x w f x wf

F F

F F

,c c c s s saf bg a f b g af bg a f b g F F F F F F

0


F

0


F

Seoul NationalUniv.

33


Theorem 1 Cosine and Sine Transforms of Derivatives

Let f (x) be continuous and absolutely integrable on the x-axis, let f ʹ(x) be piecewise

continuous on every finite interval, and let f (x) → 0 as x →∞. Then

2

0 ,c s s cf x w f x f f x w f x

F F F F

2

2

2 20 0

20

c s c

s c s

f x w f x f w f x f

f x w f x w f x wf

F F F

F F F

0

0 0

2 2 2( )cos ( )cos ( )sin (0)c sf x f x wxdxw f x wx w f x wxdx f w f x

F = - + F

0

0 0

2 2( )sin ( )sin ( )coss cf x f x wxdxw f x wx w f x wxdx -w f x

F = F

Proof)

0


F

0


F

Seoul NationalUniv.

34


Ex.3 An Application of the Operational Formula

Find the Fourier cosine transform of f (x) = e-ax (a>0)

By differentiation

ax

c eF

2 2 thus ax axe a e a f x f x

Sol)

2 2 2

2 2

2 2

2 2 0

2

2 0

c c c c

c

ax

c

a f f w f f w f a

a w f a

ae a

a w

F F F F

F

F

Seoul NationalUniv.

35

11.9 Fourier Transform. Discrete and Fast Fourier Transforms

Complex Form of the Fourier Integral

Fourier integral in Sec 11.7

0 0

1 1[cos cos sin sin ] cos( )f x f v wv wx wv wx dvdw f v wx wv dv dw

0


1 1

cos , sinA w f v wvdv B w f v wvdv

1

2

iw x vf x f v e dvdw

1

cos( )2

f v wx wv dv dw

1

sin( ) 02

f v wx wv dv dw

F(w) = even function of w

F(w) = odd function of wcos sinixe x i x

( )cos( ) sin( ) ( ) i wx wvf v wx wv dv i f v wx wv dv f v e

f(v) is independent of w & F(w) = F(-w).

(Euler formula)

Complex Fourier Integral (복소 Fourier 적분)

Seoul NationalUniv.

36


Fourier Transform and Its Inverse

Fourier Transform:

Inverse Fourier Transform:

1ˆ

2

iwxf f w f x e dx

F

1 1 ˆ

2

iwxf f x f w e dw

F

1

2

iw x vf x f v e dvdw

1 1

2 2

iwv iwxf x f v e dv e dw

Theorem 1 Existence of the Fourier Transform

If f (x) is absolutely integrable on the x-axis and piecewise continuous on every finite interval,

then the Fourier transform of given by exists. f̂ w 1ˆ

2

iwxf w f x e dx

f x

f̂ w

Seoul NationalUniv.

37


Ex. 1 Fourier Transform

Find the Fourier transform of f (x) = 1 if |x| < 1 and f (x) = 0 otherwise.

11

1 1

1 1 1ˆ

2 2 2

iwxiwx iw iwe

f w e dx e eiw iw

Sol)

2 sin 2 sin

2

i w w

wiw

cos sin , cos siniw iwe w i w e w i w

1ˆ

2

iwxf f w f x e dx

F

Seoul NationalUniv.

38


Ex. 2 Fourier Transform

Find the Fourier transform of if x > 0 and f (x) = 0 if x <0; here a >0 .

Sol)

( ) axf x e

( )

0 0

1 1 1

( )2 2 2 ( )

a iw xax ax iwx

x

ee e e dx

a iw a iw

F

1ˆ

2

iwxf f w f x e dx

F

Seoul NationalUniv.

39

11.9 Fourier Transform Physical Interpretation: Spectrum 1 ˆ( ) ( ) (7)2

i xf x f e d

The nature of the representation (7) of f (x) becomes clear if we think of it

as a superposition of sinusoidal oscillations (싸인함수 진동의 중첩) of all possible

frequencies, called a spectral representation (스펙트럼표현).

This name is suggested by optics (광학), where light is such a superposition

of colors (frequencies).

In (7), the “spectral density” measures the intensity (강도) of f (x) in

the frequency interval between ω and ω + Δ ω (Δ ω is small, fixed). We

claim that in connection with vibrations, the integral

ˆ ( )f

2ˆ ( )f d

can be interpreted as the total energy of the physical system. Hence an

integral of from a to b gives the contribution of the frequency ω

between a and b to the total energy.

2ˆ ( )f

Seoul NationalUniv.

40

11.9 Fourier Transform Physical Interpretation: Spectrum

2ˆ ( )f d

: total energy of the physical system

To make this plausible, we begin with a mechanical system giving a single

frequency, namely, the harmonic oscillator (mass on a spring)

.0 kyym (Here we denote time t by x)

Multiplication by gives y .0 ykyyym

Integrating with respect to x,

dxdx

dykyvdxv

dx

dm ,,

dx

dyyv

dx

dvy

21

22

2

1

2

1cckymv

L.H.S:

: R.H.S

1 ˆ( ) ( ) (7)2

i xf x f e d

Seoul NationalUniv.

41


2ˆ ( )f d

21

22

2

1

2

1cckymv

const2

1

2

10

22 Ekymv

kinetic energy + potential energy = total energy of the system

,0 kyym

General solution of the above ODE (See Eq. (3) with t = x) in Sec. 2.4)

0 0

1 0 1 0 1 1( ) cos sin ,i x i x

y x a x b x c e c e

2

0

k

m

2,

2

1111

111

ibacc

ibac

1 ˆ( ) ( ) (7)2

i xf x f e d

Seoul NationalUniv.

42


const2

1

2

10

22 Ekymv,0 kyym

0 0

1 0 1 0 1 1( ) cos sin ,i x i x

y x a x b x c e c e

2

0

k

m

0 0

1 1Writng simply ,i x i x

A c e B c e

,)( BAxy 0 0

0 1 0 1

0

( )

( )

i x i xy x v A B i c e i c e

i A B

Substitution of v and y on the left side of the equation for E0 gives

2 2 2

0 0

1 1( ) ( ) ( )

2 2E m i A B k A B

2,

2

1111

111

ibacc

ibac

1 ˆ( ) ( ) (7)2

i xf x f e d

2ˆ ( )f d

Seoul NationalUniv.

43


2

0

k

m

2 2 2

0 0

1 1( ) ( ) ( )

2 2E m i A B k A B

2 2 2

0

1 1( ) ( )

2 2m A B k A B

22 )(2

1)(

2

1BAkBAk

22 )()(2

1BABAk

kAB2

0 0

1 1, ,i x i x

A c e B c e

0 0

1 12i x i x

kc e c e

112 ckc

1 1c c 2

12 ck

The energy is proportional to

the square of the amplitude |c1|.

1 ˆ( ) ( ) (7)2

i xf x f e d

2ˆ ( )f d

Seoul NationalUniv.

44


2

10 2 ckE

Hence the energy is proportional to the square of the amplitude |c1|.

If a more complicated system leads to a periodic solution y = f (x) that

can be represented by a Fourier series,

we get a series of squares |cn|2 of Fourier coefficients instead of the

single energy term |c1|2.

we have a “discrete spectrum” (or “point spectrum”) consisting of

many isolated frequencies (infinitely many, in general), the

corresponding |cn|2 being the contributions to the total energy.

2

nc

2 3023

/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

Seoul NationalUniv.

45

periodic but not a sine or cosine

[Reference] Application of Fourier Series

Application 1) Express a function which is periodic but not a pure sine or cosine function

to a linear combination of sine or cosine functions.

ex) Forced damped mass-spring system

25,05.0,1 kcm

)(2505.0 tryyy

tt

tt

tr

0 if 2

0 if 2

)(

)()2( trtr

미정계수법 (Method of undetermined coefficient)을 사용하여 를 구함py

ttttr 5cos

5

13cos

3

1cos

4)(

22

Fourier Series

tFkzzczm cos0

0 cosm c k t z z z F

m z Fcz k0ks kz k kmg k

cz kkz k

cos t 0F

cos t 0F

m

Dashpot

0

z

kzks 0

mg

m

zc z

tFFext cos0extF

k z0s

(z component)

Seoul NationalUniv.

46

ˆ ( )f

1

21

2

)(tf

1

2


Application 2) Fourier transform

: Transform between time domain and frequency domain.

ex) Interpretation of the Fourier transform

t

ttf 2sin)(

ttf sin2)(

Time

Domain

Frequency

Domain

Inverse Fourier Transform

Fourier Transform

Frequency와 Amplitude로표현됨

=> 시계열의운동복원가능

* ω (angular frequency): also referred to by the terms angular speed, radial frequency, circular frequency, orbital frequency, radian frequency

22 f

T

1ˆ

2

iwxf f w f x e dx

F

1 1 ˆ

2

iwxf f x f w e dw

F

Seoul NationalUniv.

47


Application 2) Fourier transform

: Transform between time domain and frequency domain.

n

( )S

*Journee J. M. J., Massie W. W., Offshore Hydrodynamics, First Edition, Delft University of Technology, 2001, Figure 5.38

Fourier SeriesAnalysis

Superpositionn

‘Time’ domain

a

Measured Wave Record Generated Wave Record

t t

wave spectrum

Z Z

:wave amplitudea

2 ( )a S

Seoul NationalUniv.

48

[Reference] Wave Spectrum

* Journee J.M.J. and Massie W.W., Offshore Hydrodynamics, first edition, Delft University of Technology, 2001, 5-3, Figure 5.1

** http://i.pbase.com/o6/47/624647/1/71472987.Kc5MyTSE.seasurface.jpg

***Journee J.M.J. and Massie W.W., Offshore Hydrodynamics, first edition, Delft University of Technology, 2001, 5-29, Figure 5.22

* *

-Sum of many simple sine waves makes an

irregular sea*

t

z

zt

z t

-Superposition of two uni-directional

harmonic waves***

•Linear wave

amplitude

2 0

Frequency domain contains exactly the same information as that of the time domain.

Wave

spectrum

3학년 해양파 역학

2 2

2T p p

Seoul NationalUniv.

49


t

z

zt

z t

amplitude

2 0

If we know wave spectrum, we can ‘re-construct’ the original wave.

Wave

spectrum

Standard Wave Spectrum

•Modified Two-Parameter Pierson-Moskowitz Wave Spectrum*

1/3

z

H

T

: Significant Wave Height (유의파고)

: Mean Zero-Crossing Wave Period

2 2

2T p p

44542 )

2(

1exp)

2(

4

1)(

ZZ

STT

HS

The average of the highest 1/3 the waves in record

Seoul NationalUniv.

50

44542 )

2(

1exp)

2(

4

1)(

ZZ

STT

HSParameter

Input

Z

t


* Journee J.M.J. and Massie W.W., Offshore Hydrodynamics, first edition, Delft University of Technology, 2001, 5-32. Figure 5.26

How to use Standard Wave Spectrum

•Modified Two-Parameter Pierson-Moskowitz

2 2

2T p p

t

z

zt

z t

amplitude

2 0

If you know wave spectrum, can ‘re-construct’ the original wave

Wave

spectrum

•Time History of a Seaway*

Measuring

Peak-to-Peak Wave Period

Zero-Crossing Wave Period

Wave Height

Parameters

: Mean Zero-Crossing Wave PeriodzT

1/3H : Significant Have Height- The average of the highest 1/3 the waves

in record

Seoul NationalUniv.

51


( )S

*Journee J. M. J., Massie W. W., Offshore Hydrodynamics, First Edition, Delft University of Technology, 2001, Figure 5.38

Superpositionn

‘Time’ domain

a

Generated Wave Record

t

Z

: Wave amplitude

( ) :Energy density spectrum

a

S

1/3, H2 ( )a S

44542 )

2(

1exp)

2(

4

1)(

ZZ

STT

HS

Modified Two-Parameter Pierson-

Moskowitz Wave Spectrum

zT

Seoul NationalUniv.

52

11.9 Fourier Transform

Theorem 2 Linearity of the Fourier Transform

The Fourier transform is a linear operation; that is, for any functions f (x) and g (x) whose

Fourier transforms exist and any constants a and b, the Fourier transform of af + bg exists, and

Theorem 3 Fourier Transform of the Derivatives of f(x)

Let f (x) be continuous on the x-axis and f (x) → 0 as . Furthermore, let f ʹ(x) be

absolutely integrable on the x-axis. Then

af bg a f b g F F F

x

f x iw f x F F

2f x w f x F F

1

( )2

iwxf x f x e dx

FProof) 1

( ) ( ) ( )2

iwx iwxf x f x e iw f x e dx

F

Since f (x) → 0 as x

1

0 ( )2

iwxf x iw f x e dx iw f x

F F

1

2

iwxf f x e dx

F

2 2( ) ( ) { ( )}f x iw f x iw f w f x F F F F

Seoul NationalUniv.

53


Ex.3 Application of the Operational Formula

Find the Fourier transform of .2xxe

2 2 2 21 1 1

2 2 2

x x x xxe e e iw e

F F F F

Sol)

f x iw f x F F

2 2/4 /41 1

2 2 2 2

w wiwiw e e

Table III. Fourier Transforms

1

2

iwxf f x e dx

F

Seoul NationalUniv.

54

11.9 Fourier Transform – Tables of Transforms

Table I. Fourier Cosine Transforms Table II. Fourier Sine Transforms

1

2

iwxf f x e dx

F

Seoul NationalUniv.

55

11.9 Fourier Transform – Tables of Transforms

Table III. Fourier Transforms

1

2

iwxf f x e dx

F

Seoul NationalUniv.

56


Convolution (합성곱)

Convolution:

Theorem 4 Convolution Theorem

Suppose that f (x) and g (x) are piecewise continuous, bounded, and absolutely integrable on

the x-axis. Then

f g x f p g x p dp f x p g p dp

2f g f g F F F

Proof)

( )

1 1( ) ( ) ( ) ( )

2 2

,

1( ) ( )

2

i x i x

i p q

f g f p g x p dpe dx f p g x p e dxdp

x p q x p q

f p g q e dqdp

F

1

( ) ( )2

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

2

i p i qf g f p e dp g q e dq

f g f g

F

F F F F

Seoul NationalUniv.

57

11.9 Discrete Fourier Transform (DFT)

In using Fourier series, Fourier transforms, and trigonometric

approximations, we have to assume that a function f (x), to be

developed or transformed, is given on some interval.

1f2f

kf 2Nf0f

1Nf

)(xf

2

y

x0

ˆ( )f

Fourier transforms

nc

2 3023 ......

Fourier series

Discrete Fourier transform (DFT) deals with sampled values fk rather than

with functions.

Seoul NationalUniv.

58

Fourier Series


/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

nc

2 3023 ......

)(xf

x0

2 22

2f

T p p

Seoul NationalUniv.

59

Fourier Series Fourier Transform


1 ˆ( ) ( )2

i xf x f e d

1ˆ ( ) ( )2

i xf f x e dx

Fourier Transform

/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

)(xf

x0

2 22

2f

T p p

ˆ( )f

Seoul NationalUniv.

60

Fourier Series Fourier Transform Discrete Fourier Transform


1 ˆ( ) ( )2

i xf x f e d

1ˆ ( ) ( )2

i xf f x e dx

Fourier Transform

/( ) in x p

n

n

f x c e

/1( )

2

pin x p

np

c f x e dxp

nc

2 3023 ......

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

( ) ( )k kf x q x

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Discrete Fourier Transform

2 22

2f

T p p

1f2f

kf 2Nf0f

1Nf

2

y

x0

)(xf

x0

Seoul NationalUniv.

61


Let f (x) be periodic, for simplicity of

period 2π. We assume that N

measurement are taken over the interval

0≤x≤2π at regularly spaced points

2(14) where 0,1, , 1.kx k k N

N

We also say that f (x) is being sampled at these points. We now want to

determine a complex trigonometric polynomial (복소 삼각함수다항식).

1

0

( ) ( ) (15)N

inx

n

n

f x q x c e

1f2f

kf 2Nf0f

1Nf

)(xf

2

y

0kx1x

2x2Nx 1Nx

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

cos sinixe x i x (Euler formula)

Seoul NationalUniv.

62


Eq. (15) interpolates f (x) at the nodes in Eq. (14), that is,

q(xk) = f (xk), written out, with fk denoting f (xk),

)15()(1

0

N

n

inx

necxq

1,,1,0 where)16()()(1

0

NkecxqxffN

n

inx

nkkkk

Hence, we must determine the coefficients c0, …, cN-1 such

that Eq. (16) holds. (This is the Discrete Fourier Transform)

1f2f

kf 2Nf0f

1Nf

)(xf

2

y

x0

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

where 0,1, , 1.k N

2(14)kx k

N

Seoul NationalUniv.

63

We multiply Eq. (16) by :


1,,1,0 where

)16()()(1

0

Nk

ecxqxffN

n

inx

nkkkk

1f2f

kf 2Nf0f

1Nf

)(xf

2

y

x0

Sum over k from 0 to N-1:

kimxe

1

0

N

n

imxinx

n

imx

kkkk eecef

1

0

1

0

1

0

N

k

N

n

imxinx

n

N

k

imx

kkkk eecef

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Then we interchange the order of the two

summations and insert xk from Eq. (14).

1

0

1

0

)(1

0

N

n

N

k

xmni

n

N

k

imx

kkk ecef

)17(1

0

1

0

/2)(

N

n

N

k

Nkmni

n ec

2(14)kx k

N

Seoul NationalUniv.

64


1 1 1( )2 /

0 0 0

(17)k

N N Nimx i n m k N

k n

k n k

f e c e

kNmniNkmni ee /2)(/2)( Nmnik err /2)( where

1

0

1

0

/2)(N

k

kN

k

Nkmni re

,1 ,For 0 ermn① NrN

k

N

k

k

1

0

1

0

1then

② ,1 ,integer For rmnr

rr

NN

k

k

1

11

0

101

)(2sin)(2cos

2)(

mnkimnk

er kmniN

01

11

r

(Geometrical Series)

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Seoul NationalUniv.

65


Nmnier /2)( ,1

0

1

0

/2)(

N

k

kN

k

Nkmni re

)17(1

0

1

0

/2)(1

0

N

n

N

k

Nkmni

n

N

k

imx

k ecef k

NrmnN

k

k

1

0

,For ①

0 ,integer For 1

0

N

k

krmn ②

Ncec m

N

n

N

k

Nkmni

n

1

0

1

0

/2)(

from ① and ②

Ncef m

N

k

imx

kk

1

0

Writing n for m and dividing by N,

*)18(1 1

0

N

k

inx

knkef

Nc

1,,1,0),( Nnxff kk

It is practical to drop the factor 1/N from

cn and define the discrete Fourier

transform of the given signal.

)18(ˆ1

0

N

k

inx

knnkefNcf

1,,1,0),( Nnxff kk

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Seoul NationalUniv.

66


1

0

ˆN

k

nk

kn wff

1,,1,0),( Nnxff kk

)18(ˆ1

0

N

k

inx

knnkefNcf

2

(19)i

NNw w e

Define:

then kinxe

2(14)kx k

N

2 2

where , 0,1, , 1i

in k nknkN Ne e w n k N

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Seoul NationalUniv.

67


1

0

ˆN

k

nk

kn wff

1,,1,0, where,

22

Nknweeeeww nknk

N

i

inx

nkN

i

Nk

)1()1(

1

2)1(

2

1)1(

1

0)1(

01

)1(2

1

22

2

12

1

02

02

)1(1

1

21

2

11

1

01

01

)1(0

1

20

2

10

1

00

00

ˆ

ˆ

ˆ

ˆ

NN

N

NNN

N

N

N

N

N

N

N

wfwfwfwff

wfwfwfwff

wfwfwfwff

wfwfwfwff

In vector notation T

10 ][ Nff f

T

10 ]ˆˆ[ˆ Nff f

ˆNf = F f

or nk nke wNF

1

0

( ) ( )N

inx

n

n

f x q x c e

1

0

1k

Ninx

n k

k

c f eN

1

0

ˆ k

Ninx

n n k

k

f Nc f e

Seoul NationalUniv.

68

0 0 0 1 0 2 0 3

1 0 11 1 2 1 3

4 2 0 2 1 2 2 2 3

3 0 31 3 2 3 3

f̂ F f f fnk

w w w w

w w w ww

w w w w

w w w w


Let N = 4 measurements (sample values) be given. Let the

sample values be, say . Find the discrete Fourier

Transform of f.

2 2 cos sin2 2

i N i

Nw w e e i i

nknk iw )(

T9 4 1 0f

0 0 0 0

0 1 2 3

4 0 2 4 6

0 3 6 9

f̂ F f f

w w w w

w w w w

w w w w

w w w w

9

4

1

0

11

1111

11

1111

ii

ii)20(

84

6

84

14

i

i

Ex 4 Discrete Fourier Transform (DFT). Sample N = 4 of Values fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

k

n

Sol)

Seoul NationalUniv.

69

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

11.9 DFT: Fast Fourier Transform (FFT)

pN

ip

N

i

eei

22

)2sin2(cos

2( )

ip N

p N New

ipN

i

e

2

2

p

N

i

i ee

2

2

pw

Property 1)

3 3 9 1 8 1 2 1Nw w w w w Ex. N = 4)

∴ Need the values only from to .1 Nw 1N

Nw

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)

n

k Ex 4 Discrete Fourier Transform (DFT). Sample N=4 of Values

Seoul NationalUniv.

70

1 0 11 1 2 1 3

1 0 1 2 3f̂ w f w f w f w f

1 0 1 2 11 1 3

1 0 2 1 3f̂ w f w f w f w f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i


Property 2)


n

k

1 0 1 2 11 1 0 1 2

1 0 2 1 3f̂ w f w f w w f w f

even signal odd signal

same coefficients!

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

71

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1ˆ 1 0 ( ) 1 ( 1) 4 9f i i

1ˆ 1 0 ( 1) 4 ( ) 1 9f i i

Property 2) 1 0 11 1 2 1 3

1 0 1 2 3f̂ w f w f w f w f

1 0 1 2 11 1 3

1 0 2 1 3f̂ w f w f w f w f

1 0 1 2 11 1 0 1 2

1 0 2 1 3f̂ w f w f w w f w f


1ˆ 1 0 ( 1) 4 ( ) 1 1 ( 1) 9f i

same coefficients!same coefficients!

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

72

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i


fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N


n

k

Property 2) 1ˆ 1 0 ( ) 1 ( 1) 4 9f i i

1ˆ 1 0 ( 1) 4 ( ) 1 9f i i

1 0 11 1 2 1 3

1 0 1 2 3f̂ w f w f w f w f

1 0 1 2 11 1 3

1 0 2 1 3f̂ w f w f w f w f

1 0 1 2 11 1 0 1 2

1 0 2 1 3f̂ w f w f w w f w f


1ˆ 1 0 ( 1) 4 ( ) 1 1 ( 1) 9f i

same coefficients!same coefficients!

0 2 4 ( 2) 1 2 4 ( 2)

0 2 4 2 1 3 5 1ˆ n n n n N n n n n N

n N Nf f w f w f w f w w f f w f w f w

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

In general,


same coefficients

Ex 4 N=4

Seoul NationalUniv.

73

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i


fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N


n

k

Property 2) 1 1

2 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M


same coefficients

Number of signal: N

Number of signal: N/2 Number of signal: N/2

Ex 4 N=4

Seoul NationalUniv.

74

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆM M

kn n kn

n N k N N k

k k

f w f w w f


pN 2

12 p 12 p

1

0

ˆN

nk

n k

k

f f w

22 p 22 p 22 p 22 p

2 2 2 2 2 2 2 2

similarly

until signal length is 2

For N=2p this breakdown can be repeated p-1 times in order to

finally arrive at N/2 problems of size 2 each, so that the number

of multiplications is reduced as indicated.

using these properties

,p N pw w

FFT is a computational method for the DFT that needs only

Instead of . It makes the DFT a practical tool for large N*

2( ) logNO N2( )O N

*Kreyszig E., Advanced Engineering Mathematics, 9th edition, Wiley, 2006, p526

(N by N)

matrix

(2 by 2) x p

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

75

1 12 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 2 2 2 2

2 4 2 4 4 (2 1)

0 0

1 12 3 3 2 3

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

-i

i

1

-1

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i


Multiplication can be done in a very computationally efficient manner :Fast Fourier Transform (FFT)

n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1

4w

0

4w

2

4w

3

4w

0

4w

1

4w

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

2 2 2 20 1 2 3

0 1 2 34 4 4 44 4 4 41, , 1,

i i i i

w e w e i w e w e i

Seoul NationalUniv.

76

-i

i

1

-1

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 2 2 2 2

2 4 2 4 4 (2 1)

0 0

1 12 3 3 2 3

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

1

4w

0

4w

2

4w

3

4w

0

4w

1

4w

1x0+1x4

12 0

4 2

0

k

k

k

w f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

2 2 2 20 1 2 3

0 1 2 34 4 4 44 4 4 41, , 1,

i i i i

w e w e i w e w e i

Seoul NationalUniv.

77

-i

i

1

-1

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 2 2 2 2

2 4 2 4 4 (2 1)

0 0

1 12 3 3 2 3

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

1

4w

0

4w

2

4w

3

4w

0

4w

1

4w

1x0+1x4

12 2

4 2

0

k

k

k

w f

1x0+1x4

12 0

4 2

0

k

k

k

w f

same

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

2 2 2 20 1 2 3

0 1 2 34 4 4 44 4 4 41, , 1,

i i i i

w e w e i w e w e i

Seoul NationalUniv.

78

-i

i

1

-1

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 2 2 2 2

2 4 2 4 4 (2 1)

0 0

1 12 3 3 2 3

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

1

4w

0

4w

2

4w

3

4w

0

4w

1

4w sa

me

1 12 0 2 0

4 2 4 (2 1)

0 0

1 12 1 2 1

4 2 4 (2 1)

0 0

1 12 2 2 2

4 2 4 (2 1)

0 0

1 12 3 2 3

4 2 4 (2 1)

0 0

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

w f w f

w f w f

w f w f

w f w f

sa

me

sam

esa

me

In similar way,

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

2 2 2 20 1 2 3

0 1 2 34 4 4 44 4 4 41, , 1,

i i i i

w e w e i w e w e i

Seoul NationalUniv.

79

1 12 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 2 2 2 2

2 4 2 4 4 (2 1)

0 0

1 12 3 3 2 3

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

So, 1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

80

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 2f̂

0 2f̂

1f̂

0f̂

1 12 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

81

1 12 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 2f̂

0 2f̂

1f̂

0f̂ ev,0f̂ od,0f̂

ev,1f̂ od,1f̂

ev,0f̂ od,0f̂

ev,1f̂ od,1f̂

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

82

1 12 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i



n

k

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 2f̂

0 2f̂

1f̂

0f̂

ev, od,ˆ ˆ ˆn

n n N nf f w f ev,0f̂ od,0f̂

ev,1f̂ od,1f̂

ev,0f̂ od,0f̂

ev,1f̂ od,1f̂ev, od,

ˆ ˆ ˆn

n nM N nf f w f

In general,

10

2

Nn

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

83


1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

1 2f̂

0 2f̂

1f̂

0f̂

ev, od,ˆ ˆ ˆn


ev,1f̂ od,1f̂

ev,0f̂ od,0f̂


ˆ ˆ ˆn

n nM N nf f w f

In general,

Proof

Of 22a)

2 2 22 2

2 2 ,i i i

N M MN Mw e e e w

nk

M

nk

N ww 2

1

ev, ev,

0

1

od, od,n

0

ˆ

ˆ

Mkn

M k n

k

Mn kn n

N M k N

k

w f f

w w f w f

1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

ev, od,

0 0

ˆ (23a)M M

kn n kn

n M k N M k

k k

f w f w w f

(22a)

(22b)

ev, od,ˆ ˆ ˆ (22 )n

n n N nf f w f a 1

02

Nn

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

84


1 12 2

2 (2 1)

0 0

ˆ ( )2

M Mkn n kn

n N k N N k

k k

Nf w f w w f M

1 1

2 0 0 2 0

0 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

1 4 2 4 4 (2 1)

0 0

1 12 0 0 2 0

2 4 2 4 4 (2 1)

0 0

1 12 1 1 2 1

3 4 2 4 4 (2 1)

0 0

ˆ

ˆ

ˆ

ˆ

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

k k

f w f w w f

f w f w w f

f w f w w f

f w f w w f

1 2f̂

0 2f̂

1f̂

0f̂

ev, od,ˆ ˆ ˆn


ev,1f̂ od,1f̂

ev,0f̂ od,0f̂


ˆ ˆ ˆn

n nM N nf f w f

In general,

Proof

Of 22b)

1

ev, ev,

0

1

odd, od,n

0

ˆ

ˆ

Mkn

M k n

k

Mn kn n

N M k N

k

w f f

w w f w f

)a23(ˆ1

0

odd,

1

0

ev,

M

k

k

kn

M

n

N

M

k

k

kn

Mn fwwfwf

(22a)

(22b)

)b23(ˆ1

0

odd,

1

0

ev,

M

k

k

kn

M

n

N

M

k

k

kn

MMn fwwfwf

2 / 2 /2

cos sin 1

M iM N i i

Nw e e e

i

2 / 2

cos 2 sin 2 1

M iM M i

Mw e e

i

In (23a) we have n+M instead of n1 1

ev, od,

0 0

ˆM M

kn kM M n kn kM

n M M M k N N M M k

k k

f w w f w w w w f

)b22(ˆˆˆod,ev, n

n

NnMn fwff

10

2

Nn

2

NM

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

85


1

0

ˆN

nk

n k

k

f f w

1 1

ev, od,

0 0

ˆM M

kn n kn

n M k N M k

k k

f w f w w f

Fast Fourier Transform (FFT)

ev evˆ

Mf F f od odˆ

Mf F f

Divide even and odd

ev, od,ˆ ˆ ˆn

n n N nf f w f (22a)

ev, od,ˆ ˆ ˆn

n M n N nf f w f (22b)

find even and odd solution

find original solution

2

NM

10

2

Nn

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

86


T9 4 1 0f T

0 4ev f

T

1 9od fevev

ˆ fFf Mod od

ˆ, Mf F f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

1

0

ˆN

nk

n k

k

f f w

1 1

ev, od,

0 0

ˆM M

kn n kn

n M k N M k

k k

f w f w w f

ev evˆ

Mf F f od odˆ

Mf F f

Divide even and odd

ev, od,ˆ ˆ ˆn


ev, od,ˆ ˆ ˆn




2

NM

10

2

Nn

0 0 0 1

2 2

2 1 0 11

2 2

1 1

1 1M

w w

w w

F F

2 20 1

0 12 22 21, cos sin 1

i iiw e w e e i

Divide even and odd


,0

ev 2 ev

,1

ˆ 1 1 0 4f̂ F f

ˆ 1 1 4 4

ev

ev

f

f

,0

od 2 od

,1

ˆ 1 1 1 10f̂ F f

ˆ 1 1 9 8

od

od

f

f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

87


fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

T9 4 1 0fev

4f̂

4

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

1

0

ˆN

nk

n k

k

f f w

1 1

ev, od,

0 0

ˆM M

kn n kn

n M k N M k

k k

f w f w w f

ev evˆ

Mf F f od odˆ

Mf F f

Divide even and odd

ev, od,ˆ ˆ ˆn


ev, od,ˆ ˆ ˆn




2

NM

10

2

Nn

od

10f̂

8

0

0 ev,0 od,0ˆ ˆ ˆ

Nf f w f

0n

0

0 2 2 ev,0 od,0ˆ ˆ ˆ ˆ

Nf f f w f

1

1 ev,1 od,1ˆ ˆ ˆ

Nf f w f

1n

1

1 2 3 ev,1 od,1ˆ ˆ ˆ ˆ

Nf f f w f


Ex 4 N=4

Seoul NationalUniv.

88


T9 4 1 0f

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

1

0

ˆN

nk

n k

k

f f w

1 1

ev, od,

0 0

ˆM M

kn n kn

n M k N M k

k k

f w f w w f

ev evˆ

Mf F f od odˆ

Mf F f

Divide even and odd

ev, od,ˆ ˆ ˆn


ev, od,ˆ ˆ ˆn




2

NM

10

2

Nn

0

0 ev,0 od,0ˆ ˆ ˆ

Nf f w f

0

2 ev,0 od,0ˆ ˆ ˆ

Nf f w f

1

1 ev,1 od,1ˆ ˆ ˆ

Nf f w f

1

3 ev,1 od,1ˆ ˆ ˆ

Nf f w f

4 1 10

4 ( ) ( 8) 4 8i i

4 1 10

4 ( ) ( 8) 4 8i i

14

6

2 20 1

0 14 4 24 41, cos sin

2 2

i i i

w e w e e i i

ev

4f̂

4

ev,0f̂

ev,1f̂od

10f̂

8

od,0f̂

od,1f̂


fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

89


0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

In summary

T9 4 1 0f

0

0 ev,0 od,0ˆ ˆ ˆ

Nf f w f

0

2 ev,0 od,0ˆ ˆ ˆ

Nf f w f

1

1 ev,1 od,1ˆ ˆ ˆ

Nf f w f

1

3 ev,1 od,1ˆ ˆ ˆ

Nf f w f

4 1 10

4 ( ) ( 8) 4 8i i

4 1 10

4 ( ) ( 8) 4 8i i

14

6

1

0

ˆN

nk

n k

k

f f w

1 1

ev, od,

0 0

ˆM M

kn n kn

n M k N M k

k k

f w f w w f

ev evˆ

Mf F f od odˆ

Mf F f

Divide even and odd

ev, od,ˆ ˆ ˆn


ev, od,ˆ ˆ ˆn




2

NM

10

2

Nn

Divide even and odd



,0

ev 2 ev

,1

ˆ 1 1 0 1 0 1 1 4f̂ F f

ˆ 1 1 4 1 0 ( 1) 4 4

ev

ev

f

f

,0

od 2 od

,1

ˆ 1 1 1 1 1 1 9 10f̂ F f

ˆ 1 1 9 1 1 ( 1) 9 8

od

od

f

f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

90

,0

ev 2 ev

,1

ˆ 1 1 0 1 0 1 1 4f̂ F f

ˆ 1 1 4 1 0 ( 1) 4 4

ev

ev

f

f

,0

od 2 od

,1

ˆ 1 1 1 1 1 1 9 10f̂ F f

ˆ 1 1 9 1 1 ( 1) 9 8

od

od

f

f


0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

Comparison of calculation time

T9 4 1 0f

0

0 ev,0 od,0ˆ ˆ ˆ

Nf f w f

0

2 ev,0 od,0ˆ ˆ ˆ

Nf f w f

1

1 ev,1 od,1ˆ ˆ ˆ

Nf f w f

1

3 ev,1 od,1ˆ ˆ ˆ

Nf f w f

4 1 10

4 ( ) ( 8) 4 8i i

4 1 10

4 ( ) ( 8) 4 8i i

14

6

Divide even and odd



4 x 4 = 16

3 x 4 = 12

No. of multiplication

No. of addition

(2x2)x2+4x1= 12

(2)x2+4 = 8

FFTOrdinary matrix

calculation

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

91

④

,0

ev 2 ev

,1

ˆ 1 1 0 1 0 1 1 4f̂ F f

ˆ 1 1 4 1 0 ( 1) 4 4

ev

ev

f

f

,0

od 2 od

,1

ˆ 1 1 1 1 1 1 9 10f̂ F f

ˆ 1 1 9 1 1 ( 1) 9 8

od

od

f

f


0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

T9 4 1 0f

0

0 ev,0 od,0ˆ ˆ ˆ

Nf f w f

0

2 ev,0 od,0ˆ ˆ ˆ

Nf f w f

1

1 ev,1 od,1ˆ ˆ ˆ

Nf f w f

1

3 ev,1 od,1ˆ ˆ ˆ

Nf f w f

4 1 10

4 ( ) ( 8) 4 8i i

4 1 10

4 ( ) ( 8) 4 8i i

14

6

Divide even and odd



①

②

②

③

③

0

1

4

9

0

4

1

9

① ②

4

4

10

8

1

i

③

4

4

10

i8

14

i84

6

i84

④ ⑤

⑥

⑤ ⑥

Illustrated graphically

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Ex 4 N=4

Seoul NationalUniv.

92


3

2

1

0

f

f

f

f

1010

1010

0101

0101

1,

0,

1,

0,

ˆ

ˆ

ˆ

ˆ

od

od

ev

ev

f

f

f

f 004

214

024

234

1 0 0

1 0 0( )

0 1 0

0 1 0

fw

fwa

fw

fw

2 3

0 1 2 32 24 4 4 4, 1, , 1,

i p i ip iNNw e w w e i w e w e i

3

2

1

0

2

4

0

4

2

4

0

4

3

4

2

4

1

4

0

4

3

2

1

0

010

010

001

001

010

001

010

001

ˆ

ˆ

ˆ

ˆ

f

f

f

f

w

w

w

w

w

w

w

w

f

f

f

f

Example FFT in the matrix form

T T

0 1 2 3 0 1 4 9f f f f f

- divide even and odd

2

0

1,

0,

11

11

ˆ

ˆˆ

f

f

f

f

ev

ev

evf

3

1

1,

0,

11

11

ˆ

ˆˆ

f

f

f

f

od

od

odf

- find even and odd solution

0 ev,0 od,0ˆ ˆ ˆ1f f f

2 ev,0 od,0ˆ ˆ ˆ( 1)f f f

1 ev,1 od,1ˆ ˆ ˆ( )f f i f

3 ev,1 od,1ˆ ˆ ˆf f i f

- find original solution

4N

0 0 0 1 0 2 0 3 0 0 0 0

1 2 31 0 11 1 2 1 3 0 1 2 3

2 0 12 0 2 1 2 2 2 3 0 2 4 6

3 1 23 0 31 3 2 3 3 0 3 6 9

1 1 1 1

1

1

1

w w w w w w w w

w w ww w w w w w w w



1,

0,

1,

0,

ˆ

ˆ

ˆ

ˆ

od

od

ev

ev

f

f

f

f

3

2

1

0

ˆ

ˆ

ˆ

ˆ

f

f

f

f 1 0 1 0

0 1 0

1 0 1 0

0 1 0

i

i

0

4

1

4

2

4

3

4

1 0 0

0 1 0

1 0 0

0 1 0

w

w

w

w

,0

,1

,0

,1

ˆ

ˆ( )

ˆ

ˆ

ev

ev

od

od

f

fb

f

f

From (a) and (b)

Seoul NationalUniv.

93


2 3

0 1 2 32 24 4 4 4, 1, , 1,

i p i ip iNNw e w w e i w e w e i

T T

0 1 2 3 0 1 4 9f f f f f 4N

0 0 0 1 0 2 0 3 0 0 0 0

1 2 31 0 11 1 2 1 3 0 1 2 3

2 0 12 0 2 1 2 2 2 3 0 2 4 6

3 1 23 0 31 3 2 3 3 0 3 6 9

1 1 1 1

1

1

1

w w w w w w w w




0 00 04 4

1 21 14 4

2 024 42

3 234 4

3

ˆ1 0 0 1 0 0

ˆ 0 1 0 1 0 0

ˆ 1 0 0 0 1 0

0 1 0 0 1 0ˆ

f fw w

f fw w

fw wf

fw wf

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

Example FFT in the matrix form1 0 1 0 1 0 1 0 1 1 1 1

0 1 0 1 0 1 0 1 1

1 0 1 0 0 1 0 1 1 1 1 1

0 1 0 0 1 0 1 1 1

i i i

i i i

same!

Seoul NationalUniv.

94


0 00 04 4

1 21 14 4

2 024 42

3 234 4

3

ˆ1 0 0 1 0 0

ˆ 0 1 0 1 0 0

ˆ 1 0 0 0 1 0

0 1 0 0 1 0ˆ

f fw w

f fw w

fw wf

fw wf

Example FFT in the matrix form

In the signal flow graph*

*Brigham E.O., The Fast Fourier Transform, Prentice-Hall, 1974, p153

0f

2f

1f

3f

0

4w0f̂

1f̂

2f̂

3f̂

2

4w

0

4w

2

4w

0

4w

1

4w

2

4w

3

4w

3

2

1

0

2

4

0

4

2

4

0

4

1,

0,

1,

0,

010

010

001

001

ˆ

ˆ

ˆ

ˆ

f

f

f

f

w

w

w

w

f

f

f

f

od

od

ev

ev

1,

0,

1,

0,

3

4

2

4

1

4

0

4

3

2

1

0

ˆ

ˆ

ˆ

ˆ

010

001

010

001

ˆ

ˆ

ˆ

ˆ

od

od

ev

ev

f

f

f

f

w

w

w

w

f

f

f

f

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

95


Example 4N

0 0 0 1 0 2 0 30 0

1 0 11 1 2 1 31 1

4 2 0 2 1 2 2 2 322

3 0 31 3 2 3 33

3

ˆ0 1 1 1 1 0 14

ˆ 1 1 1 1 4F

ˆ 4 1 1 1 1 4

9 1 1 9ˆ

f f w w w w

f f i iw w w w

f w w w wf

f i iw w w wf

8

6

4 8

i

i

Brigham E.O., The Fast Fourier Transform, Prentice-Hall, 1974, p153

2

4w

In the signal flow graph*

0f

2f

1f

3f

0

4w0f̂

1f̂

2f̂

3f̂

2

4w

0

4w

0

4w

1

4w

2

4w

3

4w

0

1

4

9

0

4

1

9

① ②

4

4

10

8

1

i

③

4

4

10

i8

14

i84

6

i84

④ ⑤

⑥

Illustrated graphically

FFT 1 12 2

2 (2 1)

0 0

ˆM M

kn n kn

n N k N N k

k k

f w f w w f


1

0

ˆN

nk

n k

k

f f w

(until signal length is 2)

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

96


Example N=8

0

11 1 2 1 3 1 4 1 5 1 6 1 71

2 1 2 2 2 3 2 4 2 5 2 6 2 72

3 1 3 2 3 3 3 4 3 5 3 6 3 7

3

4 1 4 2 4 3 4 4 4 5 4 6 4 7

45 1 5 2 5 3 5 4 5 5

5

6

7

ˆ1 1 1 1 1 1 1 1

ˆ1

ˆ1

ˆ 1

ˆ 1

1ˆ

ˆ

ˆ

f

fw w w w w w w

f w w w w w w w

f w w w w w w w

w w w w w w wf

w w w w wf

f

f

0

1 2 3 4 5 6 7

1

2 4 6 8 10 12 14

2

3 6 9 12 15 18 21

3

4 8

4

5 6 5 7

5

6 1 6 2 6 3 6 4 6 5 6 6 6 7

6

7 1 7 2 7 3 7 4 7 5 7 6 7 7

7

1 1 1 1 1 1 1 1

1

1

1

1

1

1

f

f w w w w w w w

f w w w w w w w

f w w w w w w w

f w w

fw w

fw w w w w w w

fw w w w w w w

0

1

2

3

12 16 20 24 28

4

5 10 15 20 25 30 35

5

6 12 18 24 30 36 42

6

7 14 21 28 35 42 49

7

1

1

1

f

f

f

f

fw w w w w

fw w w w w w w

fw w w w w w w

fw w w w w w w

0

0

1 2 3 4 5 6 711

2 4 6 2 4 62 2

3 6 1 4 7 2 5

3 3

4 4 4 4

445 2 7 4 1 6 3

56 4 2 6 4 2

6 7 6 5 4 3 2 1

7

ˆ1 1 1 1 1 1 1 1

ˆ1

ˆ1 1

ˆ 1

ˆ 1 1 1 1

1ˆ

1 1ˆ1

ˆ

ff

ffw w w w w w w

f fw w w w w w

f fw w w w w w w

fw w w wf

fw w w w w w wfw w w w w w

fw w w w w w w

f

7

5

6

7

1 1 1 1 1 1 1 1

1 1 1 11 1

2 2 2 2

1 1 1 1

1 1 11 1

2 2 2

1 1 1 1 1 1 1 1

1 1 1 11 1

2 2 2 2

1 1 1 1

1 1 1 11 1

2 2 2 2

i i i ii i

i i i i

i i ii w i

i i i ii i

f

i i i if

i i i ii i

4

3

2

0

1

2

3

1

T

4 3 2 0 1 2 3 1 f

20

0 88 1

i

w e

21

1 88

1cos sin

4 4 2

i iw e i

22

2 88 cos sin

2 2

i

w e i i

23

3 88

3 3 1cos sin

4 4 2

i iw e i

24

4 88 cos sin 1

i

w e i

25

5 88

5 5 1cos sin

4 4 2

i iw e i

26

6 88

6 6cos sin

4 4

i

w e i i

27

7 88

7 7 1cos sin

4 4 2

i iw e i

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

97


Example N=8 T

4 3 2 0 1 2 3 1 f

Find: by using FFT f̂

4

3

2

0

1

2

3

1

4

2

1

3

3

0

2

1

4

1

2

3

3

2

0

1

②

0 0 0 1

0 2 0 2 1

1 0 11

1 2 0 2 1

ˆ

ˆ

f w f w f

f w f w f

4 ( 1) 3

4 ( 1) 5

2 3 5

2 3 1

3 ( 2) 5

3 ( 2) 1

0 1 1

0 1 1

③

0

0 ev,0 4 od,0ˆ ˆ ˆf f w f

1


0


1


0

4w1

4w

0

4w1

4w

① Even/Odd

1

i

1

i

3

5

5

i

5

1

1

i

3 5 8

5 i

3 5 2

5 i

5 1 4

1 i

5 1 6

1 i

until signal length is 2

2 20 1

0 12 22 21, 1

i i

w e w e

2 2

0 10 14 44 4, 1,

i i

w e w e i

n

n

Nnn fwff od,ev,ˆˆˆ n

n

NnMn fwff od,ev,ˆˆˆ

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

98


Example N=8 T

4 3 2 0 1 2 3 1 f


8

5 i

2

5 i

4

1 i

6

1 i

④

0


1


2


3


0


1


2


3


0

8w1

8w2

8w3

8w

1

(1 ) / 2ii

( 1 ) / 2i

8

5 i

2

5 i

4

2i

6i

2i

8 ( 4) 4

5 2 5 (1 2)i i i

2 6i

5 2 5 (1 2)i i i

8 ( 4) 12

5 2 5 (1 2)i i i

2 6i

5 2 5 (1 2)i i i

n

n

Nnn fwff od,ev,ˆˆˆ n

n

NnMn fwff od,ev,ˆˆˆ

20

0 88 1

i

w e

21

1 88

1cos sin

4 4 2

i iw e i

22

2 88 cos sin

2 2

i

w e i i

23

3 88

3 3 1cos sin

4 4 2

i iw e i

fFf Nˆ

2

2

,

k

i

NN

ink

inx nkNnk

w w e

e e e w

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N

Seoul NationalUniv.

99


Example N=8fFf Nˆ

2 i

NNw w e

1

0

ˆN

nk

n k

k

f f w

where , 0,1, , 1n k N T

4 3 2 0 1 2 3 1 f


Even/Odd until signal length is 2

8

5 i

2

5 i

4

1 i

6

1 i

8

5 i

2

5 i

4

2i

6i

2i

4

5 (1 2)i

2 6i

5 (1 2)i

12

5 (1 2)i

2 6i

5 (1 2)i

4

3

2

0

1

2

3

1

4

2

1

3

3

0

2

1

4

1

2

3

3

2

0

1

3

5

5

1

5

1

1

i

1

i

1

i

3

5

5

i

5

1

1i

1

(1 ) / 2ii

( 1 ) / 2i

② ③ ④

0 0

0 2 0 2 1

0 1

1 2 0 2 1

ˆ

ˆ

f w f w f

f w f w f

0


1


0


1


0


1


2


3


0


1


2


3


①

Documents

Ch. 11 Fourier Analysisocw.snu.ac.kr/sites/default/files/NOTE/Ch11 Pt II.pdf · 2019. 3. 15. · Seoul National Univ. 2 11.6 Orthogonal Series. Generalized Fourier Series Standard