ch. 11 fourier analysisocw.snu.ac.kr/sites/default/files/note/ch11 pt ii.pdf · 2019. 3. 15. ·...
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Ch. 11 Fourier AnalysisPart II
서울대학교
조선해양공학과
서유택
2018.10
※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다.
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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
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11.6 Orthogonal Series. Generalized Fourier Series
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
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[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
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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
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11.6 Orthogonal Series. Generalized Fourier Series
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
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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
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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
[Reference] 5.4 Bessel’s Equation. Bessel Functions Jv(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
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
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[Reference] 5.4 Bessel’s Equation. Bessel Functions Jv(x)
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
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11.6 Orthogonal Series. Generalized Fourier Series
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
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11.6 Orthogonal Series. Generalized Fourier Series
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
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11.6 Orthogonal Series. Generalized Fourier Series
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
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11.7 Fourier Integral
Equalizer
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11.7 Fourier Integral
/( ) 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
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11.7 Fourier Integral
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
recall, an ‘integral’ is a ‘limit of a sum’
Fourier Series Fourier Integraln
recall, an ‘integral’ is a ‘limit of a sum’
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
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Ex. 1 Rectangular Wave
Consider the periodic rectangular wave fL(x) of period 2L > 2 given by
11.7 Fourier Integral
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
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11.7 Fourier Integral
Ex. 1 Rectangular Wave
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
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[Reference] Wave Characteristics
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11.7 Fourier Integral
Ex. 1 Rectangular Wave
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
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11.7 Fourier Integral
Ex. 1 Rectangular Wave
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
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11.7 Fourier Integral
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.
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11.7 Fourier Integral
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
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11.7 Fourier Integral
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
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11.7 Fourier Integral
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
cos sinf x A w wx B w wx dw
디리클레
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11.7 Fourier Integral
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
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11.7 Fourier Integral
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
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11.7 Fourier Integral
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
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11.7 Fourier Integral
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
cos sinf x A w wx B w wx dw
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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
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11.8 Fourier Cosine and Sine Transforms
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
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11.8 Fourier Cosine and Sine Transforms
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
2ˆ cosc cf f w f x wxdx
F
0
2ˆ sins sf f w f x wxdx
F
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11.8 Fourier Cosine and Sine Transforms
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
2ˆ cosc cf f w f x wxdx
F
0
2ˆ sins sf f w f x wxdx
F
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11.8 Fourier Cosine and Sine Transforms
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
2ˆ cosc cf f w f x wxdx
F
0
2ˆ sins sf f w f x wxdx
F
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11.8 Fourier Cosine and Sine Transforms
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
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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
cos sinf x A w wx B w wx dw
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)
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11.9 Fourier Transform. Discrete and Fast Fourier Transforms
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
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11.9 Fourier Transform. Discrete and Fast Fourier Transforms
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
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11.9 Fourier Transform. Discrete and Fast Fourier Transforms
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
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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
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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
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11.9 Fourier Transform Physical Interpretation: Spectrum
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
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11.9 Fourier Transform Physical Interpretation: Spectrum
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
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11.9 Fourier Transform Physical Interpretation: Spectrum
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
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11.9 Fourier Transform Physical Interpretation: Spectrum
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
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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)
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ˆ ( )f
1
21
2
)(tf
1
2
[Reference] Application of Fourier Series
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
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[Reference] Application of Fourier Series
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
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[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
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[Reference] Wave Spectrum
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
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44542 )
2(
1exp)
2(
4
1)(
ZZ
STT
HSParameter
Input
Z
t
[Reference] Wave Spectrum
* 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
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[Reference] Wave Spectrum
( )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
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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
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11.9 Fourier Transform
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
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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
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11.9 Fourier Transform – Tables of Transforms
Table III. Fourier Transforms
1
2
iwxf f x e dx
F
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11.9 Fourier Transform
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
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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.
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Fourier Series
11.9 Discrete Fourier Transform (DFT)
/( ) 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
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Fourier Series Fourier Transform
11.9 Discrete Fourier Transform (DFT)
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
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Fourier Series Fourier Transform Discrete Fourier Transform
11.9 Discrete Fourier Transform (DFT)
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
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11.9 Discrete Fourier Transform (DFT)
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)
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11.9 Discrete Fourier Transform (DFT)
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
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We multiply Eq. (16) by :
11.9 Discrete Fourier Transform (DFT)
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
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11.9 Discrete Fourier Transform (DFT)
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
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11.9 Discrete Fourier Transform (DFT)
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
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11.9 Discrete Fourier Transform (DFT)
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
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11.9 Discrete Fourier Transform (DFT)
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
11.9 Discrete Fourier Transform (DFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
Property 2)
Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)
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
even signal odd signal
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
11.9 DFT: Fast Fourier Transform (FFT)
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
Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)
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
even signal odd signal
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,
even signal odd signal
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
11.9 DFT: Fast Fourier Transform (FFT)
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
Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)
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
even signal odd signal
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
11.9 DFT: Fast Fourier Transform (FFT)
Multiplication can be done in a very computationally efficient manner. :Fast Fourier Transform (FFT)
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
even signal odd signal
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
11.9 DFT: Fast Fourier Transform (FFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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
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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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 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
11.9 DFT: Fast Fourier Transform (FFT)
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
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,
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
11.9 DFT: Fast Fourier Transform (FFT)
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
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,
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
11.9 DFT: Fast Fourier Transform (FFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
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
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
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
find even and odd solution
,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
11.9 DFT: Fast Fourier Transform (FFT)
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
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
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
find original solution
Ex 4 N=4
Seoul NationalUniv.
88
11.9 DFT: Fast Fourier Transform (FFT)
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
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
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
find original solution
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
11.9 DFT: Fast Fourier Transform (FFT)
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
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
Divide even and odd
find even and odd solution
find original solution
,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
11.9 DFT: Fast Fourier Transform (FFT)
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
find even and odd solution
find original solution
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
11.9 DFT: Fast Fourier Transform (FFT)
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
find even and odd solution
find original solution
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
11.9 DFT: Fast Fourier Transform (FFT)
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
w w ww 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
11.9 DFT: Fast Fourier Transform (FFT)
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
w w ww w w w w w w w
w w ww w w w w w w w
w w ww 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
11.9 DFT: Fast Fourier Transform (FFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
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
even signal odd signal
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
11.9 DFT: Fast Fourier Transform (FFT)
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
11.9 DFT: Fast Fourier Transform (FFT)
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
1 ev,1 4 od,1ˆ ˆ ˆf f w f
0
2 ev,0 4 od,0ˆ ˆ ˆf f w f
1
3 ev,1 4 od,1ˆ ˆ ˆf f w f
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
11.9 DFT: Fast Fourier Transform (FFT)
Example N=8 T
4 3 2 0 1 2 3 1 f
Find: by using FFT f̂
8
5 i
2
5 i
4
1 i
6
1 i
④
0
0 ev,0 8 od,0ˆ ˆ ˆf f w f
1
1 ev,1 8 od,1ˆ ˆ ˆf f w f
2
2 ev,2 8 od,2ˆ ˆ ˆf f w f
3
3 ev,3 8 od,3ˆ ˆ ˆf f w f
0
4 ev,0 8 od,0ˆ ˆ ˆf f w f
1
5 ev,1 8 od,1ˆ ˆ ˆf f w f
2
6 ev,2 8 od,2ˆ ˆ ˆf f w f
3
7 ev,3 8 od,3ˆ ˆ ˆf f w f
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
11.9 DFT: Fast Fourier Transform (FFT)
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
Find: by using FFT f̂
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
0 ev,0 4 od,0ˆ ˆ ˆf f w f
1
1 ev,1 4 od,1ˆ ˆ ˆf f w f
0
2 ev,0 4 od,0ˆ ˆ ˆf f w f
1
3 ev,1 4 od,1ˆ ˆ ˆf f w f
0
0 ev,0 8 od,0ˆ ˆ ˆf f w f
1
1 ev,1 8 od,1ˆ ˆ ˆf f w f
2
2 ev,2 8 od,2ˆ ˆ ˆf f w f
3
3 ev,3 8 od,3ˆ ˆ ˆf f w f
0
4 ev,0 8 od,0ˆ ˆ ˆf f w f
1
5 ev,1 8 od,1ˆ ˆ ˆf f w f
2
6 ev,2 8 od,2ˆ ˆ ˆf f w f
3
7 ev,3 8 od,3ˆ ˆ ˆf f w f
①