FFT(Fast Fourier Transform)
20
FFT(Fast Fourier Transform)
-
Upload
claudia-morton -
Category
Documents
-
view
22 -
download
0
description
FFT(Fast Fourier Transform). FFT. Coefficient representation:. How to evaluate A(x 0 )?. Horner’s rule: Point-value representation:. Thm1. Pf:. n-point interpolation: Lagrange’s formula:. Cof. rep. Point-value rep. Thm2. Lemma 3 (Cancellation Lemma) - PowerPoint PPT Presentation
Transcript of FFT(Fast Fourier Transform)
FFT(Fast Fourier Transform)
p2.
FFT1
0 1 1
10 1 1
2 2 3 2 20 1 2 2 3 2 2
0
( ) .....
( ) .....
( ) ( ) ( )
= c .....
nn
nn
n nn n
j
j k j kk
A x a a x a x
B x b b x b x
C x A x B x
c x c x c x c x
c a b
•Coefficient representation: How to evaluate A(x0)?
p3.
Horner’s rule:
Point-value representation:
0 0 0 1 0 2 0 2 0 1( ) ( ( ..... ( )...))
(n)n nA x a x a x a x a x a
0 0
1 1 n-1 1 k
A point-value representation of a polynomial A(x) of
degree-bound n is a set of n point-value pairs {(x , ),
(x , ),.....(x , )}, where y ( )n k
y
y y A x
0 0 1 1 2n-1 2 1
' ' '0 0 1 1 2n-1 2 1
' ' '0 0 0 1 1 1 2n-1 2 1 2 1
:{(x , ), (x , ),.....(x , )}
:{(x , ), (x , ),.....(x , )}
:{(x , ), (x , ),.....(x , )}
n
n
n n
A y y y
B y y y
C y y y y y y
p4.
Thm1
0 0 1 1 n-1 1
k
For any set {(x , ), (x , ),.....(x , )} of n point-value
pairs, there is a unique poly A(x) of degree n-1, such that
y ( ) for k=0, 1,...., n-1
n
k
y y y
A x
Pf:2 1
0 0 0
2 11 1 1
2 11 1 1
0 0
1 1
1 1
1
1 if X = then
1
y
y
n
n
nn n n
n n
x x x
x x x
x x x
a
aX
a y
p5.
n-point interpolation:Lagrange’s formula:
0 0
1 11
1 1
det( ) ( ) 0 if 's are distinct
Thus, 's can be solved uniquely.
y
y
k jk
n
kj
j
n
a
aX
X x x x
a
a y
1
0
( )
( )( )
jnj k
kk k j
j k
x x
A x yx x
p6.
0 1 1
0 1 1
, ,....,
, ,....,n
n
a a a
b b b
0 02 2
0 02 2
2 1 2 12 2
( ), ( )
( ), ( )
,
( ), ( )
n n
n n
n nn n
A B
A B
A B
02
02
2 12
( )
( )
( )
n
n
nn
C
C
C
0 1 2 2, ,...., nc c c
Evalation
( log )n n
2
Ordinary Mult
( )n
Pointwise Mult
( )n
Interpolation
( log )n n
Point-valuerep.
Cof.rep.
p7.
Thm2The product of 2 polynomials of deg-bound n can
be computed in time (nlog n), with both the input
and output representation in coefficient form
0
( )mod
n 2 ik/n iu
2 i/n
0 1 2 1
Complex roots of unity:
1, e for k=0, 1, ..., n-1 e cos sin
e , the principal n-th root of unity
, , ,..
=?, =
.,
1
=
n
n
nn n
j k j k nn n
n
n
n n
n
n
u i u
1 1nn
p8.
Lemma 3 (Cancellation Lemma)n, k, d: non-negative integers,
Cor. 4 n: even positive integer
dk kdn n
Pf:
2 / 2 /( ) ( )dk i dn dk i n k kdn ne e
/ 22 1n
n
p9.
Lemma 5 (Halving lemma)n: even positive integer
The squares of the n complex n-th roots of unity are n/2
complex (n/2)th roots of unity.
2 2/ 2,
/ 2 2 2 2/ 2
/ 2
( ) where {0}
( )
and have the same squre
k k kn n n
k n k n k kn n n n
k k nn n
k Z
Pf:
Lemma 6 (Summation lemma)n-1
kn
j=0
, {0}, | , ( ) 0jn Z k Z n k Pf:
k nn-1k n nn k k
j=0 n n
( ) 1 ( ) 1( ) 0
1 1
n kj
p10.
DFT-1
0 1 -1
0
-1
0 1 -10
0 1 -1
0 1 n-1
Evaluate ( ) , ,....., ,
Assume n is a power of 2
Let , ,...., , and ( )
, ,...., is the DFT of the coefficient
vector a=<a ,a ,...., a
nj n
j n n nj
nk kj
n k n j nj
n
A x a x at
a a a a y A a
y y y y
,
DFT ( )ny a
p11.
Interpolation at the complex roots of unity:
02 1
1
2 4 2( 1)2
1 2( 1)1
1 1 1 1 y
1 y
y = 1
1
nn n n
nn n n
n nn n n ny
0
1
2
( 1)( 1)1
,
1
, ( )
n nn
kjn n k j n
n
a
a
a
a
y V a V
a V y
p12.
Thm 7 1
,, 0,1,....., 1, ( ) /kjn j k nj k n V n
11 1 1
, ' , , '0
1'
0
1( ' )
0
, ( ) ( ) ( )
1 =
1 = -------(*)
if j=j', then (*)=1, if j
n
n n n n n j j n j k n k jk
nkj kjn n
k
nk j jn
k
V V I V V V V
n
n
1 1
0 0
j', then by lemma 6, (*)=0
-(n-1) < j'-j <n-1, and n
1,
|
(j'-j
)n n
k j k jj k n k j n
k j
a y y an
Pf:
p13.
FFT[0] 2 / 2 1
0 2 4 2
[1] 2 / 2 11 3 5 1
[0] [12 2]
( ) .....
( ) .....
(*) ( ) ( ) ( )
nn
nn
A x a a x a x a x
A x a a x a x a x
A x x A xA x
[0] [1]
0 2
0 1
1 2 1 2
1
1. evaluating ( ) and ( ) at
Thus ev
( ) , ( ) ,....., (
aluating A(x) at , ,....., reduce
)
2. combining the results according to (*)
to
nn n n
nn n n
A x A x
p14.
[0] [0]/ 2
[1] [1]/ 2
[0] 2 [1] 2
[0] [1
/ 2 [0] 2 / 2 [1]/
]/ 2 / 2
[
2
[0] 1]
( )
( )
( ) ( ) ( )
= ( ) ( )
Le
( )
=
t
) (
(
kk n
kk n
k k k kk n n n n
k k
k n k n k nk
kn n
n n n
n
k
n
kk n
y A A
y A
y A
y A A A
A
A
A
y y
[
2
[0] / 2 [1]/ 2 / 2
[0] / 2 [ 0] [1] ]1
)
= ( ) ( )
=
k nn
k k n kn n n
k nk k
kk n kn y
A
y
A
y y
2 /n
[0]0 2 2
[1]1 3 1
[0] [0]
[1]
Recursive-FFT(a)
{ n=length[a]; /* n: power of 2 */
if n=1 the return a;
;
=1
a ( , ,....., );
a ( , ,....., );
y Recursive-FFT(a );
y Recurs
i n
n
n
e
a a a
a a a
[1]
[0] [1]k k k
[0] [1]k+n/2 k k
n
ive-FFT(a );
for k=0 to (n/2 - 1) do
{
y y y ;
y y y ;
= ; }
}
( ) 2 ( / 2) ( )
= (nlog n)
T n T n n
p16.
Thm 8 (Convolution thm)
-12 2 2
: power of 2
, : vectors of length
=DFT (DFT ( ) DFT ( ))n n n
n
a b n
a b a b
0 0 0 1 1 0 0 2 1 1 2 0
k=0
, , ,......
j-th elt: j
k j k
a b a b a b a b a b a b a b
a b
Componentwise product
Efficient FFT implement[0]ky
[1]ky
kn
[0] [1]kk n ky y
[0] [1]kk n ky y
0 1 2 3 4 5 6 7( , , , , , , , )a a a a a a a a
0 2 4 6( , , , )a a a a 1 3 5 7( , , , )a a a a
2 6( , )a a1 5( , )a a 3 7( , )a a0 4( , )a a
3
011
a 7
111
a5
101
a1
001
a6
110
a2
010
a4
100
a0
000
a
s
s-1
s-1 s-1 s-1
:
for s=1 to (lg n) do
for k=0 to n-1 by 2
combine the two 2 element DFT's in
A[k..k+2 1] and A[k+2 ..k+2 1]
i
Idea
s snto one 2 element DFT in A[k..k+2 1]
[0]y
[1]y
s
2 /m
FFT-Base(a)
{ n = length[a];
for s=1 to (lg n) do
{ m=2 ;
;
for k=0 to n-1 by do
{ =1;
for j=0 to (m/2 - 1) do
{ t= [ / 2];
m
i me
A k j m
u=A[k+j];
A[k+j]=u+t;
A[k+j+m/2]=u-t;
; }
} } }m
p19.
s 2 /
Iterative-FFT(a)
{ Bit-Reverse-Copy(a, A);
n=length[a];
for s=1 to (lg n) do
{ m=2 ; ; 1;
for j=0 to (m/2 -1) do
{ for k=j to n-1 by m do
i mm e
{ t = [ / 2];
u = A[k];
A[k]=u+t;
A[k+m/2]=u-t; }
}
}
}
m
A k m
k
Bit-Reverse-Copy(a,A)
{ n=length[a];
for k=0 to n-1 do
eg
rev(011
A[rev(k
)=110, rev(00
)] = a
}
1)=100
0a
1a
2a
4a
3a
5a
6a
7a
02
02
02
02
04
14
04
14
08
18
38
28
0y
1y
2y
4y
3y
5y
6y
7y
FFT circuit
8
In general, depth: (lg n)
size : (nlg n)
n
S=1 S=2 S=3
