1

7. Discrete Fourier Transform

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⋅

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−−−−

−

−

− 1

2

1

0

)1)(1()1(21

)1(242

12

1

2

1

0

1

11

1111

1

nnnnn

n

n

n v

vvv

n

c

ccc

M

L

MMMM

L

L

L

M

ωωω

ωωωωωω

)2exp()( niconj πωω −==

1,...,1,0,1 1

0−== ∑

−

=

nkvn

c jkn

jjk ω

c = DFT ( v ). Vector c is the Discrete Fourier Transform of vector v.

2

Discrete Fourier Transform

1,...,1,0,1 1

0−== ∑

−

=

nkvn

c jkn

jjk ω

DFT is nothing else than matrix times vector or n inner products.

Therefore, costs sequentially: O(n2)in parallel: n*log(n) processors, log(n) time steps by

n fan-in processes.

DFT is very important in many applications.

Therefore, fast algorithms have been developed by divide-and-conquer

FFT = Fast Fourier Transform

3

Odd – even Partitioning

∑∑

∑∑

∑

−

=+

−

=

−

=+

−

=

−

=

⎟⎠⎞

⎜⎝⎛⋅⋅+⎟

⎠⎞

⎜⎝⎛⋅=

⎟⎠⎞

⎜⎝⎛ +

⋅+⎟⎠⎞

⎜⎝⎛⋅=

=⎟⎠⎞

⎜⎝⎛⋅=

1

012

1

02

12/

012

12/

02

1

0

2exp2exp

)12(2exp22exp

2exp

m

kk

jm

kk

n

kk

n

kk

n

kkj

mijkc

mijkc

nkijc

nkijc

nijkcv

πωπ

ππ

π

aj aj + ωj bjButterfly:

bj aj - ωj bj

ωj

∑∑

∑∑−

=+

−

=

−

=+

+−

=

+

⎟⎠⎞

⎜⎝⎛⋅⋅−⎟

⎠⎞

⎜⎝⎛⋅=

⎟⎠⎞

⎜⎝⎛ +

⋅⋅+⎟⎠⎞

⎜⎝⎛ +

⋅=

=

1

012

1

02

1

012

1

02

2exp2exp

)(2exp)(2exp

m

kk

m

k

jk

m

kk

mjm

kk

jm

mijkc

mijkc

mkmjic

mkmjic

v

πωπ

πωπ

4

FFT: Recursive Algorithm

FUNCTION(v0,...,vn-1) = IDFT(c0,…,cn-1,n)IF n==1

v0 = c0 ;ELSE

m=n/2 ;(g0,…, gm-1) = IDFT(c0, c2,…, cn-2,m) ;(u0,…, um-1) = IDFT(c1, c3,…, cn-1,m) ; ω = exp(2iπ/n) ;FOR j=0:m-1

vj = gj + ωj uj ;vj+m = gj - ωj uj ;

ENDEND

Partitioning the sums in the IDFT in odd and even coefficients deliversthe first and the second part of v = IDFT(c):

5

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4)

(c0)

6

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4)

(c0) (c4)

7

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4)

(c0) (c4)

8

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2)

9

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

10

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

11

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

12

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5)

(c1)

13

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5)

(c1) (c5)

14

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5)

(c1) (c5)

15

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3)

16

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3) (c7)

17

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3) (c7)

18

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3) (c7)

19

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3) (c7)

20

(c0, c1, c2, c3, c4, c5, c6, c7)

(c0, c2, c4, c6)

(c0, c4) (c2, c6)

(c0) (c4) (c2) (c6)

(c1, c3, c5, c7)

(c1, c5) (c3, c7)

(c1) (c5) (c3) (c7)

21

c0

c1

c2

c3

c4

c5

c6

c7

c0

c2

c4

c6

c1

c3

c5

c7

c0

c4

c2

c6

c1

c5

c3

c7

c0

c2

c1

c3

c4

c6

c5

c7

g0

g1

g2

g3

u0

u1

u2

u3

v0

v1

v2

v3

v4

v5

v6

v7

g0

g0

u0

g0

u0

g0

u0

u0

g0

g1

u0

u1

g0

g1

u0

u1

22

FFT sequentiallyThe recursive formulation of the FFT can be written by log(n) simple loops.

Thereby, the first step is the reordering of the variables Bitreversal

Index k = (kp, … , k1)2 (k1, … , kp )2 ,

e.g. 5 = (0 0 1 0 1)2 (1 0 1 0 0)2 = 16 + 4 = 20, c5 c20

After permutation, the butterflies have to be applied between elements of

certain distance.

aj aj + ωj bjButterfly:

bj aj - ωj bj

ωj

23

c0

c1

c2

c3

c4

c5

c6

c7

c0

c2

c1

c3

c4

c6

c5

c7

g0

u0

g0

u0

g0

u0

g0

u0

v0

v1

v0

v1

v0

v1

v0

v1

24

c0

c1

c2

c3

c4

c5

c6

c7

c0

c2

c1

c3

c4

c6

c5

c7

g0

u0

g0

u0

g0

u0

g0

u0

g0

g1

u0

u1

g0

g1

u0

u1

u0 v2

g0 v0

u1 v3

g1 v1

g0 v0

u0 v2

g1 v1

u1 v3

25

c0

c1

c2

c3

c4

c5

c6

c7

c0

c2

c1

c3

c4

c6

c5

c7

g0

u0

g0

u0

g0

u0

g0

u0

g0

g1

u0

u1

g0

g1

u0

u1

g0

g1

g2

g3

u0

u1

u2

u3

g0 v0

u0 v4

g3 v3

u3 v7

g1 v1

u1 v5

g2 v2

u2 v6

26

FFT in ParallelCosts in parallel: n processors k = 0, 1, 2, …, 2p

each processor computes Bitreversal k and sendsits entry to the resulting processor π(k)

Then each two neighbouring processors compute butterfly,

log(n) times with growing distance

n processors, log(n) time steps.

Advantage over trivial parallelization only in number of processors

27

c0

c1

c2

c3

c4

c5

c6

c7

c0

c2

c1

c3

c4

c6

c5

c7

g0

g1

g2

g3

u0

u1

u2

u3

v0

v1

v2

v3

v4

v5

v6

v7

g0

g0

u0

g0

u0

g0

u0

u0

g0

g1

u0

u1

g0

g1

u0

u1

Bitreversal

28

FFT on HypercubeDistribute entries on vertices of hypercube.

Butterfly has to been applied always between neighbors in distance 1,2,4,…

Hence, the binary indices differ only at one position.

Therefore, butterflies have to computed only between neighboring vertices

c6 c7

c2 c3

2c4 c5

3c0 1 c1

110 111

010 011

2100 101

3000 1 001