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Fast Fourier Transform (FFT) Problem : we need an efficient way to compute the DFT. The answer is the FFT. Consider a data sequence and its DFT:. - PowerPoint PPT Presentation
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Fast Fourier Transform (FFT)
Problem: we need an efficient way to compute the DFT. The answer is the FFT.
Consider a data sequence and its DFT:x x x x N [ ( ), ( ),..., ( )]0 1 1
1,...,0,)()(1
0
NkwnxkXN
n
knN
We can always break the summation into two summations: one on even indices (n=0,2,4,…) and one on odd indeces (n=1,3,5,…), as
1,...,0,)()()( NkwnxwnxkXevenn oddn
knN
knN
X k x n w x n wNkn
n evenN
kn
n odd
( ) ( ) ( )
wN /2wN /2
since w e e wNj N j N
N2 2 2 2 2
2 / ( / )/
Let us assume that the total number of points N is even, ie N/2 is an integer. Then we can write the DFT as
x m w x m w
x m w w x m w
Nk m
m
N
Nk m
m
N
N
km
m
N
Nk
N
km
m
N
( ) ( )
( ) ( )
( ) ( )2 2 1
2 2 1
2
0
21
2 1
0
21
2
0
21
2
0
21
The two summations are two distinct DFT’s, as we can see below
X k x m w w x m wNkm
m
N
Nk
Nkm
m
N
( ) ( ) ( )/ /
2 2 120
21
20
21
N-point DFT N/2-point DFT N/2-point DFT wNk
X k X k w X kNe
N Nk o
N( ) ( ) ( )/ / 2 2
for k=0,…N-1, where
X DFT x x x x x N
X DFT x x x x x N
eN
even even
oN
odd odd
/
/
, ( ), ( ),..., ( ) ;
, ( ), ( ),..., ( ) .
2
2
0 2 2
1 3 1
X k X k w X kNe
N Nk o
N( ) ( ) ( )/ / 2 2
The problem is that in the expression
the N-point DFT and the N/2 point DFT’s have different lengths, since we define them as
1,...,0),( NkkX N
12/,...,0),(2/ NkkX N
For example if N=4 we need to compute
)]3(),2(),1(),0([ 44444 XXXXX
from two 2-point DFT’s
)]1(),0([
)]1(),0([
22
22
ooo
eee
XXX
XXX
So how do we compute and ?)2(4X )3(4X
We use the periodicity of the DFT, and relate the N-point DFT with the two N/2-point DFT’s as follows
X k X k w X k
X kN
X k w X k
Ne
N Nk o
N
Ne
N Nk o
N
( ) ( ) ( )
( ) ( ),
/ /
/ /
2 2
2 22 k
N 0
21,...,
where we used the facts that
• the DFT is periodic and in particular X k X kN
N N/ /( ) ;2 2 2
w w e wNk N
Nk j N N
Nk / ( / ) /2 2 2•
General Structure of the FFT (take, say, N=8):
x( )0
x( )2
x( )4
x( )6
x( )1
x( )3
x( )5
x( )7
X e ( )0
X e ( )1
X e ( )2
X e ( )3
X o ( )0
X o ( )1
X o ( )2
X o ( )3
w80
w81
w82
w83
4-DFT
4-DFT
X ( )0
X ( )1
X ( )2
X ( )3
X ( )4
X ( )5
X ( )6
X ( )7
where a
b
a b
a b
is called the butterfly.
Same for the 4-DFT:
2-DFT
2-DFT
2-DFT
2-DFT
x( )0
x( )4
x( )2
x( )6
x( )1
x( )5
x( )3
x( )7
w40
w41
w41
w40
X e ( )0
X e ( )1
X e ( )2
X e ( )3
X o ( )0
X o ( )1
X o ( )2
X o ( )3
4-DFT
4-DFT
Finally the 2-DFT’s have a simple expansion:
X x x
X x x2
2
0 0 1
1 0 1
( ) ( ) ( )
( ) ( ) ( ),
x( )1 X ( )1
X ( )0
2-DFT
x( )0
since
1
0222 )1()0()()(
n
kkn xwxwnxkX
with kkjk ew 12
Put everything together:
x( )0
x( )4
x( )2
x( )6
x( )1
x( )5
x( )3
x( )7
)0(X
N/2 mult/stage
stagesN2log N
N2 2log
)1(X
)2(X
)3(X
)4(X
)5(X
)6(X
)7(X
04w
14w
04w
14w
08w
18w
28w
38w
N adds/stage
mult
NN 2log adds
We say that, for a data set of length ,
bNNaoperationsofnumberei 2log..
complexity of the FFT is NNO 2log
for some constants a,b.
On the other hand, for the same data of length N ,
LN 2
complexity of the DFT is 2NO
Since, from the formula, 1,...,0,)()(1
0
NkwnxkXN
n
knN
N ops/term N terms N ops2
This is a big difference in the total number of computations, as shown in this graph:
N
complexity
N0 2000 4000 6000 8000 1000010
2
103
104
105
106
107
108
DFT
FFT
