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Course by: Thad Welch. E-mail the author

# Fast fourier transform (FFT)

Module by: Richard Baraniuk. E-mail the author

The fast fourier transform is an efficient computational algorithm for the DFT. The DFT can be expensive to compute,

k ,0kN1:Xk= n =0N1Xkei2πNkn k 0 k N 1 X k n 0 N 1 X k 2 N k n
(1)
For each kk, we must execute: NN complex multiplications and N1 N 1 complex additions. There are NN kk values, which means that the total cost of direct computations of a DFT is N2 N 2 complex multiplications and N2N N 2 N complex adds.

## Exercise 1

How many multiplies and adds are there for reals numbers?

This O( N2 N 2 ) computation readily gets out of hand as NN gets large. This is illustrated by Figure 1 and Table 1:

Table 1
N N N2 N 2
1 1
10 100
100 10000
1000 106 10 6
106 10 6 1012 10 12
The FFT is an efficient algorithm for computing the DFT. Efficient? We will see that it requires only O( N⁢logN N N ) coputations to compute NN DFT samples. This is illustrated in Figure 2 and Table 2:
Table 2
N N N2 N 2 NlogN N 10 N
10 100 1
100 10000 200
1000 106 10 6 3000
106 10 6 1012 10 12 6×106 6 10 6

## Exercise 2

How long is 1012 10 12 μsecμsec? 6×106 6 10 6 μsecμsec?

The FFT and digital computer were almost completely responsible for the 'Golden Age of DSP' (1960-1980). The FFT exploits the symmetries of the twiddle factors W N =ei2πN W N 2 N .

Symmetry 1: Complex conjugate symmetry: W N k ( N - n ) = W N - k n = W N kn ¯ W N k ( N - n ) W N - k n W N kn e(i)2πNk(Nn)=e(i)2πNkn=ei2πNkn¯ 2 N k N n 2 N k n 2 N k n

Symmetry 2: Periodicity in nn and kk: W N kn = W N k ( n - N ) = W N ( k + N ) n W N kn W N k ( n - N ) W N ( k + N ) n e(i)2πNkn=e(i)2πNk(n+N)=e(i)2πN(k+N)n 2 N k n 2 N k n N 2 N k N n

## Decimation in time FFT

Decimation is just one of many different FFT schemes. The idea is to build a DFT out of smaller and smaller DFTs by decompsing xn x n into smaller and smaller sequences. A condition for this scheme is that we must assume that N=2V N 2 V (it is a power of 2).

### Derivation

NN is even, so we can compute Xk X k by separating xn x n into two subsequences each of length N2 N 2 .

Xk= n =0N1xn W N kn = n =evenxn W N kn + n =oddxn W N kn X k n 0 N 1 x n W N kn n even x n W N kn n odd x n W N kn
(2)
For 0rN21 0 r N 2 1 , let n=even=2r n even 2 r and n=odd=2r+1 n odd 2 r 1 :
Xk= r =0N21x2r W N 2 k r + r =0N21x2r+1 W N ( 2 r + 1 ) k = r =0N21x2r W N 2 kr + W N k r =0N21x(2r+1) W N 2 kr X k r 0 N 2 1 x 2 r W N 2 k r r 0 N 2 1 x 2 r 1 W N ( 2 r + 1 ) k r 0 N 2 1 x 2 r W N 2 kr W N k r 0 N 2 1 x 2 r 1 W N 2 kr
(3)
But since W N 2 =e(i)2πN2=e(i)2πN2= W N / 2 W N 2 2 N 2 2 N 2 W N / 2 , then for 0kN1 0 k N 1 :
Xk= r =0N21x2r W N / 2 kr + W N k r =0N21x(2r+1) W N / 2 kr =Gk+ W N k Hk X k r 0 N 2 1 x 2 r W N / 2 kr W N k r 0 N 2 1 x 2 r 1 W N / 2 kr G k W N k H k
(4)

### Example 1

A a B (B=aA) A a B B a A

X0=G0+ W 8 0 H0 X 0 G 0 W 8 0 H 0

X1=G1+ W 8 1 H1 X 1 G 1 W 8 1 H 1

X2=G2+ W 8 2 H2 X 2 G 2 W 8 2 H 2

X3=G3+ W 8 3 H3 X 3 G 3 W 8 3 H 3

X4=G4+ W 8 4 H4=G0+ W 8 4 H0 X 4 G 4 W 8 4 H 4 G 0 W 8 4 H 0

X5= X 5

X6= X 6

X7= X 7

#### Problem 1

Why is X4=G4+ W 8 4 H4=G0+ W 8 4 H0 X 4 G 4 W 8 4 H 4 G 0 W 8 4 H 0 true?

Why would we want to do this? Because its more efficient. Recall that the cost to compute an N-ponit DFT is approximately N2 N 2 complex multiplications and additions. But decomposition into 2 N-point DFTS and combination requires only N22+N22+N=N22+N N 2 2 N 2 2 N N 2 2 N complex multiplications and additions. For N>2 N 2 , N22+N<N2 N 2 2 N N 2 (less computations).

So why stop heer? Break each of the N2 N 2 point DFTs into two N4 N 4 point DFTs, etc...

### Example 2

then that becomes Figure 5, which becomes Figure 6.

But what is an N4=2 N 4 2 point DFT?

Yk= m =01yme(i)2π2km=y01+y1-1=y0 W N 0 +y1 W N 4 Y k m 0 1 y m 2 2 k m y 0 1 y 1 -1 y 0 W N 0 y 1 W N 4
(5)
For example, in Figure 7, we have: y0=x0 y 0 x 0 , y1=x4 y 1 x 4 . For large NN, we keep dividing up by factors of 2. After splitting pp times, we need approximately N22p+pN N 2 2 p p N complex multiplies and adds to compute the NN point DFT.

### Exercise 3

How far can this go on? p=? p ?

Then what is the total number of complex multiplies and adds in this FFT?

The total number can be reduced even further using other tricks, but for N=2 N 2 , it is always O( N⁢logN N N ).

### The final picture

N=8 N 8

#### Notes

1. There are logN N stages. Why?
2. Total number of multiplies and adds is... Recall that replacing an NN point DFT with two N2 N 2 point DFTs reduced the cost from N2 N 2 to 2N22+N 2 N 2 2 N so, replacing each N2 N 2 point DFT with two N4 N 4 point DFTs will reduce cost to 2(2N42+N2)+N=4N42+2N 2 2 N 4 2 N 2 N 4 N 4 2 2 N .
• Other algorithms - eg: Decimation in frequency.
• Other structures - Note the weird ordering of xn x n on pg.C54.
• N2V N 2 V - Prime factor algorithms (Chinese remainder theorem).
• Chirp z-transform - For zooming in on DTFT.

#### Note:

CSBurrus wrote the book (literally!) on FFTs.

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