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The FFT Algorithm

Module by: Robert Nowak. E-mail the author

Summary: The FFT, an efficient way to compute the DFT, is introduced and derived throuhgout this module.

Note: You are viewing an old version of this document. The latest version is available here.

Definition 1: FFT
(Fast Fourier Transform) An efficient computational algorithm for computing the DFT.

The Fast Fourier Transform FFT

DFT can be expensive to compute directly k,0kN1:Xk=n=0N1xne(i2πkNn) k 0 k N 1 X k n 0 N 1 x n 2 k N n

For each kk, we must execute:

  • NN complex multiplies
  • N1 N 1 complex adds
The total cost of direct computation of an NN-point DFT is
  • N2 N 2 complex multiplies
  • N(N1) N N 1 complex adds
How many adds and mults of real numbers are required?

This " ON2 O N 2 " computation rapidly gets out of hand, as NN gets large:

Table 1
NN 1 10 100 1000 106 106
N2 N2 1 100 10,000 106 106 1012 1012
Figure 1
Figure 1 (figure1.png)

The FFT provies us with a much more efficient way of computing the DFT. The FFT requires only " ONlogN O N N " computations to compute the NN-point DFT.

Table 2
NN 10 100 1000 106 106
N2 N2 100 10,000 106 106 1012 1012
NlogN N 10 N 10 200 3000 6×106 66

How long is 1012μsec 10 12 μsec ? More than 10 days! How long is 6×106μsec 66 μsec ?

Figure 2
Figure 2 (figure2.png)

The FFT and digital computers revolutionized DSP (1960 - 1980).

How does the FFT work?

  • The FFT exploits the symmetries of the complex exponentials WN kn=e(i2πNkn) WN k n 2 N k n
  • WN kn WN k n are called "twiddle factors"

Symmetry 1: Complex Conjugate Symmetry

WN k(Nn)= WN (kn)= WN kn¯ WN k N n WN k n WN k n e(i2πkN(Nn))=ei2πkNn=e(i2πkNn)¯ 2 k N N n 2 k N n 2 k N n

Symmetry 2: Periodicity in n and k

WN kn= WN k(N+n)= WN (k+N)n WN k n WN k N n WN k N n e(i2πNkn)=e(i2πNk(N+n))=e(i2πN(k+N)n) 2 N k n 2 N k N n 2 N k N n WN =e(i2πN) WN 2 N

Decimation in Time FFT

  • Just one of many different FFT algorithms
  • The idea is to build a DFT out of smaller and smaller DFTs by decomposing xn x n into smaller and smaller subsequences.
  • Assume N=2m N 2 m (a power of 2)


NN is even, so we can complete Xk X k by separating xn x n into two subsequences each of length N2 N 2 . xn{N2  if  n=evenN2  if  n=odd x n N 2 n even N 2 n odd k,0kN1:Xk=n=0N1xn WN kn k 0 k N 1 X k n 0 N 1 x n WN k n Xk=n=2rxn WN kn+n=2r+1xn WN kn X k n 2 r x n WN k n n 2 r 1 x n WN k n where 0rN21 0 r N 2 1 . So

Xk=r=0N21x2r WN 2kr+r=0N21x2r+1 WN (2r+1)k=r=0N21x2r WN 2kr+ WN kr=0N21x2r+1 WN 2kr X k r 0 N 2 1 x 2 r WN 2 k r r 0 N 2 1 x 2 r 1 WN 2 r 1 k r 0 N 2 1 x 2 r WN 2 k r WN k r 0 N 2 1 x 2 r 1 WN 2 k r
where WN 2=e(i2πN2)=e(i2πN2)= W N 2 WN 2 2 N 2 2 N 2 W N 2 . So Xk=r=0N21x2r W N 2 kr+ WN kr=0N21x2r+1 W N 2 kr X k r 0 N 2 1 x 2 r W N 2 k r WN k r 0 N 2 1 x 2 r 1 W N 2 k r where r=0N21x2r W N 2 kr r 0 N 2 1 x 2 r W N 2 k r is N2 N 2 -point DFT of even samples ( Gk G k ) and r=0N21x2r+1 W N 2 kr r 0 N 2 1 x 2 r 1 W N 2 k r is N2 N 2 -point DFT of odd samples ( Hk H k ). k,0kN1:Xk=Gk+ WN kHk k 0 k N 1 X k G k WN k H k Decomposition of an NN-point DFT as a sum of 2 N2 N 2 -point DFTs.

Why would we want to do this? Because it is more efficient!


Cost to compute an NN-point DFT is approximately N2 N 2 complex mults and adds.
But decomposition into 2 N2 N 2 -point DFTs + combination requires only N22+N22+N=N22+N N 2 2 N 2 2 N N 2 2 N where the first part is the number of complex mults and adds for N2 N 2 -point DFT, Gk G k . The second part is the number of complex mults and adds for N2 N 2 -point DFT, Hk H k . The third part is the number of complex mults and adds for combination. And the total is N22+N N 2 2 N complex mults and adds.

Example 1: Savings

For N=1000 N 1000 , N2=106 N 2 10 6 N22+N=1062+1000 N 2 2 N 10 6 2 1000 Because 1000 is small compared to 500,000, N22+N1062 N 2 2 N 10 6 2

So why stop here?! Keep decomposing. Break each of the N2 N 2 -point DFTs into two N4 N 4 -point DFTs, etc, ....

We can keep decomposing: N21=N2N4N8N2m1N2m=1 N 2 1 N 2 N 4 N 8 N 2 m 1 N 2 m 1 where m=log2N= times m 2 N  times

Computational cost: N N-pt DFT  two N2 N 2 -pt DFTs. The cost is N22N22+N N 2 2 N 2 2 N . So replacing each N2 N 2 -pt DFT with two N4 N 4 -pt DFTs will reduce cost to 2(2N42+N2)+N=4N42+2N=N222+2N=N22p+pN 2 2 N 4 2 N 2 N 4 N 4 2 2 N N 2 2 2 2 N N 2 2 p p N As we keep going p=34m p 3 4 m , where m=log2N m 2 N . We get the cost N22log2N+Nlog2N=N2N+Nlog2N=N+Nlog2N N 2 2 2 N N 2 N N 2 N N 2 N N N 2 N N+Nlog2N N N 2 N is the total number of complex adds and mults.

For large NN, costNlog2N cost N 2 N or " ONlog2N O N 2 N ", since Nlog2NN N 2 N N for large NN.

Figure 3: N=8 N 8 point FFT. Summing nodes Wn k Wn k twiddle multiplication factors.
Figure 3 (figure3.png)


Weird order of time samples

Figure 4: This is called "butterflies."
Figure 4 (figure4.png)

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