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The FFT Algorithm

Module by: Robert Nowak

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

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,0≤k≤N-1:Xk=∑n=0N-1xnⅇ-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
  • N-1 N 1 complex adds
The total cost of direct computation of an NN-point DFT is
  • N2 N 2 complex multiplies
  • NN-1 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:
NN 1 10 100 1000 106 106
N2 N2 1 100 10,000 106 106 1012 1012
figure1.png
Figure 1
The FFT provides 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.
NN 10 100 1000 106 106
N2 N2 100 10,000 106 106 1012 1012
Nlog10N 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 ?
figure2.png
Figure 2
The FFT and digital computers revolutionized DSP (1960 - 1980).

How does the FFT work?

  • The FFT exploits the symmetries of the complex exponentials WNkn=ⅇ-ⅈ2πNkn WN k n 2 N k n
  • WNkn WN k n are called "twiddle factors"
Symmetry 1: Complex Conjugate Symmetry 
WNkN-n=WN-kn=WNkn¯ WN k N n WN k n WN k n ⅇ-ⅈ2πkNN-n=ⅇⅈ2πkNn=ⅇ-ⅈ2πkNn¯ 2 k N N n 2 k N n 2 k N n
Symmetry 2: Periodicity in n and k 
WNkn=WNkN+n=WNk+Nn WN k n WN k N n WN k N n ⅇ-ⅈ2πNkn=ⅇ-ⅈ2πNkN+n=ⅇ-ⅈ2πNk+Nn 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)

Derivation

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→N2ifn=evenN2ifn=odd x n N 2 n even N 2 n odd ∀k,0≤k≤N-1:Xk=∑n=0N-1xnWNkn k 0 k N 1 X k n 0 N 1 x n WN k n Xk=∑xnWNkn+∑xnWNkn X k n 2 r x n WN k n n 2 r 1 x n WN k n where 0≤r≤N2-1 0 r N 2 1 . So
Xk=∑r=0N2-1x2rWN2kr+∑r=0N2-1x2r+1WN2r+1k=∑r=0N2-1x2rWN2kr+WNk∑r=0N2-1x2r+1WN2kr 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 (1)
where WN2=ⅇ-ⅈ2πN2=ⅇ-ⅈ2πN2= W N 2 WN 2 2 N 2 2 N 2 W N 2 . So Xk=∑r=0N2-1x2r W N 2 kr+WNk∑r=0N2-1x2r+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=0N2-1x2r 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=0N2-1x2r+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,0≤k≤N-1:Xk=Gk+WNkHk 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!
Recall: 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+N≈1062 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=N2N4N8…N2m-1N2m=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 N2→2N22+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 22N42+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=34…m 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, cost≈Nlog2N cost N 2 N or " ONlog2N O N 2 N ", since Nlog2N≫N ≫ N 2 N N for large NN.
figure3.png
Figure 3: N=8 N 8 point FFT. Summing nodes Wnk Wn k twiddle multiplication factors.
Note: Weird order of time samples
figure4.png
Figure 4: This is called "butterflies."

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