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The FFT Algorithm

Module by: Robert Nowak. E-mail the author

Summary: This module introduces CTFT, DTFT, DFT and frequency domain.

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

The Fast Fourier Transform FFT

An effieient computational algorithm for comuting the DFT.

DFT can be expensive to compute directly ∀k,0≤k≤N−1:X⁢k=∑n=0N−1x⁢n⁢e−(i⁢2⁢π⁢kN⁢n) k 0 k N 1 X k n 0 N 1 x n 2 k N n

For each k, we must excute N complex multiplies and N-1 complex adds. The total cost of direct computation of an N-point DFT is N2 N 2 complex mults and N⁢(N−1) N N 1 complex adds. How many adds and mults of real #'s?

This " N2 N 2 " compution rapidly gets out of hand, as N 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 is much more efficient for computing the DFT. FFT requires only " O⁢N⁢logN O N N " computations to compute the N-point DFT.

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

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

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 k⁢n=e−(i⁢2⁢πN⁢k⁢n) WN k n 2 N k n
  • WN k⁢n WN k n are called "twiddle factors"

Symmetry 1: Complex Conjugate Symmetry

WN k⁢(N−n)= WN −(k⁢n)= WN k⁢n¯ WN k N n WN k n WN k n e−(i⁢2⁢π⁢kN⁢(N−n))=ei⁢2⁢π⁢kN⁢n=e−(i⁢2⁢π⁢kN⁢n)¯ 2 k N N n 2 k N n 2 k N n

Symmetry 2: Periodicity in n and k

WN k⁢n= WN k⁢(N+n)= WN (k+N)⁢n WN k n WN k N n WN k N n e−(i⁢2⁢πN⁢k⁢n)=e−(i⁢2⁢πN⁢k⁢(N+n))=e−(i⁢2⁢πN⁢(k+N)⁢n) 2 N k n 2 N k N n 2 N k N n WN =e−(i⁢2⁢πN) WN 2 N

Decimation in Time FFT

  • Just one of many different FFT algorithms
  • IDEA: Build a DFT out of smaller and smaller DFTs by decomposing x⁢n x n into smaller and smaller subsequences
  • Assume N=2m N 2 m (a power of 2)

Derivation

N is even, so we can complete X⁢k X k by separating x⁢n x n into two subsequences each of length N2 N 2 . x⁢n⇒{N2  if  n=evenN2  if  n=odd x n N 2 n even N 2 n odd ∀k,0≤k≤N−1:X⁢k=∑n=0N−1x⁢n⁢ WN k⁢n k 0 k N 1 X k n 0 N 1 x n WN k n X⁢k=∑n=2⁢rx⁢n⁢ WN k⁢n+∑n=2⁢r+1x⁢n⁢ WN k⁢n 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 X⁢k=∑r=0N2−1x⁢2⁢r⁢ WN 2⁢k⁢r+∑r=0N2−1x⁢2⁢r+1⁢ WN (2⁢r+1)⁢k=∑r=0N2−1x⁢2⁢r⁢ WN 2k⁢r+ WN k⁢∑r=0N2−1x⁢2⁢r+1⁢ WN 2k⁢r 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−(i⁢2⁢πN⁢2)=e−(i⁢2⁢πN⁢N2)= W N 2 WN 2 2 N 2 2 N N 2 W N 2 , So X⁢k=∑r=0N2−1x⁢2⁢r⁢ W N 2 k⁢r+ WN k⁢∑r=0N2−1x⁢2⁢r+1⁢ W N 2 k⁢r 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=0N2−1x⁢2⁢r⁢ W N 2 k⁢r r 0 N 2 1 x 2 r W N 2 k r is N2 N 2 -point DFT of even samples G⁢k G k and ∑r=0N2−1x⁢2⁢r+1⁢ W N 2 k⁢r r 0 N 2 1 x 2 r 1 W N 2 k r is N2 N 2 -point DFT of odd samples H⁢k H k . ∀k,0≤k≤N−1:X⁢k=G⁢k+ WN k⁢H⁢k k 0 k N 1 X k G k WN k H k Decomposition of an N-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 N-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 G⁢k G k . The second part is the number of complex mults and adds for N2 N 2 -point DFT H⁢k 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 compare 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: N2N4N8…N2m−1N2m=1 N 2 N 4 N 8 … N 2 m 1 N 2 m 1 where m=log2N m 2 N times.

Computational cost: N N-pt DFT  two N2 N 2 -pt DFTs. The cost is N2 N 2 2⁢N22+N 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⁢(2⁢N42+N2)+N=4⁢N42+2⁢N=N222+2⁢N=N22p+p⁢N 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+N⁢log2N=N2N+N⁢log2N=N+N⁢log2N N 2 2 2 N N 2 N N 2 N N 2 N N N 2 N N+N⁢log2N N N 2 N is the total number of complex adds and mults.

For large N, cost≃N⁢log2N cost N 2 N or " O⁢N⁢log2N O N 2 N ", since N⁢log2N≫N ≫ N 2 N N for large N.

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

Note:

Weird order of time samples

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

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