The FFT Algorithm 2.5 2002/12/12 2005/07/22 14:30:00.545 GMT-5 Rob "The Kid" Nowak nowak@rice.edu Liqun Wang liqun@rice.edu Rob "The Kid" Nowak nowak@rice.edu CTFT DFT DTFT fast fourier transform FFT frequency domain The FFT, an efficient way to compute the DFT, is introduced and derived throughout this module. 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 X k n 0 N 1 x n 2 k N n For each k, we must execute: N complex multiplies N 1 complex adds The total cost of direct computation of an N-point DFT is N 2 complex multiplies N N 1 complex adds How many adds and mults of real numbers are required? This " O N 2 " computation rapidly gets out of hand, as N gets large: N 1 10 100 1000 106 N2 1 100 10,000 106 1012
The FFT provides us with a much more efficient way of computing the DFT. The FFT requires only " O N N " computations to compute the N-point DFT. N 10 100 1000 106 N2 100 10,000 106 1012 N 10 N 10 200 3000 66
How long is 10 12 μsec ? More than 10 days! How long is 66 μsec ?
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 2 N k n WN k n are called "twiddle factors" Complex Conjugate Symmetry WN k N n WN k n WN k n 2 k N N n 2 k N n 2 k N n Periodicity in n and k WN k n WN k N n WN k N n 2 N k n 2 N k N n 2 N k N n 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 x n into smaller and smaller subsequences. Assume N 2 m (a power of 2)
Derivation N is even, so we can complete X k by separating x n into two subsequences each of length N 2 . x n N 2 n even N 2 n odd k 0 k N 1 X k n 0 N 1 x 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 N 2 1 . So 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 2 N 2 2 N 2 W N 2 . So 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 0 N 2 1 x 2 r W N 2 k r is N 2 -point DFT of even samples ( G k ) and r 0 N 2 1 x 2 r 1 W N 2 k r is N 2 -point DFT of odd samples ( 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 N 2 -point DFTs. Why would we want to do this? Because it is more efficient! Cost to compute an N-point DFT is approximately N 2 complex mults and adds. But decomposition into 2 N 2 -point DFTs + combination requires only N 2 2 N 2 2 N N 2 2 N where the first part is the number of complex mults and adds for N 2 -point DFT, G k . The second part is the number of complex mults and adds for N 2 -point DFT, H k . The third part is the number of complex mults and adds for combination. And the total is N 2 2 N complex mults and adds. Savings For N 1000 , N 2 10 6 N 2 2 N 10 6 2 1000 Because 1000 is small compared to 500,000, N 2 2 N 10 6 2 So why stop here?! Keep decomposing. Break each of the N 2 -point DFTs into two N 4 -point DFTs, etc., .... We can keep decomposing: N 2 1 N 2 N 4 N 8 … N 2 m 1 N 2 m 1 where m 2 N  times Computational cost: N -pt DFT  two N 2 -pt DFTs. The cost is N 2 2 N 2 2 N . So replacing each N 2 -pt DFT with two N 4 -pt DFTs will reduce cost to 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 3 4 … m , where m 2 N . We get the cost N 2 2 2 N N 2 N N 2 N N 2 N N N 2 N N N 2 N is the total number of complex adds and mults. For large N, cost N 2 N or " O N 2 N ", since ≫ N 2 N N for large N.
N 8 point FFT. Summing nodes Wn k twiddle multiplication factors.
Weird order of time samples
This is called "butterflies."