Fast Fourier Transform ( FFT )

The DFT can be reduced from exponential time with the Fast Fourier Transform algorithm. Fast Fourier Transform (FFT) One wonders if the DFT can be computed faster: Does another computational procedure an algorithm exist that can compute the same quantity, but more e ciently. We could seek methods that reduce the constant of proportionality, but do not change the DFT's complexity O ( N ) . Here, we have something more dramatic in mind: Can the computations be restructured so that a smaller complexity results? In 1965, IBM researcher Jim Cooley and Princeton faculty member John Tukey developed what is now known as the Fast Fourier Transform (FFT). It is an algorithm for computing that DFT that has order O (N logN) for certain length inputs. Now when the length of data doubles, the spectral computational time will not quadruple as with the DFT algorithm; instead, it approximately doubles. Later research showed that no algorithm for computing the DFT could have a smaller complexity than the FFT. Surprisingly, historical work has shown that Gauss1 in the early nineteenth century developed the same algorithm, but did not publish it! After the FFT's rediscovery, not only was the computation of a signal's spectrum greatly speeded, but also the added feature of algorithm meant that computations had exibility not available to analog implementations. Exercise 1: Before developing the FFT, let's try to appreciate the algorithm's impact. Suppose a short-length transform takes 1 ms. We want to calculate a transform of a signal that is 10 times longer. Compare how much longer a straightforward implementation of the DFT would take in comparison to an FFT, both of which compute exactly the same quantity. Solution: If a DFT required 1ms to compute, and signal having ten times the duration would require 100ms to compute. Using the FFT, a 1ms computing time would increase by a factor of about log210 = 3.3, a factor of 30 less than the DFT would have needed. ∗http://creativecommons.org/licenses/by/1.0 1http://www-groups.dcs.st-and.ac.uk/∼history/Mathematicians/Gauss.html http://cnx.rice.edu/content/m10250/latest/ Connexions module: m10250 2 To derive the FFT, we assume that the signal's duration is a power of two: N = 2. Consider what happens to the even-numbered and odd-numbered elements of the sequence in the DFT calculation. S (k) = s (0) + s (2) e(−i) 2π2k N + · · · + s (N − 2) e(−i) 2π(N−2)k N + s (1) e(−i) 2πk N + s (3) e(−i) 2π(2+1)k N + · · · + s (N − 1) e(−i) 2π(N−(2−1))k N = s (0) + s (2) e 2πk 2 + · · · + s (N − 2) e 2π(2 −1)k 2 + s (1) + s (3) e(−i) 2 π + · · · + s (N − 1) e 2π(2 −1)k 2  e−(i2πk) N (1) Each term in square brackets has the form of a N2 -length DFT. The rst one is a DFT of the even-numbered elements, and the second of the odd-numbered elements. The rst DFT is combined with the second multiplied by the complex exponential e−( i2πk N ). The half-length transforms are each evaluated at frequency indices k = 0, . . ., N−1. Normally, the number of frequency indices in a DFT calculation range between zero and the transform length minus one. The computational advantage of the FFT comes from recognizing the periodic nature of the discrete Fourier transform. The FFT simply reuses the computations made in the half-length transforms and combines them through additions and the multiplication by e−( i2πk N ), which is not periodic over N2 , to rewrite the length-N DFT. Figure 1 illustrates this decomposition. As it stands, we now compute two length-N2 transforms (complexity 2O ( N2 4 ) ), multiply one of them by the complex exponential (complexity O (N)), and add the results (complexity O (N)). At this point, the total complexity is still dominated by the half-length DFT calculations, but the proportionality coe cient has been reduced. Now for the fun. Because N = 2, each of the half-length transforms can be reduced to two quarter-length transforms, each of these to two eighth-length ones, etc. This decomposition continues until we are left with length-2 transforms. This transform is quite simple, involving only additions. Thus, the rst stage of the FFT has N2 length-2 transforms (see the bottom part of Figure 1). Pairs of these transforms are combined by adding one to the other multiplied by a complex exponential. Each pair requires 4 additions and 4 multiplications, giving a total number of computations equaling 8 · N4 = N 2 . This number of computations does not change from stage to stage. Because the number of stages, the number of times the length can be divided by two, equals log2N , the complexity of the FFT is O (N log2N). Doing an example will make computational savings more obvious. Let's look at the details of a length-8 DFT. As shown on Figure 2, we rst decompose the DFT into two length-4 DFTs, with the outputs added and subtracted together in pairs. Considering Figure 2 as the frequency index goes from 0 through 7, we recycle values from the length-4 DFTs into the nal calculation because of the periodicity of the DFT output. Examining how pairs of outputs are collected together, we create the basic computational element known as a butter y (Figure 2). By considering together the computations involving common output frequencies from the two half-length DFTs, we see that the two complex multiplies are related to each other, and we can reduce our computational work even further. By further decomposing the length4 DFTs into two length-2 DFTs and combining their outputs, we arrive at the diagram summarizing the length-8 fast Fourier transform (Figure 1). Although most of the complex multiplies are quite simple (multiplying by e−(iπ) means negating real and imaginary parts), let's count those for purposes of evaluating the complexity as full complex multiplies. We have N2 = 4 complex multiplies and 2N = 16 additions for each stage and log2N = 3 stages, making the number of basic computations 3N 2 log2N as predicted. Exercise 2: http://cnx.rice.edu/content/m10250/latest/ Connexions module: m10250 3 Length-8 DFT decomposition s0 s2 s4 s6 s1 s3 s5 s7 S0 S1 S2 S3 S4 S5 S6 S7 e–j0 e–j2π/8 e–j2π2/8 e–j2π3/8 e–j2π4/8 e–j2π5/8 e–j2π6/8 e–j2π7/8 Length-4 DFT Length-4 DFT