4.3 The Discrete Fourier Transform (DFT) and the Fast Fourier Transform (FFT)

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1 CHAPTER. TIME-FREQUECY AALYSIS: FOURIER TRASFORMS AD WAVELETS.3 The Discrete Fourier Transform (DFT and the Fast Fourier Transform (FFT.3.1 Introduction In this section, we discuss some of the mathematics involved in the DFT and the FFT. After this section, the reader should be able to understand and explain the Cooley and Tukey algorithm behind the FFT..3. The DFT The reader will recall that the Fourier transform is defined f (w = 1 π f (x e iωx dx (.11 It was also mentioned that there were other definitions in which the constant 1 π was not present. Indeed, some texts use the following definition for the Fourier transform f (w = f (x e iωx dx (.1 The integral is the important part and we will focus on that. For certain functions f, this computation is not practical or even sometimes impossible. In this case we need to approximate the Fourier transform. We first remark that for suffi ciently large a < 0 and b > 0, b a f (x e iωx dx is a good approximation to f (x e iωx dx, for any function f L 1 (R that is functions satisfying f (x dx <. We then approximate this new integral, b a f (x e iωx dx as follows: 1. We sample the signal at a finite number of equally spaced times x 0 = a < x 1 < x <... < x = b and we call f (k to be the value of the sampled signal at x k for k = 0, 1,,..., 1.. Let x = b a then x k = a + k x for k = 0, 1,,..., Remembering the definition of the integral, we can now approximate f (w = b a f (x e iωx dx by φ (w = = f (k e iwx k x f (k e b a iw(a+k x

2 .3. THE DISCRETE FOURIER TRASFORM (DFT AD THE FAST FOURIER TRASFORM (FFT3 that is φ (w = e iωa b a iwk x f (k e (.13. We neglect the constant e iωa x, to have φ (w = b a iwk f (k e (.1 5. The discrete Fourier transform treats the data as if it were periodic over the interval [a, b] that is the data f ( to f ( 1 is the same as the data f (0 to f ( As we will see below, computing this Discrete Fourier transform amount to a system of equations. With equations, we can only find unknowns. So, if we have sampled the signals with values, we can only hope to find frequencies which make up the signal, though the Fourier transform in theory finds the continuous values of ω. We compute the DFT for the fundamental frequency and its multiples, that is frequencies of the form ω n = πn n (in rad/s or in Hz (.15 b a b a for n an integer (in fact, n = 0, 1,..., 1. In this case, n is called the wavenumber. You will also recall that for a Fourier series over an interval of width L, we used frequencies of the form πn L = nπ L. Here, the width of the interval is b a. 7. Hence, or φ (ω n = φ (ω n = We will call φ (ω n = Df (n b a iwnk f (k e πnk i f (k e This is how we define the discrete Fourier transform. Definition.3.1 The discrete Fourier transform of f, denoted Df (n is Df (n = πnk i f (k e (.16

3 CHAPTER. TIME-FREQUECY AALYSIS: FOURIER TRASFORMS AD WAVELETS Remark.3. In the above formula, think of k as related to time (the time values are x k for k = 1,, 3,..., 1 and n to frequencies (the fundamental frequencies are ω n. So, Df (n finds the contribution of the nth fundamental frequency to the signal. Remark.3.3 As for the definition of the Fourier transform, which is up to a constant, there are other possible definitions for the DFT. They include Df (n = 1 πnk i f (k e and Df (n = 1 πnk i f (k e. These are mostly to achieve certain symmetry between a signal and its transform. Remark.3. If we define w = e πi, the th root of unity (that is w = 1, then we can rewrite Df (n = = f (k w nk (.17 f (k w nk (.18 In the second equation, w is the complex conjugate of w that is since w = e πi then w = e πi Definition.3.5 The inverse discrete Fourier transform is f (k = f (k = n=0 n=0.3.3 The DFT in Matrix Form Df (n e i πnk (.19 Df (n w nk (.0 In order to better understand the FFT algorithm, we rewrite the DFT (equation.18 in matrix form. Df (0 Df (1 Df (. Df ( Df ( 1 = w w w 3 w w 1 w w w 6 w w 1 w 3 w 6 w 9 w 6 w w w w 6 w w 1 w w w 3 w w (.1 f (0 f (1 f (. f ( f ( 1

4 .3. THE DISCRETE FOURIER TRASFORM (DFT AD THE FAST FOURIER TRASFORM (FFT5 1. We will call M the matrix in equation.1 that is w w w 3 w w 1 w w w 6 w w M = 1 w 3 w 6 w 9 w 6 w w w w 6 w w 1 w w w 3 w w (.. ote that for each entry in the vector on the left in equation.1, it takes multiplications and additions. Since that vector has entries, it is easy to see why the DFT requires O ( operations..3. The FFT In 196, Cooley and Tukey drastically improved the algorithm to compute the DFT. Their algorithm is commonly known as the Fast Fourier Transform (FFT algorithm. This algorithm is so good that it is considered one of the top ten algorithms of the twentieth century. It reduced the number of operations needed to compute the DFT from O ( to O ( log. In this section we briefly describe the main idea behind this algorithm. We give enough information so that the reader will be able to understand and explain the Cooley and Tukey algorithm. The idea behind the FFT algorithm is that the computation of the standard DFT involves a lot of redundant calculations. Let us change our notation slightly. Let W = w = e πi where w is the th root of unity, that is w = 1. ote that W = 1 as well. W k are the numbers which appear in the matrix representation of the DFT. We use the subscript because we are going to be looking at roots of unity for various values of. Let us assume that is a power of, hence is even. We begin by rewriting: Df (n = f (k w nk = It is easy to see that the same values of W nk for several reasons: f (k W nk are calculated many times, this 1. The integer product nk repeats itself for different combinations of n and k.. W nk is a periodic function with only distinct values. Example.3.6 Find all the values of W nk for = 8 that is find all the values of W nk 8.

5 6CHAPTER. TIME-FREQUECY AALYSIS: FOURIER TRASFORMS AD WAVELETS First, we note that W nk = e πink = cos ( ( πnk +i sin πnk ( = cos πnk i sin ( πnk since cos x is even and sin x is odd. Hence, W nk 8 = cos ( πnk 8 i sin ( ( πnk 8 = cos πnk ( πnk. When nk = 0, W8 nk = W8 0 = 1. ( π ( π When nk = 1, W8 1 = cos = 1 i. ( π ( π When nk =, W8 = cos = i. ( 3π When nk = 3, W8 3 = cos ( π When nk =, W8 = cos ( 5π When nk = 5, W8 5 = cos ( 6π When nk = 6, W8 6 = cos ( 7π When nk = 7, W8 7 = cos ( 3π ( π = 1. ( 5π ( 6π = i. ( 7π = 1 i = iw 1 8. = 1 + i = W 1 8. = 1 + i = iw 1 8 = W 1 8. When nk = 8, W 8 8 = 1 by definition. We can also compute it. When nk falls outside the range 0 7, well get one of the above values. For example, if n = 5 and k = 7 then W8 35 = W8 8+3 = ( W8 8 W 3 8 = W8 3 (since W8 8 = 1. In conclusion, we have W 8 = W 0 8 W 5 8 = W 1 8 W 6 8 = W 8 W 7 8 = W 3 8 Hence, we see that knowing half of the coeffi cients gives us the other half. If we repeat the process, we can write half of the first half in terms of the other half of the first half. W 8 = iw 0 8 W 3 8 = iw 1 8 Finally, W 1 8 = 1 i W 0 8

6 .3. THE DISCRETE FOURIER TRASFORM (DFT AD THE FAST FOURIER TRASFORM (FFT7 This is the main idea of the FFT algorithm, to realize that to know coefficients, it is enough to know of them. In turn, to know then coeffi cients, it is enough to know of them, and so on. To generalize this, we begin by splitting the summation over samples into two summations over samples. One summation will be for even k, and one for odd k. Df (n = Then, we note that 1 f (k W kn + 1 f (k + 1 W (k+1n W kn = e πi = e nk πi nk = W kn Hence Df (n = 1 f (k W kn + W n 1 f (k + 1 W kn (.3 We can write it as where and Df (n = E (n + W n H (n E (n = H (n = 1 1 f (k W kn f (k + 1 W kn So, the points DFT can be obtained from two transforms, one on even input data, E (n, one on odd input data, H (n. Both E (n and H ( are periodic with period. In the case of our example above, with = 8, the input data will be: Even input data: f (0, f (, f (, f (6

7 8CHAPTER. TIME-FREQUECY AALYSIS: FOURIER TRASFORMS AD WAVELETS Odd input data: f (1, f (3, f (5, f (7 Df (0 = E (0 + W8 0 H (0 Df (1 = E (1 + W8 1 H (1 Df ( = E ( + W8 H ( Df (3 = E (3 + W8 3 H (3 Df ( = E (0 + W8 H (0 = E (0 W8 0 H (0 Df (5 = E (1 + W8 5 H (1 = E (1 W8 1 H (1 Df (6 = E ( + W8 6 H ( = E ( W8 H ( Df (7 = E (3 + W8 7 H (3 = E (3 W8 3 H (3 Everything can be obtained by using the information/data of the first points. Assuming is a power of, we can repeat the process on these two points by breaking them down to point transforms until we come down to point transforms. One final step to compute these will be needed. This is an outline of the FFT algorithm, enough for the reader to be able to understand the Cooley and Tukey algorithm. To better appreciate what is gained with this algorithm, we finish this section with a table counting the number of operations for various values and the gain obtained..3.5 Exercises log Saving 5 = % 8 = % 10 = % 15 = % For some of these problems, you may need to review (or learn roots of unity. 1. Without doing a formal proof, explain why b a f (x e iωx dx is a good approximation to f (x e iωx dx, for any function f L 1 (R that is functions satisfying f (x dx <. (hint: think of the geometric meaning of the integral of a positive function.. How can we justify neglecting the constant e iωa x? (hint: think of the meaning of the Fourier transform. 3. If w = e πi, the th root of unity ( > 1 (that is w = 1, then prove that:

8 .3. THE DISCRETE FOURIER TRASFORM (DFT AD THE FAST FOURIER TRASFORM (FFT9 (a 1 w = w (b ww = 1 (c w k = w k = w +k = w k for any integer k. (d 1 + w + w + w w = 0. (hint: note that 1 + x + x x = 1 x 1 x (e If a b (mod that is if a = b + k for some integer k then w a = w b.. Derive the matrix equation shown in equation.1. (hint: you will need the properties derived in the previous problem. 5. Vandermonde Matrix: The Vandermonde matrix is defined to be z 0 z 1 z V (z 0, z 1,..., z = z0 z1 z z0 z1 z It can be shown that its determinant det V (z 0, z 1,..., z = Show that: i,j=0 j<i (z i z j. (a The matrix in equation. can be written as V ( 1, w, w,..., w where w = e πi. (b det M Read, study and be ready to explain the paper on the Cooley and Tukey algorithm.