Jim Lambers MAT 773 Fall Semester 2018-19 Lecture 11 and 12 Notes These notes correspond to Section 3.1 in the text. The Discrete Fourier Transform By the Composite Trapezoidal Rule, Z 2π 1 1 2π Y0 Yn F (t) dt ≈ + Y1 + Yn−1 + ; 2π 0 2π n 2 2 where Yj = F (hj) = F (2πj=n), for j = 0; 1; : : : ; n. If F (t) is 2π-periodic, then Y0 = Yn and we then have n−1 1 Z 2π 1 X F (t) dt ≈ Y : 2π n j 0 j=0 Thus complex Fourier coefficients can be approximated by n−1 n−1 1 Z 2π 1 X 2πj 1 X α = f(t)e−ikt dt ≈ f e−2πijk=n = y wjk; k 2π n n n j 0 j=0 j=0 where 2πj y = f ; w = e2πi=n: j n Definition of Discrete Fourier Transform Let Sn be the set of n-periodic sequences of complex numbers. 1 Definition 1 Let y = fyjgj=−∞ 2 Sn. The discrete Fourier transform (DFT) of y is the sequence (Fn[y])k =y ^k, where n−1 X 2πi y^ = y wjk; w = exp : k j n j=0 It follows that the complex Fourier coefficients and discrete Fourier transform are related by 1 α ≈ y^ : k n k The discrete Fourier transform can also be expressed as a matrix-vector multiplication Fn[y] =y ^ = F ny; where 2 1 1 1 ··· 1 3 2 n−1 6 1 w w ··· w 7 6 7 6 1 w2 w4 ··· w2(n−1) 7 Fn = 6 7 : 6 . .. 7 4 . 5 1 wn−1 w2(n−1) ··· w(n−1)2 1 Properties of the Discrete Fourier Transform Lemma 2 Fn is a linear operator from Sn to Sn. This can be proved by showing directly thaty ^k+n =y ^k. Theorem 3 Let y 2 Sn and let n−1 X jk y^k = Fn[y]k = yjw : j=0 Then n−1 1 X y = F −1[^y] = y^ wjk: j n j n k k=0 Equivalently, Theorem 3 states that 1 y = F −1y^ = F y;^ n n n which, in view ofy ^ = F ny, yields 1 F F = I : n n n n H p Because Fn is complex symmetric, we have F n = Fn , and therefore 1= nFn is a unitary matrix. To show that this is in fact the case, we can show directly that n−1 ( 1 H 1 X `k kj 1 if j = `; FnF = w w = n n n 0 otherwise: `j k=0 Theorem 4 The following properties hold for the discrete Fourier transform. k • Shifts or translations: If y; z 2 Sn and zj = yj+1, then Fn[z]k = w Fn[y]k, where w = e2πi=n. • Convolutions: If y; z 2 Sn, then the sequence y ∗ z defined by n−1 X [y ∗ z]j = ykzj−k k=0 is also in Sn. • Convolution Theorem: Fn[y ∗ z]k = Fn[y]kFn[z]k. • If y 2 Sn is a sequence of real numbers, then Fn[y]n−k = Fn[y]k: That is,y ^n−k = y^k. 2 The Fast Fourier Transform Consider the discrete Fourier transform with 2N nodes: 2N−1 X jk y^k = yjw j=0 Split into sums with even- and odd-numbered terms: N−1 N−1 X 2jk X (2j+1)k y^k = y2jw + y2j+1w j=0 j=0 Let W = w2: N−1 0N−1 1 X jk k X jk y^k = y2jW + w @ y2j+1W A : j=0 j=0 That is, k y^k = FN [fy0; y2; : : : ; y2N−2g]k + w FN [fy1; y3; : : : ; y2N−1g]k: Note that •FN [yeven], FN [yodd] 2 SN • wk+N = wkwN = wke−iπ = −wk Then, for k = 0; 1;:::;N − 1, we have k y^k = FN [fy0; y2; : : : ; y2N−2g]k + w FN [fy1; y3; : : : ; y2N−1g]k; k y^k+N = FN [fy0; y2; : : : ; y2N−2g]k − w FN [fy1; y3; : : : ; y2N−1g]k: Similarly, for the inverse transform, we have 1 y = F −1[fy^ ; y^ ;:::; y^ g] + wjF −1[fy^ ; y^ ;:::; y^ g] ; j 2 N 0 2 2N−2 j N 1 3 2N−1 j 1 y = F −1[fy^ ; y^ ;:::; y^ g] − wjF −1[fy^ ; y^ ;:::; y^ g] : j+N 2 N 0 2 2N−2 j N 1 3 2N−1 j In terms of matrix-vector products, we have IN DN F N 0 yeven F2N y = F 2N y = ; IN −DN 0 F N yodd 2 N−1 where DN = diag(1; w; w ; : : : ; w ). Example 5 Let f(t) = t + t2; and lety ^k, k = 0; 1; : : : ; n, denote the (n + 1)-point DFT of f. It can be seen from a graph thaty ^k is symmetric around k = n=2. Furthermore, jy^kj decreases with k until k = n=2, only to increase after that due to this symmetry. 2 3 Example 6 Consider the signal 2 y(t) = e−(cos t) (sin 2t + 2 cos 4t + 0:4 sin t sin 50t): This signal has noise with a frequency greater than 5, so if we compute its (n + 1)-point DFT, where n is even, we can remove the noise from the signal by settingy ^k = 0 for 6 ≤ k ≤ n=2, and then by symmetry,y ^k = 0 for n=2 ≤ k ≤ n − 6. 2 Example 7 Consider the signal 2 y(t) = e−t =10(sin 2t + 2 cos 4t + 0:4 sin t sin 10t) on [0; 2π]. We can compress this signal by zeroingy ^k that are smaller than some threshold. However, because the 2π-periodic extension of y(t) is not continuous, Gibbs' phenomenon appears in the compressed signal. 2 The FFT Approximation to the Fourier Transform Let f be defined on [a; b] from samples taken at n equally spaced nodes a + jT , j = 0; 1; : : : ; n − 1 where T = (b − a)=n. We assume f(t) = 0 outside [a; b], continuous on [a; b), and periodic (that is, f(b) = f(a)). Then 1 Z b f^(!) = p f(t)e−i!t dt: 2π a t−a Let θ = 2π b−a to map [a; b] to [0; 2π]. Then b − a Z 2π f^(!) = f(a + (b − a)θ=(2π))e−i!(a+(b−a)θ=(2π) dθ 3=2 (2π) 0 Z 2π b − a −i!a −i (b−a)! θ = e f(a + (b − a)θ=(2π))e 2π dθ: 3=2 (2π) 0 Let 2π g(θ) = f(a + (b − a)θ=(2π));! = k: k b − a Then b − a 1 Z 2π ^ −i!ka −ikθ f(!k) = p e g(θ)e dθ 2π 2π 0 b − a −i!ka = p e g^k 2π We then define 2π b − a y = g j = g(jh) = f a + j = f(a + jT ); j = 0; 1; : : : ; n − 1: j n n Thus the FFT and (continuous) Fourier transform are related by b − a T ^ −i!ka −i!ka f(!k) ≈ p e y^k = p e y^k: n 2π 2π 4 Application: Parameter Identification Consider the ODE au00 + bu0 + cu = f(t); 2 where b −4ac < 0 and f(t) = f0δ(t). Using the Laplace transform, and assuming initial conditions u(0) = u0(0) = 0, we obtain 8 <0 t < 0; u(t) = f 0 sin(!t)e−µt t ≥ 0; :!a p 4ac − b2 b where ! = and µ = . 2a 2a In the case where the coefficients a; b; c are unknown, but the solution u(t) is known, we can use the DFT to obtain approximate values for ! and µ from the frequency and amplitude of the solution, respectively. Then, values for a, b and c can be obtained from the formulas for ! and µ. Application: Discretization of Ordinary Differential Equations Let h = 2π=n, and suppose that we approximate derivatives of u(t) using the finite differences u(t) − u(t − h) u0(t) ≈ ; h u(t + h) − 2u(t) + u(t − h) u00(t) ≈ : h2 Let tk = kh and uk = u(tk) for k = 0; 1; : : : ; n. Since u is periodic, un = u0. The preceding difference formulas then become, for k = 1; 2; : : : ; n − 1, u − u u0(t ) ≈ k k−1 ; k h u − 2u + u u00(t ) ≈ k+1 k k−1 : k h2 Substituting these approximations into the ODE au00 + bu0 + cu = f(t) yields 2 auk+1 + βuk + γuk−1 = h fk; k = 1; 2; : : : ; n − 1; 2 where fk = f(kh), β = ch + bh − 2a, and γ = a − bh. Let u 2 Sn be the solution of this difference equation, and letu ^ = Fn[u], f^ = Fn[f], and w = eih. From Theorem 4, we obtain 2 j j −1 u^j = h (aw + β + γw ) f^j: We can then obtain u via the inverse DFT. Exercises Chapter 3: Exercises 2, 10, 11, 12 5.