1.10 trigonometric interpolation and FFT (CSE) convention: In this chapter, i = √ 1 with i2 = 1, that is, not an index. Numbering of indices of vectors starts at 0. − − 1.10.1 trigonometric interpolation n 1 − ijx Definition 1.38 The polynomials p : R C, p(x)= cje , cj C are called trigono- → j=0 ∈ metric polynomials of degree n 1. P − goal: given distinct knots xj [0, 2π), j =0,...,n 1 and values yj, j =0,...,n 1 solve: ∈ − − find trigonometric polyomial p of degree n 1 s.t.p(xj)= yj, j =0,...,n 1 (1.33) − − Remark 1.39 (i) The trigonometric polynomial x p(x) is a 2π-periodic function. The 7→ coefficients cj are its Fourier coefficients. (ii) The (continuous) Fourier transform is an important tool in signal processing, e.g., when analyzing audio signals. In the simplest setting, a signal is assumed to be periodic (over a given time interval (0, T )) and writing it as a Fourier series decomposes the signal into different frequency components. These components are then analyzed or modified (e.g., low pass or high pass filters). For T =2π, the Fourier series is simply the representation 2π ∞ ixj 1 ixj f(x)= a e , a = f(x)e− dx, (1.34) j j 2π j= 0 X−∞ Z and aj are the Fourier coefficients. In order to avoid evaluating the integrals, one could proceed as follows: 1) sample the signal in the points xj; 2) approximate f by its trigono- metric interpolant p; 3) interpret the Fourier coefficients of p as (good) approximations to the Fourier coefficients of f.3 (iii) For convenience, we work in the present section with the complex exponentials eixj. How- ever, the Euler identities eix = cos x + i sin x show eixj = cos(jx)+ i sin(jx) so that one could also formulate the problem in terms of “true” trigonometric polynomials involving sin(kx) and cos(kx). Also the FFT discussed below can be formulated for these functions (“discrete cosine transform” and “discrete sine transform”). 3 The procedure described only suggests how to get the coefficients aj for j 0 since the interpolating n−1 ijx ≥ polynomial is of the form p(x) = j=0 cj e . It would therefore be more natural to seek the interpolating n−1 ijx c e real valued a− a polynomial of the form j=−(n−1)Pj . For signals, one has j = j so that one is lead to seek the interpolating polynomial of the form 1 (p(x)+ p(x)) for a trigonometric polynomial p. For real-valued f, P 2 this p is the solution of (1.33). 21 Theorem 1.40 Let xj [0, 2π), j = 0,...,n 1 be distinct. Then (1.33) is uniquely ∈ n 1 n − solvable for each sequence (yj) − C . j=0 ∈ n 1 ixj − ijx Proof: Set zj := e , j =0,...,n 1. Then the zj are distinct. The ansatz p(x)= ci e − j=0 yields the linear system of equations: P 0 1 n 1 z0 z0 ... z0 − c0 y0 0 1 n 1 z1 z1 ... z1 − c1 y1 . . = . (1.35) . 0 1 n 1 zn 1 zn 1 ... zn−1 cn 1 yn 1 − − − − − e =:V =:c =:y | {z } | {z } | {z } V is a so-called Vandermonde matrix with det V = (zk zj) =0 ✷ 0 j<k n 1 − 6 ≤ ≤ − Q Ine the remainder of the chapter, we consider thee uniform knot distribution 2πj xj = , j =0,...,n 1. (1.36) n − It is expedient to introduce the root 2πi ωn := e− n , (1.37) n j ixj which satisfies ωn = 1. J note: ωn = e− K The matrix V of (1.35) is easily inverted under the assumption (1.36): n Theorem 1.41 Assume (1.36) and (1.37). Let y := (y0,...,yn 1)⊤ C be given and n 1 − ∈ − ijx p(x)= cj e be the solution to (1.33). Set j=0 P j k n 1 · − Vn := ωn j,k=0 J “ DFT matrix” K (1.38) Then: n 1 1 1 − j k (i) n Vn y = c J i.e., ck = n ωn· yj K j=0 P 1 1 1 − 1 H 1 (ii) √n Vn symmetric and unitary i.e., √n Vn = √n Vn = √n Vn n 1 n 1 jk − jk − (iii) Vn = ωn j,k=0 = ωn− j,k=0 Proof: ad (iii): X 1 ad (ii): Let vj, j =0,...,n 1 be the columns of Vn. Then: − √n n 1 H 1 − jk jk vk vk = n ωn− ωn = 1 • j=0 P 22 k = l : • 6 n 1 n 1 H 1 − jk lj 1 − l k j geometr. vk vl = ωn− ωn = ωn− = n n series j=0 j=0 X X l k n nl k 1 1 ω − 1 1 (ω ) − (1.37) = − n = − n = 0 since ωn = 1 n 1 ωl k n 1 ωl k n − n− − n− ad(i): For the equidistant points xj, j = 0,...,n 1, given by (1.36), the linear system of − 1 (ii) 1 1 1 − 1 1 equations (1.35) has the form Vnc = y c = Vn− y = Vn y = Vny = ⇒ √n √n √n √n 1 ✷ n Vny. Exercise 1.42 Show the formula for the geometric series by mulitplying out the right-hand side: n 1 xn+1 = (1 x) xi, x =1. − − 6 i=0 X Definition 1.43 The linear map n n n : C C F → y0 . y = . Vn y . 7→ yn 1 − is called the discrete Fourier transform (DFT) of length n. 1 The inverse − is called IDFT (inverse discrete Fourier transform). Fn Remark 1.44 Theorem 1.41 yields 1 1 1 1 − y = Vny = Vny = n(y) (1.39) Fn n n nF 1.10.2 FFT observation: The matrix Vn is fully populated. A naive realization of the DFT therefore 2 requires O(n ) arithmetic operations. Exploiting the special structure of Vn leads to the Fast Fourier transform (FFT), which only needs O(n log n) arithmetic operations. 2 i π n Lemma 1.45 Let n =2m, ω = e± n . Let (y0,...,yn 1) C . Then the terms − ∈ n 1 − kj αk := yj ω k =0,...,n 1 − j=0 X 23 with ξ := ω2 and l =0,...,m 1 can be computed as follows: − m 1 − jl α2l = gj ξ with gj := yj + yj+m j=0 X m 1 − jl j α l = hjξ with hj := (yj yj m) ω j =0,...,m 1 2 +1 − + − j=0 X Proof: Since ωnl = 1 we get n 1 n 1 2 − − lj lj l j n 2 2 n 2 ( + 2 ) α2l = yj ω = yj ω + yj+ 2 ω = j=0 j=0 X X n 1 n 1 2 − 2 − lj ln lj 2 n 2 n = ω (yj + yj+ 2 ω )= ω (yj + yj+ 2 ) j=0 j=0 X X n Since ω 2 = 1 we have: − n 1 − (2l+1)j α2l+1 = yj ω = j=0 X n 1 2 − l j l j n (2 +1) n (2 +1)( + 2 ) = ω yj + yj+ 2 ω = j=0 Xn 2 1 − n 2lj j ln j 2 = ω yj ω + yj n ω ω ω = + 2 · j=0 X n 1 2 − j 2lj = (yj yj n ) ω ω − + 2 j=0 X ✷ Lemma 1.45 shows that the computation of yˆ = (ˆy0,..., yˆn 1)⊤ := n(y) can be reduced to − n n F the computation of 2 (g) and 2 (h). With n =2m we haveF F m 1 (ˆy0, yˆ2,..., yˆn 2)⊤ = m(g) , g =(yj + yj+m) − − j=0 F j m 1 (ˆy1, yˆ3,..., yˆn 1)⊤ = m(h) , h = ((yj yj+m) ωn) − − F − j=0 This yields Algorithm 1.46 (FFT) p n Input: n =2 , p N0, y =(y0,...,yn 1)⊤ C ∈ − ∈ Output: yˆ = (ˆy0,..., yˆn 1)= n(y) − F if n =1 then 24 yˆ0 := y0 else −2πi ω := e n n m := 2 m 1 m 1 − − (gj)j=0 := (yj + yj+m)j=0 m 1 j m 1 (hj) − := ((yj yj m) ω ) − j=0 − + j=0 (ˆy0, yˆ2,..., yˆn 2) := F F T (m, g) − (ˆy1, yˆ3,..., yˆn 1) := F F T (m, h) end if − return (yˆ) 1 Similarly for the Inverse Fourier Transform: (ˇy0,..., yˇn 1) := n− (y) (cf. first equation in − (1.39)): F 1 1 m 1 (ˇy0, yˇ2,..., yˇn 2)⊤ = −n (g) , g =(yj + yj+m) − − 2 F 2 j=0 1 1 j m 1 (ˇy1, yˇ3,..., yˇn 1)⊤ = −n (h) , h = (yj yj+m)ωn − − 2 F 2 − j=0 Algorithm 1.47 (IFFT) p n Input: n =2 , p N0, y =(y0,...,yn 1)⊤ C 1 ∈ − ∈ Output: yˇ = − (y) Fn if n =1 then yˇ0 := y0 else 2πi ω := e n n m := 2 m 1 1 m 1 − − (gj)j=0 := 2 (yj + yj+m)j=0 m 1 1 j m 1 (hj) − := ((yj yj m) ω ) − j=0 2 − + j=0 (ˇy0, yˇ2,..., yˇn 2) := IFFT (m, g) − (ˇy1, yˇ3,..., yˇn 1) := IFFT (m, h) end if − return (yˆ) p Cost of the FFT: Denote by A(n) the cost of the call of F F T (n, y) and let n =2 , p N0. Then: ∈ A(n) 2A(n/2) + C n (1.40) ≤ computation of g, h |{z} 25 and thus: (1.40) n A(n) 2 A + C n = ≤ 2 (1.40) p 1 p = 2 A 2 − + C 2 ≤ (1.40) p 2 p 1 p 2 2A 2− + C 2 − + C 2 = ≤ (1.40) 2 p 2 p = 2 A 2 − +2 C 2 ≤ (1.40) 2 p 3 p 2 p 2 2A 2− + C 2 − +2 C 2 = ≤ 3 p 3 p = 2 A 2 − +3 C 2 ..