Lecture 10. Digital Signal Processing. Chapter 7. Discrete Fourier transform DFT. Mikael Swartling Nedelko Grbic Bengt Mandersson. rev.

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1 Lecture 10 Digital Signal Processing Chapter 7 Discrete Fourier transform DFT Mikael Swartling Nedelko Grbic Bengt Mandersson rev. 016 Department of Electrical and Information Technology Lund University

2 The Discrete-Time Fourier Transform (DTFT) The Fourier transform of a discrete signal: X(ω) = x(n)e jωn (1) n= x(n) = = π X(ω)e jωn dω () π π 0 X(ω)e jωn dω (3) Convergence if x(n) is stable: x(n) < n Weaker convergence if x(n) n (4) (5) but x(n) < n (6) The z-transform Let h(n) be a causal impulse response. Causal means that h(n) = 0 for n < 0. The z-transform of the impulse response is defined as H(z) = h(n)z n n=0 (7) where z = r e jω is a complex number that we often write as a magnitude and a phase. H(z) is thus a complex function of a complex variable. If h(n) is causal and stable then H(ω) = H(z z = e jω ) (8) The Discrete Fourier Transform (DFT) Let x(n) = { (9)

3 Choose a length N and calculate the normal DTFT at the frequencies ω = π { 0 1 N N 3 N... N 1 N This yields the Discrete Fourier Transform (DFT) N 1 X DFT (k) = x(n)e jπ kn n for k = 0,1,...,N 1 (11) n=0 and the inverse transform (IDFT) x IDFT (n) = 1 N 1 N X(k)e jπ kn n for n = 0,1,...,N 1 (1) Periodicity The Fourier transform (DTFT) X(ω) is periodic in ω since e jωn = e j(ω+π)n. The Discrete Fourier transform (DFT) Both x(n) and X(k) are periodic with period N since n = n + pn and k = k + pn for p integers and yields the same numerical values. e jπ kn (n+pn) = e jπ kn n e jπkp (13) k+pn jπ e N n = e jπ kn n e jπnp (14) Indices are calculated modulo-n. If x(n) is defined only for 0 n < N (length N) we get X DFT (k) = X(ω ω = π k N ) (15) which is the same as sampling X(ω) in N uniformly distributed frequencies. If N is an even power-of-, the calculations can be made efficiently, on the order of O(NlogN) instead of O(N ) for the direct DFT implementation. The algorithm is caleld the Fast Fourier Transform (FFT) and is described in chapter 8 but is not a part of the course. (10) Roots of unity N 1 e jπ kn (n l) N if n l = 0 + pn = 0 if n l 0 + pn (16) = N δ(n l mod N) (17) N δ((n l)) N (18) The sum of the points evenly distributed around the unit circle is 0. 3

4 C D B E A F H G The sum of the values at the points are: A + A + A + A + A + A + A + A = 8 for n l = 0 A + B + C + D + E + F + G + H = 0 for n l = 1 A + C + E + G + A + C + E + G = 0 for n l =, and so on Compare to the integral of cos(ωt) over a whole number of periods that is 0 except when Ω = 0. Special properties for the DFT Both x(n) and X(k) are periodic, which imposes the property that all indices are calculated modulo N. x(n) = { (19) x(n 1) = { Circular shift becomes (0) x(n n 0 mod N) X(k) e jπ kn n 0 (1) Example of shift for the DFT: x(n) = { x(n 1) = { () Circular convolution and the DFT, length N (page ) Multiplication in the DFT-domain is circular convolution in the time domain. N 1 X(k) = X 1 (k) X (k) x(n) = x 1 (n) x (n) = x 1 (k) x (n k mod N) (3) 4

5 Example Given: x(n) = { (4) h(n) = { 1 1 (5) Find: y C (n) = x(n) 4 h(n) (6) Graphical solution (with x(n) repeated and h(n) time reversed): h(k) 1 1 x(k) y C (k) Equivalent solution with a convolution table For linear convolution the length of this convolution is 7, but in circlular convolution the indices wrap around modulo-4 instead. >> x = [1 3 4]; >> h = [ 1 1]; >> yc = ifft ( fft (x ).* fft (h)) yc =

6 Linear convolution and the DFT The convolution between x(n) and h(n) yields y(n) of length Choose a DFT length of N = 8. h(k) 1 1 x(k) y L (k) >> x = [1 3 4]; >> h = [ 1 1]; >> yl = ifft ( fft (x,8).* fft (h,8)) yl = The linear convolution is closely related to the circular convolution. y C (n) = { y L (0) + y L (4) y L (1) + y L (5) y L () + y L (6) y L (3) + y L (7) = { = { (7) (8) (9) The linear convolution wrap around modulo-4 and sum to form the circular convolution. Sampling the spectrum Let x(n) = a n u(n) X(ω) = 1 1 ae jω (30) Read X(ω) at N frequencies and form X(k) = X(ω ω = πk/n). The signal x(n) is an infinitely long sequence, but the invers-dft of X(k) yields a finitely long sequence of length N. What is x DFT (n) = IDFT{X(k)? X(k) = 1 jπ 1 ae kn (31) 6

7 Inverse-DFT yields: x DFT (n) = 1 N 1 N X(k)e jπ nn k (3) = 1 N 1 N = 1 N = 1 N = m= m= m= m= The signal x(n) extends to +. x(m)e jπ mn k e jπ nn k (33) N 1 n m jπ x(m) e N k (34) x(m) N δ(n m mod N) (35) x(n mn) (36) Time [n] The signal x DFT (n) is the sum of all shifted and repeated x(n). The DFT size is N = x DFT (n) Time [n] Periodicity in time A square pulse of length L is x(n) = { (37) 7

8 and its DTFT and DFT are L 1 X(ω) = x(n)e jωn = sin( ) ω L n=0 sin ( L 1 jω ) e ω 1 (38) and L 1 X(k) = n=0 x(n)e jπ kn n = sin( π L N k ) sin ( π 1 N k ) e jπ L 1 k (39) Let Y 1 (k) = X(k) y1(n) = IDFT{Y 1 (k) (40) and Y (k) = X(k) X(k) y1(n) = IDFT{Y (k) = x(n) x(n) (41) >> N = 16; >> L = 6; % or L =10 >> k = 0:N -1; >> k(k ==0) = sqrt ( eps ); >> X = sin (* pi*l*k /(* N ))./ sin (* pi*k /(* N )).* exp (-j ** pi *(L -1)* k /(* N )); >> y1 = real ( ifft (X )); >> y = real ( ifft (X.*X)); What does y 1 (n) and y (n) look like for N = 16 and for L = 6 or L = 10? >> subplot (,1,1); stem (k, y1 ); >> subplot (,1,); stem (k, y ); A practical application Increase the resolution in frequency by zero padding or trailing zeros. Let x(n) = { (4) Calculate the DFT of length N = 8 of x(n). 4 Frequency [f]

9 Calculate the DFT of length N = 16 of x(n). 4 Frequency [f] The DFT on matrix form (page ) Define W N = e jπ/n = W (43) The DFT and the IDFT becomes N 1 X(k) = x(n)w kn n=0 (44) x(n) = 1 N 1 N X(k)W kn (45) Let and x = X = [ x(0) x(1) x(n 1)] T (46) [ T X(0) X(1) X(N 1)] (47) and W W W N 1 D = W N 1 W (N 1) W (N 1)(N 1) (48) We can now describe the DFT and the IDFT as X = Dx (49) 9

10 and x = D 1 X or x = 1 N D X (50) Some identities: D 1 = 1 N D (51) DD = N I (5) 10