Discrete Fourier transform (DFT)

Transcription

1 Discrete Fourier transform (DFT) Signal Processing 2008/9 LEA Instituto Superior Técnico Signal Processing LEA (IST) Discrete Fourier transform 1 / 34

2 Periodic signals Consider a periodic signal x[n] with period ( x[n + r] = x[n] for r Z) This signal is completely determined by samples according to x[n] = 1 m=0 x[m] p [n m] p [n] = + r= δ[n r] is a train of impulses spaced by samples Example (Discrete pulse train) The discrete impulse train p [n] is shown below for = 4 0 n Signal Processing LEA (IST) Discrete Fourier transform 2 / 34

3 Periodic signals Theorem (Harmonic representation of a discrete impulse train) The discrete impulse train p [n] can be represented as Proof 1 1 k=0 But W n Hence p [n] = 1 W kn = 1 1 k=0 1 k=0 W kn, with W = e j 2π ( W n = e j 2π (n) = 1 and W n 1 1 W kn = k=0 ) k = + r= 1 1 W n 1 W n,if W n 1 1,if W n = 1 = ej 2π n = 1 n = r (r Z). δ[n r] = p [n] Signal Processing LEA (IST) Discrete Fourier transform 3 / 34

4 Periodic signals Using this representation for p [n] in the expression of x[n] on page 2: x[n] = 1 m=0 = 1 x[m] p [n m] = 1 1 k=0 m=0 x[m]w km } {{ } X[k] 1 m=0 W kn x[m] ( 1 1 k=0 W k(n m) ) Remark Signal X[k] is also periodic, with period Signal Processing LEA (IST) Discrete Fourier transform 4 / 34

5 Discrete Fourier series (DFS) Definition (Discrete Fourier series (DFS)) Given a periodic sequence x[n] (n Z), the DFS associates it with another sequence X[k] (k Z) having the same period X[k] = 1 n=0 x[n] = 1 x[n]w kn 1 k=0 X[k]W kn [Analysis] [Synthesis] otation: x[n] DFS X[k], X[k] = DFS { x[n]}, x[n] = DFS 1 { X[k] } Signal Processing LEA (IST) Discrete Fourier transform 5 / 34

6 Interpretation of the DFS Remark (Interpretation of the DFS) [Synthesis] x[n] is a finite linear superposition of complex exponentials [Analysis] X[k] measures the importance of frequency ω k = k 2π (k = 0,1,..., 1) in x[n] Signal Processing LEA (IST) Discrete Fourier transform 6 / 34

7 Properties of the DFS Linearity: α 1 x 1 [n] + α 2 x 2 [n] DFS α 1 X1 [k] + α 2 X2 [k], α 1,α 2 C Time shift: Frequency shift: x[n n 0 ] DFS W kn 0 X[k], n 0 Z W k 0n x[n] DFS X[k k 0 ], k 0 Z Duality: X[n] DFS x[ k] Signal Processing LEA (IST) Discrete Fourier transform 7 / 34

8 Properties of the DFS Periodic convolution: 1 m=0 h[m] x[n m] }{{} ( x h)[n] DFS X[k] H[k], x, h : period Multiplication in time: x[n] h[n] DFS 1 1 l=0 }{{} 1 ( X H)[k] H[l] X[k l], x, h : period Signal Processing LEA (IST) Discrete Fourier transform 8 / 34

9 Relation between the DFS and FT: Periodic signals Question For a signal x[n] with period, how is its Fourier transform X(e jω ) related to the DFS X[k]? Intuitively, both should contain the same information a spectral representation of x[n] Signal Processing LEA (IST) Discrete Fourier transform 9 / 34

10 Relation between the DFS and FT: Periodic signals From x[n] = 1 m=0 x[m] p [n m] the FT X(e jω ) is given by X(e jω ) = = 1 m=0 1 m=0 = 2π = 2π x[m]e jωm P (e jω ) x[m]e jωm ( 2π 1 m=0 + x[m] 1 k= m=0 + k= + k= ( δ ω k 2π ) ) ( e jωm δ ω k 2π ) }{{} x[m]w km } {{ } X[k] e j 2π km δ(ω k 2π ) δ ( ω k 2π ) Signal Processing LEA (IST) Discrete Fourier transform 10 / 34

11 Relation between the DFS and FT: Periodic signals Conclusion Up to a factor 2π the DFS coefficients coincide with the magnitudes of Dirac impulses, evenly-spaced by 2π, that modulate the FT X(e jω ) Example X(e jω ) for = 3 2π 3 X[ 1] 2π 3 X[0] 2π 3 X[1] 2π 3 X[2] 2π 3 X[3] 2π 3 X[4] 0 2π 3 ω Signal Processing LEA (IST) Discrete Fourier transform 11 / 34

12 Relation between the DFS and FT: Finite-duration signals Consider a signal x[n] with length (x[n] = 0 outside the interval 0 n 1) Build its periodic extension x[n] = (x p )[n] = + r= x[n r] Example for = 3 x[n] x[n] 0 n 0 n Question How is the FT X(e jω ) related to the DFS X[k]? Signal Processing LEA (IST) Discrete Fourier transform 12 / 34

13 Relation between the DFS and FT: Finite-duration signals From x[n] = (x p )[n] the FT X(e jω ) is given by X(e jω ) = X(e jω ) P (e jω ) = X(e jω ) = 2π = 2π + k= + k= ( 2π ( X(e jω )δ ω k 2π X ) + k= ( ) ( e jk 2π δ ω k 2π ) ( δ ω k 2π ) ) Comparing with the representation of X(e jω ) on page 10 yields ( ) X[k] = X e jk 2π Signal Processing LEA (IST) Discrete Fourier transform 13 / 34

14 Relation between the DFS and FT: Finite-duration signals Conclusion The DFS coefficients X[k] of the periodic extension of x[n] are samples of the TF X(e jω ) taken along the unit circle and equispaced by 2π rad Example for = 8 k =..., 5,3,11,... k =..., 4,4,12,... k =..., 6,2,10,... 2π k =..., 7,1,9,... k =..., 8,0,8,... k =..., 3,5,13,... k =..., 1,7,15,... k =..., 2,6,14,... Signal Processing LEA (IST) Discrete Fourier transform 14 / 34

15 Relation between the DFS and FT: Arbitrary signals Consider an arbitrary signal x[n] with FT X(e jω ) Build the periodic signal, with period, ( ) X[k] = X e jk 2π, k Z Question { X[k] } How is x[n] = DFS 1 related to x[n]? Remark We know the answer if the duration of x[n] is smaller than : x[n] is the periodic extension of x[n] Signal Processing LEA (IST) Discrete Fourier transform 15 / 34

16 Relation between the DFS and FT: Arbitrary signals Computing x[n]: x[n] = 1 = 1 (p. 3) = 1 k=0 1 k=0 + m= X[k]W kn ( + = (x p )[n] = + r= m= x[m] x[n r] = 1 1 k=0 X x[m]e jk 2π m ) 1 1 W k(n m) k=0 }{{} p [n m] ( ) e jk 2π W kn W kn Signal Processing LEA (IST) Discrete Fourier transform 16 / 34

17 Relation between the DFS and FT: Arbitrary signals Conclusion x[n] is a sum of replicas of x[n], spaced by samples Corollary (Regeneration of signals) If x[n] has finite duration, then it can be recovered from x[n] (take one period) If x[n] has duration > then there is aliasing in the time domain and in general it cannot be recovered from x[n] If x[n] has finite duration, aliasing can be avoided by choosing sufficiently large Essentially, this is a sampling theorem in the frequency domain Signal Processing LEA (IST) Discrete Fourier transform 17 / 34

18 Discrete Fourier transform (DFT) Definition (Discrete Fourier transform (DFT)) Given a finite-duration signal x[n] (n = 0,1,..., 1), the DFT associates it with the finite-duration sequence X[k] (k = 0, 1,..., 1) X[k] = 1 n=0 x[n] = 1 x[n]w kn 1 k=0 X[k]W kn [Analysis] [Synthesis] otation: x[n] DFT X[k], X[k] = DFT {x[n]}, x[n] = DFT 1 {X[k]} Signal Processing LEA (IST) Discrete Fourier transform 18 / 34

19 Interpretation of the DFT Remark (Interpretation of the DFT) [Synthesis] x[n] is a finite linear superposition of complex exponentials with finite duration [Analysis] X[k] measures the importance of frequency ω k = k 2π (k = 0,1,..., 1) in x[n] Signal Processing LEA (IST) Discrete Fourier transform 19 / 34

20 Relation between the DFT and DFS Consider a signal x[n] with finite duration and its periodic extension x[n] = (x p )[n] x[n] can also be represented as x[n] = x[ n ] otation: n = n mod Examples: 3 3 = 0, 2 3 = 1, 1 3 = 2, 1 3 = 1, 3 3 = 0, 4 3 = 1 Relation between the DFT and DFS x[n] DFT X[k] truncate 1 period truncate 1 period x[n] periodical extension DFS X[k] periodical extension Signal Processing LEA (IST) Discrete Fourier transform 20 / 34

21 Properties of the DFT Remark Below, denotes the DFT length is not smaller than the original signal length (typically equal) Linearity: Circular shift: α 1 x 1 [n] + α 2 x 2 [n] DFT α 1 X 1 [k] + α 2 X 2 [k], α 1,α 2 C, max{length x 1, length x 2 } x[ n n 0 ] DFT W kn 0 X[k], n 0 Z Signal Processing LEA (IST) Discrete Fourier transform 21 / 34

22 Properties of the DFT Example (Circular shift for = 4) x[n] x[ n 1 4 ] n 0 0 n x[ n 2 4 ] x[ n ] 0 n 0 n Signal Processing LEA (IST) Discrete Fourier transform 22 / 34

23 Properties of the DFT Frequency shift: W k 0n x[n] DFT X[ k k 0 ], k 0 Z Duality: X[n] DFT x[ k ] Signal Processing LEA (IST) Discrete Fourier transform 23 / 34

24 Properties of the DFT Circular convolution: 1 m=0 Multiplication in time: h[m]x[ n m ] } {{ } (x h)[n] DFT X[k]H[k] x[n]h[n] DFT 1 1 H[l]X[ k l ] l=0 }{{} 1 (X H)[k] Signal Processing LEA (IST) Discrete Fourier transform 24 / 34

25 Relation between the DFT and FT: Finite-duration signals Consider a signal x[n] with length (x[n] = 0 outside the interval 0 n 1) Question How is the FT X(e jω ) related to the DFT X[k]? We have: Conclusion X[k] = 1 n=0 x[n]w kn = + = X(e jω ) ω=k 2π n= x[n]w kn = + n= x[n]e jk 2π n The DFT coefficients X[k] are samples of the TF X(e jω ) taken along the unit circle and equispaced by 2π rad Signal Processing LEA (IST) Discrete Fourier transform 25 / 34

26 Relation between the DFT and FT: Arbitrary signals Consider an arbitrary signal x[n] with FT X(e jω ) Define a finite duration signal x [n] from the set of DFT coefficients X [k] = X(e jω ) ω=k 2π, k = 0,1,..., 1 Question How is x [n] = DFT 1 {X [k]} related to x[n]? Remark We know the answer if x[n] is a finite duration signal whose length does not exceed (hence x [n] = x[n]) Signal Processing LEA (IST) Discrete Fourier transform 26 / 34

27 Relation between the DFT and FT: Arbitrary signals Using the diagram of page 20 and the result of page 17 x [n] = + r= x[n r], n = 0,1,..., 1 Conclusion x [n] is a sum of replicas of x[n], spaced by samples, and restricted to the time interval n = 0,1,..., 1 Signal Processing LEA (IST) Discrete Fourier transform 27 / 34

28 Linear and circular convolution 1 Consider finite-duration signals x[n] with length L and h[n] with length P 2 Let y[n] = (x h)[n] be their linear convolution, which has length L + P 1 3 ow choose max{p,l} and define the circular convolution y [n] = (x h)[n] Question How is y [n] related to y[n]? Motivation Very efficient algorithms are available (FFT) for computing y [n] Signal Processing LEA (IST) Discrete Fourier transform 28 / 34

29 Linear and circular convolution ote that Y [k] = DFT {y [n]} = X[k]H[k] = X(e jω )H(e jω ) ω=k 2π Using the result of page 27 we have = Y (e jω ) ω= y [n] = DFT 1 {Y [k]} = Conclusion (Important!) If L + P 1 then + r= y[n r] (x h)[n] = (x h)[n] for n = 0,1,..., 1 If = L P then (x h)[n] = (x h)[n] for n = P,P + 1,..., 1 Signal Processing LEA (IST) Discrete Fourier transform 29 / 34

30 Conclusion (Linear convolution from circular convolution) Choosing P + L 1 the linear convolution y[n] = (x h)[n] can be computed as follows: 1 Compute the -point DFT of x[n]: X[k] = DFT {x[n]} 2 Compute the -point DFT of h[n]: H[k] = DFT {h[n]} 3 Multiply in the frequency domain: Y [k] = X[k]H[k], k = 0,1,..., 1 4 Invert the DFT: y[n] = DFT 1 {Y [k]}, n = 0,1,..., 1 Remark Very efficient algorithms are available for computing DFT { } and DFT 1 { } When L P (or L = + ) use time-recursive versions of the above procedure: overlap-add and overlap-save Signal Processing LEA (IST) Discrete Fourier transform 30 / 34

31 Overlap-add In the overlap-add method x[n] is viewed as a concatenation of disjoint blocks x[n] = L/M 1 r=0 x r [n rm] where x r [n] denotes the signal in one of the blocks (of size M) Compute each convolution y r [n] = (x r h)[n] by the method of page 30 Combine the partial results to get y[n] = (x h)[n] = L/M 1 r=0 y r [n rm] Signal Processing LEA (IST) Discrete Fourier transform 31 / 34

32 Overlap-add Example (Overlap-add) h[n] has length P = 3 and x[n] has length L = 18 The block size is chosen as M = 6 h[0] = 1, h[1] = 1, h[2] = 1 The DFT size must satisfy M + P 1 = 8. It is chosen as = 8 in this example Signal Processing LEA (IST) Discrete Fourier transform 32 / 34

33 Overlap-add Example (Overlap-add (cont.)) 3 x[n] 3 x 0 [n] x 1 [n] x 2 [n] Signal Processing LEA (IST) Discrete Fourier transform 33 / 34

34 Overlap-add Example (Overlap-add (cont.)) y 0 [n] y 1 [n M] y 2 [n 2M] y[n] Signal Processing LEA (IST) Discrete Fourier transform 34 / 34