EE518 Digital Signal Processing University of Washington Autumn 2001 Dept. of Electrical Engineering Lecture 19: Discrete Fourier Series Dec 5, 2001 Prof: J. Bilmes <bilmes@ee.washington.edu> TA: Mingzhou Song <msong@u.washington.edu> 19.1 Discrete Fourier Series DTFT x[n] 1 On a computer (MATLAB), how dow we represent? n π x[n]e jωn π X(e jω )e jωn dω X(e jω ) is defined on continuous domain. So can not be represent exactly in general. x[n] is infinite in length so it is impossible to deal with. Must do: 1) assume x[n] is finite length (as typically occurs in practice) 2) x[n] is periodic extension of x[n], so we can analyze x[n] to gain properties of x[n]. DFS applies to x[n]. DFT applies to x[n]. But they are basically the same thing. (except that there are some circular things we need to work with. Question: (to ponder) Could we perhaps sample X(e jω ) at a sufficiently high enough rate so to get, say X[k] when X(e jω ) is recoverable? Answer: Yes. Ok: Consider x[n] periodic x[n] x[n + r]. Recall Fourier series For discrete time signals, period, we have f (t) F[k] 1 T F[k]e jkt/t k T /2 T /2 f (t)e jkt/t dt X[k]e j(/)kn 1 k X[k]e k [n] (19.1) k e k [n] e j(/)kn e k [n + r] (19.2) 19-1
19-2 But note: e k+l [n] e j(/)(k+l)n e j(/)kn So we need only sum over values, i.e., 1 X[k]e j(/)kn ote: 1 x[n]e j(/)rn 1 1 1 X[k]e j(/)(k r)n 1 X[k] [ 1 1 e j(/)(k r)n 1 X[k]δ[k r m] X[r] 1 X[k] ] by orthogonality of sinusoids x[n]e j(/)kn (19.3) X[k] is periodic, too. 1 X[k + ] 1 X[k] x[n]e j(/)(k+)n x[n]e j(/)kn e jn Define W e j(/) (19.4) DFS 1 Analysis Equation X[k] Synthesis Equation ote: can think of them as finite length sequence: of length, or as periodic with period. Ex: x[n] r 1 [k] 1 x[n] { 1, n r δ[n r] 0, else δ[n]w kn W 0 1 W kn 1 1 e j(/)kn x[n]e j(/)kn (19.5) 1 X[k]e j(/)kn (19.6)
19-3 Ex: x[n] 10, x[n] r x[n r] { 1, 0 n 9 0, else X[k] 9 W10 kn 9 e j(/10)kn 1 W 5k 10 j(4kπ/10) sin(πk/2) 1 W10 k e sin(πk/10) 19.2 Properties of DFS 1) Linearity x 1 [n] DFS X 1 [k], x 2 [n] DFS X 2 [k] a x 1 [n] + b x 2 [n] DFS a X 1 [k] + b X 2 [k] 2) Shift x[n m] DFS W km X [ k] ote, same for any ˆm m modulo M (reminder when divided by M) 3) Duality x[n m] DFS W k ˆm X[k] x[n] DFS X[k] But analysis and synthesis equation are almost the same (1/ and sign change) Consider Switch symbol n and k, the above can be written as 1 X[k]W kn 1 x[ n] X[k]W kn 1 x[ k] X[n]W kn So, we can say x[n] DFS X[k], X[n] DFS x[ k] 4) Periodic Convolution We can show X[k] X 1 [k] X 2 [k] 1 x[n] m0 It might spill over into next period for some n and m. x 1 [m] x 2 [n m]
19-4 Intuition X[k] 1 1 m0 1 m0 ( 1 m0 x 1 [m] x 1 [m]w km X 1 [k] X 2 [k] x 1 [m] x 2 [n m] ) W kn ( ) 1 x 2 [n m]w kn X 2 [k] Since analysis and synthesis equations are so similar, we also have Other Properties See O&S table 8.1, they are very similar to DTFT. x 1 [n] x 2 [n] DFS 1 1 X 1 [l] X 2 [k l] l0 Recall DT FT {x[n] e jω 0n }? ote x[n] is neither absolute summable nor square summable. Define r δ(ω ω 0 + r) 1 π δ(ω ω 0 )e jωn dω e jω 0n π For general periodic time signals x[n], consider representation by DFS X[k], the impulse train at harmonically related frequency starting at fundamental frequency /. Why? k 1 π X(e jω )e jωn dω 1 π π π 1 1 x[n] k X[k]δ(ω k ) k X[k] X[k]δ(ω k )e jωn dω π π X[k]e j(/)kn k δ(ω k )e jωn dω Ex: Important Interpretation of X[k]. since ρ[n] r ρ[k] 1, δ[n r] k ρ(e jω ) k k δ(ω )
19-5 Consider x[n] x[n] ρ[n] x[n] X(e jω ) ρ(e jω ) X(e jω ) k k k r δ[n r] k δ(ω ) r X(e j2(π/)k )δ(ω k ) X[k]δ(ω k ) x[n r] from definition X[k] X(e j2(π/)k ) X(e jω ) ω(/)k which is sampling of DTFT, i.e., X[k] from x[n] is equivalent to sampling of X(e jω ) from x[n], which is one period of x[n] from n 0 to n 1. ote also: (another way to look at it) n x[n]e jωn 1 x[n]e jωn So sampling the DTFT X(e j2(π/)k 1 ) x[n]e j2(π/)kn 1 x[n]w kn X[k] X(e jω ) ω(/)k X(z) ze j2(π/)k DFS analysis The sampling frequencies are points equally distributed on the unit circle in the z-plane. Question: What would x[n] be from this X[k]? How would x[n] be related to x[n] from X(e jω )? n x[n]e jωn X[k] X(e j(/)k ) Recall from previous example So x[n] 1 1 1 m m X[k]W kn [ ] x[m]e j2(π/)km W kn m ] x[m] [ 1 1 W k(n m) ρ[n m] DFS W km x[m] ρ[n m] x[n] r δ[n r] ote sampling X(e jω ) with equally spread samples. If the length of x[n] is larger than, there is time-aliasing, which is dual of frequency aliasing. Suppose x[n] of length is to be reconstructed from samples in frequency domain.
19-6 Case 1: Sample X(e jω ) with points. o time aliasing and x[n] can be reconstructed. Case 2: Sample X(e jω ) with less than points. Time aliasing will occur and x[n] can not be constructed. Case 3: Sample X(e jω ) with larger than points. o time aliasing but zero will be padded with every period of. Key points: - X(e jω ) for finite length sequence is redundant in that we can sample it and recover the full signal x[n]. Oversampling implies extra space in time. - for X( jω) finite length (finite bandwidth) can sample the time signal x(t) at a fast enough rate to get x[n] and fully recover x(t). Oversampling means space in frequency. The above forms a duality.