EECE 340 Chapter 12, Fall 2011

Chapter 12 The Discrete-Time Fourier Series (DTFS) Definition Consider a Discrete-Time signal x[n] of period N or frequency Ω0=(2π/N). The DTFS representation of x[n] is defined as: Where We say that x[n] and X[k] are a DTFS pair ∑ − = − Ω = 1 N 0 n jk 0n x[n e] N 1 [k] X ∑ − = Ω = 1 N 0 k X[k e] jk 0n x[n] Example Let x[n] be a periodic Discrete-time signal of period N=5. Over one period x[n] is defines as: Determine the Frequency-domain representation of x[n] Answer: ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎨ ⎧ = = = = − = = 4 n 5.0 3 n 0 2 n 0 1 n 5.0 0 n 1 [n] x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = = = − − = − − = − Ω ∑ ∑ 5 8 5 2 4 0 5 2 1 0 5.0 5.0 5 1 1 [ ] 5 1 [ ] 1 ] [ 0 π π π jk jk n n jk N n jk n e e x n e x n e N X k EXAMPLE Determine the DTFS coefficients of Solution: ⎟⎠ ⎞ ⎜⎝ ⎛ π + π = + 8 3 12 sin n 1 [n] x ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎨ ⎧ ≤ ≤ − = = = − − = π − π − 12 k 11 0 1 k 2j e 0 k 1 1 k 2j e X[k] / 8 3j / 8 3j EXAMPLE Find the DTFS coefficients of the N-periodic impulse train Solution: Since there is only one nonzero value in x[n] per period, it is convenient to evaluate X[k] over the interval n = 0 to n = N - 1 to obtain ∑ ∞ =−∞ − δ = � � ] N [ n [n] x N 1 [n e] N 1 X[k] 1 N 0 n jkn2 / N = δ = ∑ − = π − EXAMPLE: The Inverse DTFS One period of the DTFS coefficients of a signal is given by Find the discrete- time domain signal x[n] assuming N=10. Answer: 9 k 0 2 1 X[k] k ≤ ≤ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) / 5)n (j 10 1( / 2) 1 1/ 2 1 x[n] π − − = EXAMPLE: DTFS Representation of a Square wave Find the DTFS coefficients for the N-periodic square wave given by That is, each period contains 2M+1 consecutive ones and the remaining {N-(2M+1)} values are zero. Note that this definition requires that N > 2M + 1. ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ≤ ≤ − = Over the rest of the period 0 M n M 1 x[n] Solution The period is N, so Ω0=2π/N. It is convenient in this case to evaluate the DTFS of x[n] over indices n = -M to n = N - M - 1. We thus have We perform the change of variable on the index of summation by letting m=n+M to obtain ∑ ∑ − = Ω − − − − = Ω − = = M M n n jk M 1 N M n n jk 0 0 e N 1 x[n e] N 1 [k] X ∑ ∑ = Ω − Ω = − − Ω = = M 2 0 m 0m jk 0M jk M 2 0 m jk 0 (m M) e N e e N 1 X[k] Solution Now, for k=0, ±N, ±2N, ±3N, …, we have for k=0, ±N, ±2N, ±3N, …, we may sum the geometric series to obtain 1 e e 0 jk jk 0 = = Ω − Ω 0, N, 2N, ... for k N 1 2M 1 N 1 X[k] M 2 m 0 ± ± = + = = ∑ = ( ) ( / 2) sin k 2 /)1 (2M sin k N 1 e 1 e 1 N e X[k] 0 0 jk )1 (2M jk M jk 0 0 0 Ω + Ω ⎟ = ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = Ω − + Ω − Ω Solution Substituting Ω0=2π/N by its value in the equation we obtain ( ) ( / 2) sin k 2 /)1 (2M sin k N 1 e 1 e 1 N e X[k] 0 0 jk )1 (2M jk M jk 0 0 0 Ω + Ω ⎟ = ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = Ω − + Ω − Ω ( ) ( sin k / N) N /)1 sin k (2M N 1 X[k] π + π = Solution The above equation is also valid for k=0, ±N, ±2N, ±3N, … using L”Hopital’s rule. Convolution of periodic signals � This subsection addresses the convolution of two signals that are periodic functions of time. The convolution of periodic signals does not occur naturally in the context of evaluating the input- output relationships of systems, since any system with a periodic impulse response is unstable. However, the convolution of periodic signals often occurs in the context of signal analysis and manipulation. � The discrete-time convolution of two N-periodic sequences x[n] and z[n] is defined as: ∑ − = − = ⊗ = 1 N k 0 k] x[k [z] n [z n] x[n] [y n] Convolution of periodic signals � This is the periodic convolution of x[n] and z[n]. The signal y[n] is N periodic, so the DTFS is the appropriate representation for all three signals: x[ n], z[n], and y[n]. � Thus, convolution of time signals transforms to multiplication of DTFS coefficients X[k] Z[k] Y[k] [z n] x[n] [y n] N DTFT , 2 ⋅ = ⎯ ⎯⎯ → ← ⊗ = π