Discrete Fourier Series (DFS) 对区]=∑动 e J/(2T/Nkn .Let n- Analysis equation:]=∑p n=0 ◆ Synthesis equation:x]=∑Xk k=0 F DES 团平[] discrete<> periodic F periodic<> discrete
13 Discrete Fourier Series (DFS) ◆Let 2 N j W N e − = − = = 1 0 ~ ~ N n kn n WN X k x xn Xk DFS ~ ~ periodic discrete F discrete periodic F ◆Analysis equation: − = − = 1 0 ~ 1 ~ N k kn WN X k N ◆Synthesis equation: x n ( ) 1 0 2 N n j N kn X k x n e − = − =
Ex, 8 1 determine the des of a impulse train Consider the periodic impulse train 1, n=rN, r is any integer SIn-rN 0. otherwise N points xn -N-N+1 -2-10 2∴……N-1NN+1N+2 Solution =∑如=W 0
14 Ex. 8.1 determine the DFS of a impulse train ◆Consider the periodic impulse train = = − = =− otherwise n rN r is any integer x n n rN r 0, 1, , ~ n 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points xn ~ 1 ~ 0 1 0 = = = − = N N n kn X k n WN W Solution:
Ex.8.1 DFS of a impulse train N points -N-N+1 2-101 2 ·N-1NN+1N+2 [k]=∑0[可]=W points k -N-N+1…-2 2 N-1NN+1N+2 ]=∑[n-N n=rN, r is any integer otherwise ,(2x/N) N ∑x[ k=0 N k=0 15
15 Ex. 8.1 DFS of a impulse train 1 0 0 1 N kn N N n X k n W W − = = = = 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points X k ~ k 1 0 1 N kn N k X k W N − − = = 1, , 0, n rN r is any integer otherwise = = r x n n rN =− = − 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points xn ~ ( ) 1 2 0 1 N j N kn k e N − = =
oints x n 0 2 N-1NN+1N+2 n points N-N+1…-2 0 2……N-1NN+1N+2 16
16 n 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points 1 xn ~ k X k ~ 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points 1
Example 8.2 Duality in the Discrete Fourier series If the discrete fourier series coefficients is the periodic impulse train, determine the signal N points ∑Nk-mN]N [] r三-00 N∴-2-1012..N Solution =1∑kj kn drs N 米]∑x[ k=0 到]=∑Y=∑N收=W=1
17 Example 8.2 Duality in the Discrete Fourier Series ◆If the discrete Fourier series coefficients is the periodic impulse train, determine the signal. =− = − r Y k N k rN ~ 1 ~ 1 ~ 1 0 1 0 1 0 = = = = − = − − = − N N k kn N N k kn N N k W W N Y k W N y n − = − = 1 0 ~ 1 ~ N k kn WN X k N x n − = = 1 0 ~ ~ N n kn n WN X k x 0 1 2 … … N … -2 -1 … … -N … N points N Y k Solution: DFS