8.2 Properties of the Discrete Fourier series ◆8.2.1 Linearity two periodic sequence, both with period DES aIn]e>Xik DES x2nk x2k DES ax, In+ 饭>Xk]+bX2]
23 8.2 Properties of the Discrete Fourier Series ◆8.2.1 Linearity ◆two periodic sequence, both with period N: x n X k DFS 1 1 ~ ~ x n X k DFS 2 2 ~ ~ ax n bx n aX k bX k DFS 1 2 1 2 ~ ~ ~ ~ + +
8.2 Properties of the Discrete Fourier series 8.2.2 Shift of a sequence DES <> DES xn-m]>WwmX[k], W M=e-/(2z/N) DES W动女>[k=],Wn=e G(2/N)nl
24 8.2 Properties of the Discrete Fourier Series ◆8.2.2 Shift of a sequence xn Xk DFS ~ ~ DFS km N x n m W X k − ， − − DFS nl W x n X k l N ， j( N ) N W e − 2 = nl j N nl (2 ) W e N − =
8.2 Properties of the Discrete Fourier series ◆8.23 Duality DES DES x<> Nxl-k xn 1X[小 n 2 N-1 2 N-1 Nxl-k k N-1 2 N-1 8.2.4 Symmetry Properties 25
25 8.2 Properties of the Discrete Fourier Series ◆8.2.3 Duality xn Xk DFS ~ ~ Xn Nx k DFS − ~ ~ 0 1 2 …… N-1 X n n 1 1 X k 0 1 2 …… N-1 k xn ~ 0 1 2 …… N-1 n 1 0 1 2 …… N-1 Nx k − k N ◆8.2.4 Symmetry Properties
8.2.4 Symmetry TABLE 8.1 SUMMARY OF PROPERTIES OF THE DFS Periodic sequence(Period N) DFS Coefficients(Period N) X[一k 10.x[-n X KI 11.Re对 Problem8.53, HW Elk】=(X]+'[-k 12.过m([nl] 文。Dk1=3(文伙1-X[一k 13.en]=号([n]+x*[-n]) ReX [kl] 14. on]=Gi[n]miNd jIm(X [I Properties 15-17 apply only when x[n I is real XIk]=X[k] Re(X联k]=Re{X[一k] 15. Symmetry properties for x[n] real 工m(X]}=-mX[-k]] Xk]=|X[一kl Xk]=-∠X[-k 16.Ren]=5(ln]+x-n]) Rex [kll 17.G0l=3(xn-x-n jIm(X[kl) 26
26 8.2.4 Symmetry Problem 8.53, HW
8.25 Periodic Convolution v四andx团] are two periodic sequences, each with period n and with discrete Fourier series N andx,文]=]Fk] N then[n]=∑[m]nm∑无[m]1[n-m Proof: 0 X[]=∑[川W=∑∑无m[n-m如 n=0 n=0m=0 ∑m∑利mmW=∑年mW如m2[k m=0 n=0 m=0 N-1 ∑[m团=]] m=0 27
27 8.2.5 Periodic Convolution ◆ and are two periodic sequences, each with period N and with discrete Fourier series and , if x n 1 ~ x n 2 ~ X k 1 ~ X k 2 ~ X k X kX k 3 1 2 ~ ~ ~ = 1 1 3 1 2 2 1 0 0 N N m m x n x m x n m x m x n m − − = = = − = − 1 1 1 3 3 1 2 0 0 0 N N N kn kn N N n n m X k x n W x m x n m W − − − = = = = = − ( ) 1 1 1 1 2 1 2 0 0 0 N N N kn km N N m n m x m x n m W x m W X k − − − = = = = − = 1 1 2 1 2 0 N km N m x m W X k X k X k − = = = then Proof: