Chapter 8 The Discrete Fourier Transform ◆8.0 Introduction 8.1 Representation of Periodic Sequence:the Discrete Fourier Series(DFS) 8.2 Properties of the DFS 8.3 The Fourier Transform of Periodic Signal 8.4 Sampling the Fourier Transform(DTFT) 8.5 Fourier Representation of Finite-Duration Sequence:the Discrete Fourier Transform(DFT) 8.6 Properties of the DFT 8.7 Linear Convolution using the DFT 8.8 the discrete cosine transform (DCT
2 Chapter 8 The Discrete Fourier Transform ◆8.0 Introduction ◆8.1 Representation of Periodic Sequence: the Discrete Fourier Series (DFS) ◆8.2 Properties of the DFS ◆8.3 The Fourier Transform of Periodic Signal ◆8.4 Sampling the Fourier Transform(DTFT) ◆8.5 Fourier Representation of Finite-Duration Sequence: the Discrete Fourier Transform(DFT) ◆8.6 Properties of the DFT ◆8.7 Linear Convolution using the DFT ◆8.8 the discrete cosine transform (DCT)
8.1 Representation of Periodic Sequence: the Discrete Fourier Series Given a periodic sequence [n]with period N so that [n]=n +rN] The Fourier series representation can be written as (defined as): 网=表∑]e2wm Fourier series representation of continuous-time periodic signals require infinitely many complex exponentials, These complex exponential sequences are periodic g2k+m0n=e8e2m=e/3,k=01,2,N-1
6 ◆Fourier series representation of continuous-time periodic signals require infinitely many complex exponentials, ◆These complex exponential sequences are periodic : 8.1 Representation of Periodic Sequence: the Discrete Fourier Series ( ) 2 j k N N m n e + ◆The Fourier series representation can be written as 2 j n N k e = ( ) 2 2 j n N k j mn e e = (defined as): (2 / ) ( ) ( ) , j t k T k k x t x t a T e + =− + = = 1 (2 / ) ( ) jk T t k T T a x t e dt − =
8.1 Representation of Periodic Sequence: the Discrete Fourier Series )m) 2元 JN ,k=0,1,2,…,W-1 Due to periodicity,we only need N complex exponentials for Fourier series representation of [n]: 网∑[e祭n =之[]e: The Fourier series representation of periodic sequence [n]
7 8.1 Representation of Periodic Sequence: the Discrete Fourier Series ◆Due to periodicity,we only need N complex exponentials 1 0 2 1 [ ] j N N k kn x n X k e N − = = 2 1 [ ] N k j kn x n X k e N = , 0,1,2, , 1 k N = − ( ) 2 j k N N m n e + 2 j n N k e = ( ) 2 2 j n j m k N n e e =
Discrete Fourier Series Pair 测=太2[] 27 kn ◆The Fourier series: N k二0 To obtain the Fourier series coefficients,we multiply both sides by e-j(2/N)rn f for0sn≤N-1, and then sum both the sides we obtain e废2ee n=0 -2p4"方k位动e
8 Discrete Fourier Series Pair ◆The Fourier series: 1 0 2 [ ] N n j n N r x n e − = − 0 1 1 0 2 ( ) [ ] 1 N n N k j k r n N N X k e − − = = − = ◆To obtain the Fourier series coefficients , we multiply both sides by for 0nN-1, and then sum both the sides , we obtain j n (2 / ) N r e − 1 1 0 0 2 ( ) [ ] 1 n N k N j n N k r N X k e − − = = − = 1 0 2 1 [ ] j N N k kn x n e X k N − = = 1 0 1 0 2 2 [ 1 ] N k N n j n j nr N N k X k N e e − − = = − = X k
Proof of orthogonality of the complex exponentials ”=1 1-e2π(k-r） N 1-e() 1, k-r=mN,m is an integer Problem 0, otherwise 8.54 =∑6[k-r-mW]=p[k-r N points k-小 K-r -N-N+1·-2-101 2 N-1NN+1N+2. the periodic impulse train
9 - - m k r r k mN p =− = − = 1 2 ( ) 0 2 - ( ) 2 ( ) 1 1 1 1 N j k n r j k r N j k n r N e N N e e − = − − − = − Proof of orthogonality of the complex exponentials 1, - , is 0, k r mN m aninteger otherwise = = k-r 0 1 -N -N+1…… -2 -1 2 …… N-1 N N+1 N+2 …… N points 1 pk -r the periodic impulse train Problem 8.54
Discrete Fourier Series Pair k-r)n 1,k-r=mN,mis an integer 10,otherwise Using orthogonality of the complex exponentials: =]=[k灯 n=0 k=0 0≤r,k≤W-1 0only when k-r=0 (m=0,k=r) Xk-岁ae n=0 D5之a coefficients The Discrete Fourier Series Periodic 10
0 only when k − =r 0 10 Discrete Fourier Series Pair 1 0 2 1 ( ) N n j k r n N N e − = − = = X X [ ] [ ] r k 1 0 2 [ ] N n j n N r x n e − = − 0 1 1 0 2 ( ) [ ] 1 N n N k j k r n N N X k e − − = = − = 1, - , 0, k mN mis aninteger othe e r rwis = = (2 ) 1 / 0 1 [ ] j N N k kn x n e X N k − = = 1 0 2 [ ] N n j n N k x n e − = − X[ ] k = The Discrete Fourier Series coefficients Periodic Using orthogonality of the complex exponentials: DFS 0 1 − r k, N ( 0, ) m k = = r
8.1 Representation of Periodic Sequence:the Discrete Fourier Series a periodic sequence with period N, n =n+rN]for any integer r The Discrete Fourier Series: (defination) =∑[]e京 Synthesis N k=0 equation Coefficients:(is periodic with period N) k]=∑nle0u、 N一 Analysis equation n=0
11 8.1 Representation of Periodic Sequence: the Discrete Fourier Series ◆a periodic sequence x n with period N, x n x n rN for any integer r = + ◆The Discrete Fourier Series: 1 0 2 [ ] [ ] , N n j kn X k x n N e − = − = 1 0 2 1 [ ] , N k j kn N x n X k N e − = = Synthesis equation Analysis equation ◆Coefficients: (defination) (is periodic with period N) (
8.1 Representation of Periodic Sequence:the Discrete Fourier Series X[]=∑[n]e2xa n=0 The sequence]is periodic with period N [o]=[],[]=[N+] k+N-立远ee n=( n=0 经e如eP和-[个 = 17
12 8.1 Representation of Periodic Sequence: the Discrete Fourier Series ◆The sequence is X k periodic with period N X X X X 0 , 1 1 = = + N N ( ) 1 0 N 2 n j k n N N X k x n N e − = − + + = 1 0 2 N 2 n j kn N j n x n X k e e − = − − = = ( ) 1 0 2 N n j N kn X k x n e − = − = 1 0 N 2 2 n j k n n j N N N x n e e − = − − =
Discrete Fourier Series (DFS) []=∑n]e ◆Let Wv=e N ◆Analysis equation:[k]=∑[nW n=0 ◆Synthesis equation:]-=∑[m 三0 like CFS periodic discrete diserete periodicnk periodic discrete discrete like DTFT periodic 13
13 Discrete Fourier Series (DFS) ◆Let 2 N j W N e − = 1 0 N kn N n X k x n W − = = DFS x n X k ◆Analysis equation: 1 0 1 N kn N k x n X k W N − − = ◆Synthesis equation: = 1 0 N 2 n j kn N X k x n e − = − = periodic discrete like CFS discrete like DTFT periodic discrete periodic periodic discrete (2 / ) ( ) ( ) , j t k k T k x t x t a e T + =− + = =
Ex.8.1 determine the DFS of a impulse train Consider the periodic impulse train -之-则-6wm N points -N-N+1·-2-101 2·N-1NN+1N+2 Solution:X[附-∑sm%==for all k W-e彩 n 大名刚县*=大艺 orthogonality 14
14 Ex. 8.1 determine the DFS of a impulse train ◆Consider the periodic impulse train 1, , 0, r n rN r is any integer x n n rN otherwise =− = = − = n -N -N+1…… -2 -1 0 1 2 …… N-1 N N+1 N+2 …… N points xn ~ 1 1 0 0 1 , N kn N n N X k nW W for all k − = Solution: = = = 1 0 1 N kn N k x n X k W N − − = = 0 1 2 1 N j k N n k e N − = = orthogonality 2 N j W e N − = 1 0 1 N kn N k W N − − = =