5_HWSolution - PBM 7.1 To derive Parsevals theorem for the...

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PBM 7.1 To derive Parseval’s theorem for the DFT: 1 1 2 2 0 0 1 ( ) ( ) , N N n k x n X k N = = = () we start from the R.H.S of (7.1) and proceed as 1 1 2 * 0 0 1 1 * 0 0 1 1 ( ) ( ) ( ) 1 2 ( ) ( )exp( ), N N k k N N k n X k X k X k N N nk X k x n j N N π = = = = = = () where in (7.2), we use the relationship given by DFT. Interchanging the order of the summation, we can write 1 1 1 2 * 0 0 0 1 2 0 1 1 2 ( ) ( ) ( )exp( ) ( ) , N N N k k n N k nk X k x n X k j N N N x n π = = = = = = () where we have used the IDFT expression. This concludes the proof. PBM 7.4 We start with the DFT expression: 1 / 2 1 / 2 1 / 2 0 0 0 / 2 1 0 2 ( ) ( )exp ( ) 2 ( ) ( 1) , 2 N N N nk Nk nk N N N n n n N k nk N n j nk N X k x n x n W W x n W N N x n x n W π = = = = = = + + = − − + () where we have used 2 exp N j W N π ± and ( ) / 2 exp ( 1) Nk k N W j k π = = − . Considering only the even values of k , i.e., k =2 m for m= 0,1,…,N/2, (7.9) can be written as / 2 1 / 2 0 (2 ) ( ) , for 0,1, , / 2 1 2 N nm N n N X m x n x n W m N = = + + = . ( Similarly, for the odd values of k , (7.9) can be represented as / 2 1 / 2 0 (2 1) ( ) , for 0,1, , / 2 1 2 N n nm N N n N X m x n x n W W m N = + = + = .
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