5_HWSolution

# 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: 11 22 00 1 () (), NN nk x nX k N −− == = ∑∑ () we start from the R.H.S of (7.1) and proceed as 2 * * () () 12 ( ) ( )exp( ), kk kn Xk XkX k nk x n j π = = () where in (7.2), we use the relationship given by DFT. Interchanging the order of the summation, we can write 111 2 * 000 1 2 0 2 ( ) ( ) ( )exp( ) NNN kkn N k nk j N xn −−− === = = ⎩⎭ = ∑∑∑ () 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 /2 1 0 2 ()e x p 2 () (1 ) , 2 N nk Nk nk N nn n N k N n jn k N X kx n x n W W W N N x n W = = ⎛⎞ =− = + + ⎜⎟ ⎝⎠ ⎧⎫ + ⎨⎬ () where we have used 2 exp N j W N ± and ( ) exp ( 1) Nk k N Wj k = −= . Considering only the even values of k , i.e., k =2 m for m= 0,1,…,N/2, (7.9) can be written as 0 (2 ) ( ) , for 0,1, , / 2 1 2 N nm N n N Xm x n x nW m N = =+ + = . ( Similarly, for the odd values of k , (7.9) can be represented as 0 (2 1) ( ) 0,1, , / 2 1 2 N m n N x nWW m N = += + = . ()

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Therefore, if we define 01 () a n d 22 n N NN x nx n x n x n W ⎧⎫ ⎛⎞ ++ −+ ⎨⎬ ⎜⎟ ⎝⎠ ⎩⎭ ±± , we observe that X ( k ) can be realized as the N /2 point DFT of 0 x n and 1 x n length- N /2 sequences. Note that, N/2 complex multiplications are required to obtain 1 x n . The Flow-graph of this reduced complexity DFT implementation is shown in the Figure 7.1
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## This note was uploaded on 11/02/2009 for the course EEE DSP taught by Professor Ap during the Fall '09 term at ASU.

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

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