# HW3sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 3 Solution Prof. Bresler, Prof. Jones September 17, 2008 Problem 1 (10 points) Let X ( k ) denotes the N-point DFT of the discrete-time signal x [ n ] (0 n N - 1) (a) Show that X (0) = 0 if x [ n ] = - x [ N - 1 - n ]: X [ k ] = N - 1 X n =0 x [ n ] W nk N , 0 k N - 1 where W N = e - j 2 π N X [0] = N - 1 X n =0 x [ n ] W 0 N = ( x [0] + x [ N - 1]) | {z } 0 +( x [1] + x [ N - 2]) | {z } 0 + ... = 0 Note that according to the given condition if N is even, x £ N - 1 - N 2 / + x £ N 2 / = 0; and if N is odd, x £ N - 1 2 / = - x £ N - 1 2 / = 0. (b) Given N is an even number, show that X ( N/ 2) = 0 if x [ n ] = x [ N - 1 - n ]: X [ N 2 ] = N - 1 X n =0 x [ n ] e - jπn = ( x [0] - x [ N - 1]) | {z } 0 +( - x [1] + x [ N - 2]) | {z } 0 + ... = 0 Problem 2 (25 points) Let x [ n ] be a discrete-time sequence: x [ n ] = ( - 1) n , 0 n 3 0 , otherwise (a) The analytical expression for DTFT of x [ n ]: X d ( ω ) = 3 X n =0 ( - 1) n e - jωn 1

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When ω [0 , 2 π ] , ω 6 = π : X d ( ω ) = 1 - e - j 4 ω 1 + e - = e - j 3 2 ω · j · sin(2 ω ) cos( ω 2 ) = e - j ( 3 ω 2 - π 2 ) · sin(2 ω ) cos( ω/ 2) When ω = π : X d ( ω ) = 3 X n =0 ( - 1) n e - jπn = 4 So: X d ( ω ) = ( e - j ( 3 ω 2 - π 2 ) · sin(2 ω ) cos( ω/ 2) ω [0 , 2 π ] , ω 6 = π 4 ω = π Plot its magnitude and phase for 0 ω < 2 π : 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 1 2 3 4 ω / (2 π ) |X d ( ω )| DTFT of x[n] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -3 -2 -1 0 1 2 3 ω / (2 π ) phase of X ) in radians (b) Compute the 4-point DFT of x [ n ] , 0 n < 4 and stem plot its magnitude and phase: 2
0 0.5 1 1.5 2 2.5 3 3.5 4 0 1 2 3 4 magnitude of 4-point DFT of x m |X[m]| 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 -0.5 0 0.5 1 angle of 4-point DFT of x m radians How can you relate the plots of part (b) to the plots of part (a)? Answer: The plots in part (b) (DFT) are the sampled version of the plots in part (a) (DTFT). In particular the 4 points of the DFT correspond to digital frequencies, 0 ,π/ 2 and 3 π 2 (c) Compute the 8-point DFT of x [ n ] and stem plot its magnitude and phase: 0 1 2 3 4 5 6 7 8 0 1 2 3 4 magnitude of 4-point DFT of x m 0 1 2 3 4 5 6 7 8 -2 -1 0 1 2 angle of 4-point DFT of x m Comment on the eﬀect of zero-padding the signal on its DFT: Answer: The zero-padding operation gives a more densely spaced spectrum of x [ n ]; the DTFT 3

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sampled at digital frequencies, 0 ,π/ 4 ,π/ 2 , 3 π/ 4 ,π, 5 π/ 4 , 3 π/ 2 and 7 π/ 4. However, it does not provide any new information since adding zeros does not change the original signal x [ n ]. (d) Compute the 128-point DFT of x [ n ] , 0 n < 128 and plot its magnitude and phase: 0 20 40 60 80 100 120 0 1 2 3 4 magnitude of 4-point DFT of x m |X[m]| 0 20 40 60 80 100 120 -4 -2 0 2 4 angle of 4-point DFT of x m radians (e) Compare your result for part (d) to the plots of part (a). Answer: They are the same, as expected.
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HW3sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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