midterm_solution - Digital Signal Processing I (ECE 410,...

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Digital Signal Processing I (ECE 410, Summer II 2009) University of Illinois at Urbana-Champaign Electrical and Computer Engineering Midterm I J. W. Choi and P. Linden 1. (a) Non-causal since the output depends on the future input. (b) Time-invariant since x [ n - n 0 ] k =1 h [ k ] x [ n - n 0 - k ] = y [ n - n 0 ]. (c) Not stable since the inner most poles are ± 1 10 j and hence the ROC does not include the unit circle. 2. (a) X d (0) = X n = -∞ x [ n ] e - jωn | ω =0 = X n = -∞ x [ n ] = X n =0 ± 1 2 2 k + ± - 1 4 + 1 4 + 3 4 + 5 4 = 1 1 - 1 2 + 2 = 4 (b) X d ( π ) = X n = -∞ x [ n ] e - jωn | ω = π = X n = -∞ x [ n ]( - 1) n = X n =0 ± 1 2 2 k - ± - 1 4 + 1 4 + 3 4 + 5 4 = 2 - 2 = 0
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(c) Using Parseval’s theorem X n = -∞ | x [ n ] | 2 = 1 2 π Z π - π | X d ( ω ) | 2 Z π - π | X d ( ω ) | 2 = 2 π X n = -∞ | x [ n ] | 2 = 2 π X n = -∞ ± 1 2 2 n 2 + ± 1 16 + 1 16 + 9 16 + 25 16 = 2 π ˆ X n =0 ± 1 4 n + 1 16 9 4 # = 2 π ± 3 4 + 9 4 = 2 π × 16 + 27 12 = 2 π × 43 12 = 43 6 π (d) Z π - π Y d ( ω ) = 2 πy [0] (1) To obtain y [0], we express the output y [ n ] as y [ n ] = h [ n ] * x [ n ] = X k = -∞ x [ k ] h
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This note was uploaded on 01/22/2011 for the course ECE 410 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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midterm_solution - Digital Signal Processing I (ECE 410,...

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