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**Unformatted text preview: **Digital Signal Processing I (ECE 410, Summer II 2009) University of Illinois at Urbana-Champaign Electrical and Computer Engineering Midterm 2 Solution 1. (a) (7 points) See the attached pages. (b) (7 points) See the attached pages. (c) (6 points) First, we shift X d ( ) right by 3 8 , i.e., w [ n ] = e j 3 8 x [ n ] due to modulation property. Hence, = 3 8 . Then, we pass w [ n ] through the LPF with c = 8 , i.e., g [ n ] = sin ( 8 ) n . You can obtain the desired output after D/A conversion. 2. (a) (8 points) Since H d (0) 6 = 0, h [ n ] is type I, odd N generalized linear-phase (GLP) filter. Hence, we can write h [ n ] = { h ,h 1 ,h 2 ,h 1 ,h } . From the given conditions, h [0] = h = 1 2 Z - H d ( ) d = 1 (1) H d (0) = 4 X n =0 h [ n ] = 2 h + 2 h 1 + h 2 = 4 (2) H d (0) = 4 X n =0 h [ n ](- 1) n = 2 h- 2 h 1 + h 2 = 4 . (3) From (1), (2), and (3), we have h = 1, h 1 = 0, and h 2 = 2. The impulse response h [ n ] is { 1 , , 2 , , 1 } ....

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