114_1_sol_midterm_2008_Part1

114_1_sol_midterm_2008_Part1 - UCLA Dept. of Electrical...

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UCLA Dept. of Electrical Engineering EE 114, Winter 2008 Midterm Exam Solutions 1. ( a ) Let x ( n ) = [ a ;b;c;d ]. The 4-point DFT is determined as: X ( k ) = N ¡ 1 X n =0 x ( n ) e ¡ j 2 …kn=N (1) X (0) = N ¡ 1 X n =0 x ( n ) = 0 : 5( a + b + c + d ) (2) X (1) = N ¡ 1 X n =0 x ( n ) e ¡ j…k= 2 = 0 : 5 / ( a ¡ jb ¡ c + jd ) (3) X (2) = N ¡ 1 X n =0 x ( n ) e ¡ j…k = 0 : 5 / ( a ¡ b + c ¡ d ) (4) X (3) = N ¡ 1 X n =0 x ( n ) e ¡ j… 3 k= 4 = 0 : 5 / ( a + jb ¡ c ¡ jd ) (5) (i) If x ( n ) is assumed even and periodic with period 4, then x ( n ) = x ( N ¡ n ). This implies that b = d . This leads to: X (0) = ( a + 2 b + c ) (6) X (1) = ( a ¡ c ) (7) X (2) = ( a ¡ 2 b + c ) (8) X (3) = ( a ¡ c ) (9) (ii) If x ( n ) is assumed odd and periodic with period 4,then x ( n ) = ¡ x ( N ¡ n ). This implies b = ¡ d , c = ¡ c , and a = ¡ a . This leads to c = 0 and a = 0, and further: X (0) = 0 (10) X (1) = ¡ j 2 b (11) X (2) = 0 (12) X (3) = j 2 b (13) 1
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2. ( a ) Let x ( n ) = [3 ; ¡ 2 ; 1 ; 4] and y ( n ) = [4 ; 4 ; 2 ; 2 ; 1 ; 1]. x ( n ) / y ( n ) = [12 ; 4 ; 2 ; 22 ; 17 ; 11 ; 7 ; 5 ; 4] (14) ( b ) Let x ( n ) = [2 ; ¡ 2 ; 1 ; ¡ 1] and y ( n ) = [ ¡ 4 ; 4 ; ¡ 2 ; 2 ; ¡ 1 ; 1]. C xy ( k ) = [2 ; ¡ 4 ; 7 ; ¡ 10 ; 15 ; ¡ 20 ; 14 ; ¡ 8 ; 4] : (15) 2
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3. The given tube model can be decoupled into three separate sources of resonance: the back cavity, the front cavity, and the Helmholtz resonance due to the coupling of the two.
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This note was uploaded on 11/12/2011 for the course EE 114 taught by Professor Vanschaar during the Spring '11 term at UCLA.

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114_1_sol_midterm_2008_Part1 - UCLA Dept. of Electrical...

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