630F09_hw1_sol - ENEE 630 F’2009 Homework Solution #1...

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Unformatted text preview: ENEE 630 F’2009 Homework Solution #1 Problem 1 solution: x(n) 3 2 3 relatively prime 2 y(n) x(n) 2 3 3 2 y(n) doing nothing x(n) 2 2 y(n) y [n] = Problem 2 solution: we know Y (ω ) = X (ωL) 1 [1 + (−1)n ]x[n] 2 For y0 [n], since Hk (ω ) = π [H0 (ω − Y0 (ω ) = π [H0 (ω − 2πk L )+ H0 (ω + 2πk L )], 2πk 2πk ) + H0 (ω + )] · X (ωL) L L For y1 [n], U (ω ) = X (ωL)H0 (ω ), with the modulation process, Y1 (ω ) = π [U (ω − 2πk 2πk ) + U (ω + )] = Y0 (ω ) L L |Y(ω)| 0 2π/5 2π ω 1 |U(ω)| 0 π/5 2π ω |Y(ω)| 0 0 π/5 3π/5 2π ω Problem 3 solution: Y0 (z ) = Hk (z )Y (z ) Y1 (z ) = π [H0 (zW k )Y (zW k ) + H0 (zW −k )Y (zW −k )] They are not in general equal. If Y (z ) = X (z L ), then Y0 (z ) = Y1 (z ) Problem 4 solution: x(n) H(z) 2 H(z) 2 s(n) 4 y(n) H (ejω ) 1 H (ejω ) X (e ) jω X (e ) -π 0 jω 0.25 0.1875 |Y(ω)| -π 0 π ω -π 0 π ω π ω -π 0 π/4 πω 2 ...
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630F09_hw1_sol - ENEE 630 F’2009 Homework Solution #1...

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