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HW13f08solutions

# HW13f08solutions - Georgia Institute of Technology School...

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Georgia Institute of Technology School of Electrical and Computer Engineering ECE 2025 Fall 2008 Solutions for Problem Set #13 13.1 (a) i. Using the following trigonometric identity: cos 2 ( θ ) = 1 + cos(2 θ ) 2 x ( t ) = 3 2 (1 + cos(34 t )) X ( ) = 3 2 (2 πδ ( ω ) + πδ ( ω - 34) + πδ ( ω + 34)) ii. x ( t ) = δ ( t + 1) - 2 e - t u ( t - 1) = δ ( t + 1) - ( 2 e - 1 ) e - ( t - 1) u ( t - 1) X ( ) = e - ( 2 e - 1 ) e - 1 + = e - 2 e - (1+ ) 1 + iii. x ( t ) = d dt sin(17 π ( t - 2)) 2 πt - 4 π = 1 2 d dt sin(17 π ( t - 2)) π ( t - 2) X ( ) = 1 2 ( )( e - 2 ) [ u ( ω + 17) - u ( ω - 17)] = ωe - j (2 ω - π/ 2) [ u ( ω + 17) - u ( ω - 17)] 2 iv. x ( t ) = u ( t - 1) - u ( t - 8) = u t + 7 2 - 4 . 5 - u t - 7 2 - 4 . 5 X ( ) = e - j (4 . 5) ω sin(7 ω/ 2) ω/ 2 v. X ( ) = 14 πδ ( ω ) + 10 π e - jπ/ 3 δ ( ω - 4) + e jπ/ 3 δ ( ω + 4) (b) i. Using the following trigonometric identity: cos 2 ( θ ) = 1 + cos(2 θ ) 2 X ( ) = 1 + cos(6 ω ) 2 = 1 2 + 1 4 ( e j 6 ω + e - j 6 ω ) x ( t ) = 1 2 δ ( t ) + 1 4 ( δ ( t + 6) + δ ( t - 6)) ii. X ( ) = 2 + - 1 2 + x ( t ) = d dt ( e - 2 t u ( t ) ) - e - 2 t u ( t ) = ( - 2 e - 2 t u ( t ) + e - 2 t δ ( t ) ) - e - 2 t u ( t ) = δ ( t ) - 3 e - 2 t u ( t ) 1

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iii. X ( ) = 13 π { δ ( ω - 3 π ) - δ ( ω + 3 π ) } x ( t ) = - sin(3 πt ) π/ 13 iv. X ( ) = u ( ω + π ) u ( π - ω ) e - jω/ 10 = [ u ( ω + π ) - u ( ω - π )] e - jω/ 10 x ( t ) = sin( π ( t - 1 / 10)) π ( t - 1 / 10) = sin( πt - π/ 10) πt - π/ 10 v.
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HW13f08solutions - Georgia Institute of Technology School...

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