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HW10sol

# HW10sol - ECEN 314 Signals and Systems Solutions to HW 10...

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ECEN 314: Signals and Systems Solutions to HW 10 1. (4.3)(a) Note that the period of x 1 ( t ) = sin(2 πt + π 4 ), is T = 1. Hence the fundamental frequency, ω o = 2 π . Also x 1 ( t ) = 1 2 j ( e j (2 πt + π 4 ) - e - j (2 πt + π 4 ) ) = 1 2 j e jπ/ 4 e j (2 πt ) - 1 2 j e - jπ/ 4 e j ( - 2 πt ) Therefore, a 1 = 1 2 j e jπ/ 4 , and a - 1 = - 1 2 j e - jπ/ 4 . The corresponding Fourier transform X 1 ( ) of x 1 ( t ) is given by (Refer Sec 4.2 of the text), X 1 ( ) = 2 πa 1 δ ( ω - ω o ) + 2 πa - 1 δ ( ω + ω o ) = ( π/j ) e jπ/ 4 δ ( ω - 2 π ) - ( π/j ) e - jπ/ 4 δ ( ω + 2 π ) (b) Note that the period of x 2 ( t ) = 1 + cos(6 πt + π 8 ), is T = 1 / 3. Hence the fundamental frequency, ω o = 6 π . Also x 2 ( t ) = 1 + 1 2 ( e j (6 πt + π 8 ) + e - j (6 πt + π 8 ) ) = 1 + 1 2 e jπ/ 8 e j (6 πt ) + 1 2 e - jπ/ 8 e j ( - 6 πt ) Therefore, a 0 = 1 , a 1 = 1 2 e jπ/ 8 , and a - 1 = 1 2 e - jπ/ 8 . The corresponding Fourier transform X 2 ( ) of x 2 ( t ) is given by, X 2 ( ) = 2 πa 0 δ ( ω ) + 2 πa 1 δ ( ω - ω o ) + 2 πa - 1 δ ( ω + ω o ) = 2 πδ ( ω ) + πe jπ/ 8 δ ( ω - 6 π ) + πe - jπ/ 8 δ ( ω + 6 π ) 2. (4.4) (a) The inverse Fourier transform is x 1 ( t ) = 1 2 π Z -∞ £ 2 πδ ( ω ) + πδ ( ω - 4 π ) + πδ ( ω + 4 π ) / e jωt =

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HW10sol - ECEN 314 Signals and Systems Solutions to HW 10...

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