HW8sol - ECEN 314: Signals and Systems Solutions to HW 8 1....

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Unformatted text preview: ECEN 314: Signals and Systems Solutions to HW 8 1. Note that H ( s ) = 10 s +10 . The steady state response is then: y ( t ) = 2 | H ( j 10) | cos(10 t + ∠ H ( j 10)) = 1 . 414cos(10 t- 45 ◦ ) 2. Note that the period of x ( t ) is T = 1. The simplest way to do this problem is as follows: 2cos10 πt + sin6 πt = e j 10 πt + e- j 10 πt + 1 2 j e j 6 πt- e- j 6 πt ¶ = e j 2 π (5) t + e j 2 π (- 5) t + 1 2 j e j 2 π (3) t- e j 2 π (- 3) t ¶ One can easily see that x 5 = x- 5 = 1 and x 3 =- x- 3 =- j/ 2. By direct inspection of x ( t ) one easily gets a v = 0, a 5 = 2 and b 3 = 1. We can also verify the above result by using the standard integrals as follows: x k = 1 / 2 Z- 1 / 2 x ( t ) e- j 2 πkt dt = 1 / 2 Z- 1 / 2 e j 10 πt + e- j 10 πt + 1 2 j ( e j 6 πt- e- j 6 πt ) ¶ e- j 2 πkt dt = sin π (5- k ) π (5- k ) + sin π (5 + k ) π (5 + k ) + j 2 sin π (3 + k ) π (3 + k )- sin π (3- k ) π (3- k ) ¶ = 1 k = ± 5 j/ 2 k =- 3- j/ 2 k = 3...
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This note was uploaded on 01/01/2012 for the course ECEN 314 taught by Professor Halverson during the Spring '08 term at Texas A&M.

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HW8sol - ECEN 314: Signals and Systems Solutions to HW 8 1....

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