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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.
 Spring '08
 HALVERSON
 Steady State

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