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HW5sol - ECEN 314 Signals and Systems Solutions to HW 5...

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ECEN 314: Signals and Systems Solutions to HW 5 Problem 2.11 (a) Since h ( t ) is non-zero only for 0 t ≤ ∞ , we have y ( t ) = x ( t ) * h ( t ) = Z τ =0 e - 3 τ ( u ( t - τ - 3) - u ( t - τ - 5) ) Note that ( u ( t - τ - 3) - u ( t - τ - 5) ) is non-zero only for ( t - 5) < τ < ( t - 3). Therefore, for t 3, the above integral evaluates to zero. For 3 < t 5, the above integral is: y ( t ) = Z t - 3 0 e - 3 τ = 1 - e - 3( t - 3) 3 . For t > 5, we have: y ( t ) = Z t - 5 t - 3 e - 3 τ = (1 - e - 6 ) e - 3( t - 5) 3 . Putting the above together, we get y ( t ) = 0 -∞ < t 3 1 - e - 3( t - 3) 3 , 3 < t 5 (1 - e - 6 ) e - 3( t - 5) 3 5 < t ≤ ∞ (b) Differentiating x ( t ) we get dx ( t ) dt = δ ( t - 3) - δ ( t - 5). Therefore g ( t ) = dx ( t ) dt * h ( t ) = e - 3( t - 3) u ( t - 3) - e - 3( t - 5) u ( t - 5) (c) From (a), we compute the derivative of y ( t ) to be dy ( t ) dt = 0 -∞ < t 3 e - 3( t - 3) , 3 < t 5 ( e - 6 - 1) e - 3( t - 5) 5 < t ≤ ∞ Problem 2.17 (a) Note that y ( t ) is the sum of the particular and homogenous solution to the given differen- tial equation. We first determine the particular solution y p ( t ) by using the method specified in Example 2.14. Since the input x ( t ) = e ( - 1+3 j ) t u ( t
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