HW3_solution

# HW3_solution - MAE107 Homework#3 Solution Prof M’Closkey Problem 1 Consider ∞ −∞ Set s = t − τ to get −∞ ∞ h(t − τ)u(τ)dτ

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MAE107 Homework #3 Solution Prof. M’Closkey Problem 1 Consider, Z −∞ h ( t τ ) u ( τ ) dτ. Set s = t τ to get Z −∞ h ( s ) u ( t s )( ds )= Z −∞ h ( s ) u ( t s ) ds. Note that we can use any symbol for the variable of integration. Problem 2 The deFnition of ˜ h is ˜ h ( t Z −∞ h 2 ( t τ ) h 1 ( τ ) dτ, which, using the results of Problem 1, is also, ˜ h ( t Z −∞ h 1 ( t τ ) h 2 ( τ ) dτ, ±rom the problem statement, y 2 ( t Z −∞ h 2 ( t τ ) y 1 ( τ ) = Z −∞ h 2 ( t τ | {z } defne v ) ± Z −∞ h 1 ( τ s ) u 1 ( s ) ds ² = Z −∞ h 2 ( v ) ± Z −∞ h 1 ( t v s ) u 1 ( s ) ds ² dv = Z −∞ u 1 ( s ) Z −∞ h 1 ( t v s ) h 2 ( v ) | {z } ˜ h ( t s ) dv ds (switch order of integration) = Z −∞ u 1 ( s ) ˜ h ( t s ) ds Thus, the impulse response of the series connection of two subsystems is the convolution of their impulse responses. 1

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Another solution. Let u 1 ( t )= δ ( t ), then y 1 = h 1 .S inc e u 2 = y 1 ,then y 2 ( t Z −∞ h 2 ( t τ ) u 2 ( τ ) = Z −∞ h 2 ( t τ ) h 1 ( τ ) dτ. is the impulse response of the series connection. Problem 3 Using the deFnition of the derivative ˙ y ( t ) = lim Δ 0 1 Δ ± Z −∞ h ( t τ ) u ( τ ) Z −∞ h ( t Δ τ ) u ( τ ) ² = lim Δ 0 1 Δ Z −∞ ( h ( t τ )
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## This note was uploaded on 10/09/2009 for the course MAE 107 taught by Professor Tsao during the Winter '06 term at UCLA.

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HW3_solution - MAE107 Homework#3 Solution Prof M’Closkey Problem 1 Consider ∞ −∞ Set s = t − τ to get −∞ ∞ h(t − τ)u(τ)dτ

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