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hw3soln_f09

# hw3soln_f09 - ME 132 Fall 2009 Solutions to Homework 3...

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ME 132 Fall 2009 Solutions to Homework 3 1. ( § 3.3, Problem 3) (a) The time constant will be the same regardless of the input u ( t ). However, for the steady state gain γ , assume that the input is a constant. As given in the notes, the response of the system to a step input, u ( t ) = ¯ u for t 0, is y ( t ) = e at y (0) - b a ( 1 - e at ) ¯ u. If a< 0, then the steady-state output is ¯ y = - b a ¯ u. The time constant, τ , and the steady-state gain, γ , are given by τ = - 1 a , γ = - b a . Rearranging these yields a = - 1 τ , b = γ τ . (b) The response of the system to the given input is shown in Figure 1. The output y ( t ) always exponentially approaches a value of γu ( t ) — that is, the input multiplied by the steady-state gain. The rate of the exponential convergence is determined by the time constant, τ . During each period of one time constant, the output gets 63% closer to the steady-state value. (c) The response of the system to the given input is shown in Figure 2. The response is identical in shape to that of the previous system. This is because this system is the same as the previous system with time compressed (“sped up”) by a factor of 1000. The new time constant is 1000 times smaller than the old time constant, so the system responds 1000 times faster. Additionally, the input signal is the same except 1000 times faster. Therefore, the behaviour of the system is the same, only 1000 faster. 2. ( § 3.3, Problem 4) We will solve this problem both of the ways that the statement suggested. (Convolution integral) For the linear, time-invariant system ˙ x ( t ) = A [ x ( t ) - u ( t )] 1

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0 2 4 6 8 10 12 14 16 18 20 2 2.5 3 3.5 4 4.5 5 5.5 6 Time y(t) (b) System with τ =1, γ =2 Figure 1: System with τ = 1, γ = 2. 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 2 2.5 3 3.5 4 4.5 5 5.5 6 Time y(t) (c) System with τ =0.001, γ =2 Figure 2: System with τ = 0 . 001, γ = 2. 2
the solution (starting from t 0 = 0 with x (0) = x 0 ) is x ( t ) = e At x 0 + integraldisplay t 0 e A ( t τ ) ( - A ) u ( τ ) = e At x 0 - integraldisplay t 0 e A ( t τ ) Aτdτ = e At x 0 - e At A integraldisplay t 0 e τdτ = e At x 0 - e At A integraldisplay t 0 e τdτ = e At x 0 - e At A parenleftbigg - t A e At + 1 A integraldisplay t 0 e parenrightbigg = e At x 0 + t + 1 A - 1 A e At = ( x 0 + τ c ) e At + ( t - τ c ) (Satisfaction of ODE and initial condition) The initial condition can be verified by plugging in t = 0. x (0) = ( x 0 + τ c ) e A 0 + (0 - τ c ) = x 0 + τ c - τ c = x 0 We can verify that the given solution satisfies the ODE by directly plugging it in. First, it will be useful to compute ˙ x ( t ).

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hw3soln_f09 - ME 132 Fall 2009 Solutions to Homework 3...

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