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homework5_solution.pdf - Solutions to Homework 5 ECE 460...

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Solutions to Homework 5 ECE 460: Automatic Control Due: Monday, February 22, at 2:10pm Note 1. Your solutions should be neat and legible; always show your work. Problem 1. (30 points) Given the system shown in Figure 1, find the following. (a) The closed-loop transfer function (b) The system type (c) The steady-state error for an input of 5 u ( t ). (d) The steady-state error for an input of 5 tu ( t ). (e) The closed-loop poles. (f) Discuss the validity of your answers to part (c) and (d) . Figure 1 Solution: (a) For the inner loop: G 1 ( s ) = 1 s 2 ( s +1) 1 + 1 s 3 ( s +1) = s s 4 + s 3 + 1 The equivalent open-loop transfer function G e ( s ) is then given by G e ( s ) = 1 s 2 ( s + 3) G 1 ( s ) = 1 s ( s 5 + 4 s 4 + 3 s 3 + s + 3) The closed-loop transfer function is given by T ( s ) = G e ( s ) 1 + G e ( s ) = 1 s 6 + 4 s 5 + 3 s 4 + s 2 + 3 s + 1 (b) From inspection of G e ( s ), we can say that K p = 1; K v = constant and K a = 0. Thus the system is Type 1. 1
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(c) Since the system is type 1, for r ( t ) = 5 u ( t ), e ss = e (1) = 0. (d) From G e ( s ), K v = lim s 0 sG e ( s ) = 1 / 3. Therefore, e ss = 5 /K v = 15.
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