Solutions of Final Exam
Subject : Control System Engineering 1, Lecturer : Prof. Youngjin Choi, Date : June 17, 2015 (Contact e-mail :xxxx@xxxxang.ac.kr)
Problem 1 (20pt) Consider the following control system
(1.1) Find the position error constant Kp (1.2) Find the velocity error constant Kv (1.3) Find the acceleration error constant Ka (1.4) Find the steady-state error ess of the system when the input is X(s) =
5 2s
3 s2
+
4 s3
Solution of Problem 1 (20pt) 1. Since the forward-path TF is G(s) = we have Kp = lim G(s) = ¡Ä
s¡æ0
1 s2 4 s2 = s+2 s2 s2
Kv = lim sG(s) = ¡Ä
s¡æ0 s¡æ0
Ka = lim s2 G(s) = 2 2. The input can be decomposed as X(s) = Also, When the unit-step input is applied X1 (s) When the unit-ramp input is applied X2 (s) When the parabolic input is applied X3 (s) ¡æ ¡æ ¡æ 1 =0 1 + Kp 1 e2 (¡Ä) = =0 Kv 1 e3 (¡Ä) = = 0.5 Ka e1 (¡Ä) = 3 5 4 5 + 3 = X1 (s) 3X2 (s) + 4X3 (s) 2s s2 s 2
Thus the steady-state error becomes ess = e(¡Ä) = 2.5e1 (¡Ä¡¦(»ý·«)
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