PracticeProblemsMidterm_solu

Solution since the system is type 1 the steady state

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Unformatted text preview: and Q(s) is Type 3 30. What are the steady state errors for the step, ramp, and parabolic reference inputs? Assume that a 6= 0, a 6= 10, and the system is stable. Solution: Since the system is Type 1, the steady state errors are zero for the step input and infinity for the parabolic input . For the ramp input, e(1) = lim sQ(s) s! 0 1 10 a = 2 s 10a 31. Assume a 6= 0 and H (s) = 1, and find the error constants. Solution: The open-loop transfer function is L(s) := G(s)D(s) = a + bs . s(s + 1) Then, by definition of the error constants, we have Kp := lim L(s) = 1, s! 0 Kv := lim sL(s) = a, s! 0 Ka := lim s2 L(s) = 0. s! 0 8 32. Assume a 6= 0 and H (s) = 1, and find the steady state errors for the step, ramp, and parabolic inputs. Solution: Step: e(1) = 1/(1 + Kp ) = 0, Ramp: e(1) = 1/Kv = 1/a, Parabola: e(1) = 1/Ka = 1. 33. Assume that the closed-loop system is stable, r = v = 0, and a 6= 0. Which of the following signals converge(s) to zero in response to a step input in w? (A) u (B) y (C) x (D) None of these (E) All of these Solution: T...
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