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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 openloop 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 closedloop 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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 Winter '09
 IDAN

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