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w 21. Sketch the root locus for the characteristic equation of the system for
which
(5 + 1) s(s + 1)(s + 2)’ and determine the value of the rootlocus gain for which the complex
conjugate poles have a damping ratio of 0.5. . L(s) = I Solution: This must be a typo! The roots at —1 cancel and the second order system
will have damping of 0.5 at K = 4. A more interesting case occurs for
num = 3 +3. In this case, the roots are at —.42+j.7 and the gain is 0.47 22. For the system in Fig. 5.69: I i ’
$5 Figure 5.69: Feedback system for problem 5.22 lLUUh lUUUﬁ HILL]. ﬁhEP lﬁprﬂﬁb‘ iUl ILUULUJJJ. {LLL 23. Fer the feedback system shown in Fig. 5.70, ﬁnd the value of the gain K
that results in dominant cieeed—leep poles with a damping ratio C : 0.5. Figure 5.70: Feedback 'eystem fer Preblern 5.23 Solution:
10.3 The root locus is for 13(5) = m . the required gain is K = 0.216 28. A numerically controlled machine tool positioning servomechanism has a
normalized and scaled transfer function given by l
s(s+l)' C(s) 2 Performance speciﬁcations of the system in the unity feedback conﬁgu—
ration of Fig. 5.71 are satisﬁed if the closedloop poles are located at
s = —1 :l: NE. (a) Show that this speciﬁcation cannot be achieved by choosing propor
tional control alone, D(s) = kp. 'l“ _ 5+2 (b) Design a lead compensator D(s) = K 8 + that will meet the speci—
P ﬁcation. (a) With proportional control, the poles have real part at s = —.5. (b) To design a lead, we select the pole to be at p = —10 and compute
the zero and gain to be 2 = —3, k = 12. ...
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This note was uploaded on 03/29/2010 for the course MEM MEM355 taught by Professor Yousuff during the Spring '07 term at Drexel.
 Spring '07
 Yousuff

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