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Homework3solutions_mae143b

Homework3solutions_mae143b - MAE143B Homework Set 3...

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MAE143B Homework Set 3 - Solutions Question 1 FPE-N Question 5.30: Assume that the closed-loop system has plant transfer function P ( s ) = 1 s ( s + 2) . Design a lag compensator so that the dominant poles of the closed-loop system are located at s = - 1 ± j and the steady-state error to a unit ramp input is less than 0.2. Without compensation, the root locus of the system is as shown in the following figure. By choosing 2.5 2 1.5 1 0.5 0 0.5 1.5 1 0.5 0 0.5 1 1.5 Root Locus Real Axis Imaginary Axis a constant feedback gain of K = 2 , we can place the closed loop poles at s = - 1 ± j . However, this leaves the steady-state error to a ramp as lim t →∞ e t = lim s 0 sE ( s ) = lim s 0 s s 2 + 2 s s 2 + 2 s + 1 1 s 2 = 1 . We need to increase the low-frequency gain via a lag compensator without affecting the closed-loop poles very much. Since we need an increase in gain by a factor of five to get our error down to 0.2, let’s try C ( s ) = s +0 . 1 s +0 . 02 . This has the low-frequency gain of five and has a pole and zero which effectively cancel out when s is as big as - 1 ± j . With this new choice of compensator, we plot the root locus of C ( s ) P ( s ) = s + 0 . 1 s + 0 . 02 1 s ( s + 2) = s + 0 . 1 s 3 + 2 . 02 s 2 + 0 . 04 s . This yields Note that the gain for poles close to s = - 1 + j is given by K = - s 3 + 2 . 02 s 2 + 0 . 04 s s + 0 . 1 s = - 1+ j = 1 . 9900 - 0 . 0101 j 2 .
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2.5 2 1.5 1 0.5 0 0.5 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 Root Locus Real Axis
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