This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAE143B Homework Set 3  Solutions Question 1 FPEN Question 5.30: Assume that the closedloop system has plant transfer function P ( s ) = 1 s ( s + 2) . Design a lag compensator so that the dominant poles of the closedloop system are located at s = 1 j and the steadystate 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 0.5 1.5 1 0.5 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 steadystate error to a ramp as lim t e t = lim s sE ( s ) = lim s s s 2 + 2 s s 2 + 2 s + 1 1 s 2 = 1 . We need to increase the lowfrequency gain via a lag compensator without affecting the closedloop poles very much. Since we need an increase in gain by a factor of five to get our error down to 0.2, lets try C ( s ) = s +0 . 1 s +0 . 02 . This has the lowfrequency 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 . 0101 j 2 . 2.5 2 1.5 1 0.5 0.5 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 Root Locus Real Axis Imaginary Axis Figure 1: Root locus with lag compensator. Question 2....
View Full
Document
 Fall '09
 CALLAFON

Click to edit the document details