lecture_26 - MIT OpenCourseWare http://ocw.mit.edu 2.004...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . +- R ( s ) C ( s ) P c o n t r o l l e r p l a n t K 1 s + 3 s + 5 s + 2 3 2 Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 26 1 Reading: Nise: Chapter 6 Nise: Chapter 8 1 Determining Stability Bounds in Closed-Loop Systems Consider the closed-loop third-order system with proportional controller gain K with open- loop transfer function K G f ( s ) = s 3 + 3 s 2 + 5 s + 2 shown below. The closed loop transfer function is: N ( s ) K G cl ( s ) = = D ( s ) + N ( s ) s 3 + 3 s 2 + 5 s + (2 + K ) Lets examine the closed-loop stability by using the pzmap() function in MATLAB: sys = tf(1,[1 3 5 2]); pzmap(sys); hold on; for K = 2:2:30 sys = tf(K,[1 3 5 2+K]); pzmap(sys); end; which superimposes the closed-loop pole/zero plots for K = 0 ... 30 on a single plot: 1 copyright c D.Rowell 2008 261 Pole-Zero Map Real Axis Imaginary Axis-4-3.5-3-2.5-2-1.5-1-0.5 0.5 1-3-2-1 1 2 3 K=0 K=30 K=30 K=30 K=0. K=0 K increasing K increasing K increasing From the plot we note the following: This system always has two complex conjugate poles and a single real pole. When K = the poles are the open-loop poles. As K increases, the real pole moves deeper into the l.h. plane, and the complex con- jugate poles approach and cross the imaginary ( j ) axis, and the system becomes unstable. Close examination of the plot shows that the system becomes unstable at a value of K between K = 12 and K = 14. We now look at three methods for determining the stability limit of the proportional gain K for this system. Example 1 Use the Routh-Hurwitz method to find the range of proportional controller gain K for which the above system will be stable. The first two rows of the Routh array are taken directly from D ( s ): s 3 1 5 s 2 3 2 + K 262 and the next two rows are computed as above 1 a n a n 2 1 1 5 1 b 1 = a n 1 a n 1 a n 3 = = ( K 13) 3 3 2 +...
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.004 taught by Professor Derekrowell during the Spring '08 term at MIT.

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lecture_26 - MIT OpenCourseWare http://ocw.mit.edu 2.004...

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