MIT16_30F10_rec01

MIT16_30F10_rec01 - 16.30/31, Fall 2010 - Recitation # 1...

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± ± 16.30/31, Fall 2010 Recitation # 1 September 14, 2010 In this recitation we considered the following problem. Given a plant with open-loop transfer function 1 . 1569 s + 0 . 3435 G p ( s ) = s 2 + 0 . 7410 s + 0 . 9272 , design a feedback control system such that the closed-loop dominant poles have undamped natural frequency ω n = 3rad / s and damping ratio ζ = 0 . 6. (This problem was mentioned during lecture, see the Topic 2 notes, as a stability augmentation system for a B747.) r G p ( s ) u y G c ( s ) e - + Figure 1: The standard block diagram for a unit-feedback loop. Since we are working with the root locus method, it is convenient to rewrite the transfer function in the form z Zeroes( G p ) ( s z ) s + 0 . 2969 G p ( s ) = K p p Poles( G p ) ( s p ) = 1 . 1569 · s 2 + 0 . 7410 s + 0 . 9272 . Note that the “root-locus gain” of the plant’s transfer function K p = 1 . 1569 is negative . This is a fact that is easy to overlook, and may generate confusion if not recognized and handled properly. More on this later. The pole-zero map for the system is shown in Figure 2 , where: z 1 = 0 . 2969 is the zero (root of the numerator) of G p . p 1 , 2 = 0 . 3705 ± 0 . 8888 are the poles (roots of the denominator) of G p . p d = 1 . 8 ± 2 . 4 j are the desired closed-loop poles. (We know that the magnitude of the 1 , 2 desired poles is equal to ω n = 3, and the real part is ζω n = 1 . 8.) Q1. Can we achieve the control objective using proportional feedback? A ±rst question one may want to answer is whether a simple proportional feedback (i.e., G c ( s ) = K , for some K ) would do the trick. You
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This note was uploaded on 02/03/2012 for the course AERO 16.30 taught by Professor Ericferon during the Fall '04 term at MIT.

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MIT16_30F10_rec01 - 16.30/31, Fall 2010 - Recitation # 1...

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