lecture_21 - 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 . t y = 1 y ( t ) s t e p z < 1 d a m p e d o s c i l l a t o r y r e s p o n s e y ( t ) = 1 - s t e p e 1 - z 2- z w t n c o s ( w t- f ) d y p e a k s s T p o v e r s h o o t Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 21 1 Reading: Nise: Secs. 4.6 4.8 (pp. 168- 186) 1 Second-Order System Response Characteristics (contd.) 1.1 Percent Overshoot The height of the first peak of the response, expressed as a percentage of the steady-state response. %OS = y peak y ss 100 y ss At the time of the peak y ( T p ) y peak = y ( T p ) = 1 + e ( / 1 2 ) and since y ss = 1 %OS = e ( / 1 2 ) 100 . Note that the percent overshoot depends only on . Conversely we can find to give a specific percent overshoot from the above: ln(%OS / 100) = 2 + ln 2 (%OS / 100) 1 copyright c D.Rowell 2008 211 Example 1 Find the damping ratio that will generate a 5% overshoot in the step response of a second-order system. Using the above formula ln(%OS / 100) ln(0 . 05) = = = 0 . 69 2 + ln 2 (%OS / 100) 2 + ln 2 (0 . 05) Example 2 Find the location of the poles of a second-order system with a damping ratio = 0 . 707, and find the corresponding overshoot. The complex conjugate poles line on a pair of radial lines at an angle = cos 1 . 707 = 45 from the negative real axis. The percentage overshoot is e ( / 1 2 ) %OS =...
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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_21 - MIT OpenCourseWare http://ocw.mit.edu 2.004...

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