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# topic13 - Topic#13 16.31 Feedback Control Systems...

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Unformatted text preview: Topic #13 16.31 Feedback Control Systems State-Space Systems • Full-state Feedback Control • How do we change the poles of the state-space system? • Or, even if we can change the pole locations. • Where do we change the pole locations to? • How well does this approach work? Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Fall 2007 16.31 13–1 Full-state Feedback Controller • Assume that the single-input system dynamics are given by ˙ x ( t ) = A x ( t ) + Bu y = C x ( t ) so that D = 0 . – The multi-actuator case is quite a bit more complicated as we would have many extra degrees of freedom. • Recall that the system poles are given by the eigenvalues of A . – Want to use the input u ( t ) to modify the eigenvalues of A to change the system dynamics. r u A,B,C y x ( t ) K − • Assume a full-state feedback of the form: u = r − K x ( t ) where r is some reference input and the gain K is R 1 × n – If r = 0 , we call this controller a regulator • Find the closed-loop dynamics: ˙ x ( t ) = A x ( t ) + B ( r − K x ( t )) = ( A − BK ) x ( t ) + Br = A cl x ( t ) + Br y = C x ( t ) October 14, 2007 Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Fall 2007 16.31 13–2 • Objective: Pick K so that A cl has the desired properties, e.g., – A unstable, want A cl stable – Put 2 poles at − 2 ± 2 i • Note that there are n parameters in K and n eigenvalues in A , so it looks promising, but what can we achieve? • Example #1: Consider: ˙ x ( t ) = 1 1 1 2 x ( t ) + 1 u – Then det( sI − A ) = ( s − 1)( s − 2) − 1 = s 2 − 3 s + 1 = 0 so the system is unstable. – Define u = − k 1 k 2 x ( t ) = − K x ( t ) , then A cl = A − BK = 1 1 1 2 − 1 k 1 k 2 = 1 − k 1 1 − k 2 1 2 – which gives det( sI − A cl ) = s 2 + ( k 1 − 3) s + (1 − 2 k 1 + k 2 ) = 0 – Thus, by choosing k 1 and k 2 , we can put λ i ( A cl ) anywhere in the complex plane (assuming complex conjugate pairs of poles). October 14, 2007 Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Fall 2007 16.31 13–3 • To put the poles at s = − 5 , − 6 , compare the desired character- istic equation ( s + 5)( s + 6) = s 2 + 11 s + 30 = 0 with the closed-loop one s 2 + ( k 1 − 3) x ( t ) + (1 − 2 k 1 + k 2 ) = 0 to conclude that k 1 − 3 = 11 1 − 2 k 1 + k 2 = 30 k 1 = 14 k 2 = 57 so that K = 14 57 , which is called Pole Placement ....
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## This note was uploaded on 11/07/2011 for the course AERO 16.31 taught by Professor Jonathanhow during the Fall '07 term at MIT.

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topic13 - Topic#13 16.31 Feedback Control Systems...

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