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MIT16_30F10_lec11

# MIT16_30F10_lec11 - Topic#11 16.31 Feedback Control Systems...

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Topic #11 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? Reading: FPE 7.3

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Fall 2010 16.30/31 11–1 Full-state Feedback Controller Assume that the single-input system dynamics are given by x ˙ ( t ) = A x ( t ) + B u ( t ) y ( t ) = 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. K Assume a full-state feedback of the form: u ( t ) = 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 ) + B r = A cl x ( t ) + B r y ( t ) = C x ( t ) A, B, C x ( t ) y ( t ) r u ( t ) October 17, 2010
Fall 2010 16.30/31 11–2 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: 1 1 1 x ˙ ( t ) = x ( t ) + u 1 2 0 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 1 1 1 A cl = A BK = 1 2 0 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 17, 2010

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Fall 2010 16.30/31 11–3 To put the poles at s = 5 , 6 , compare the desired characteristic equation ( s + 5)( s + 6) = s 2 + 11 s + 30 = 0 with the closed-loop one s 2 + ( k 1 3) s + (1 2 k 1 + k 2 ) = 0 to conclude that k 1 3 = 11 k 1 = 14 1 2 k 1 + k 2 = 30 k 2 = 57 so that K = 14 57 , which is called Pole Placement . Of course, it is not always this easy, as lack of controllability might be an issue. Example #2: Consider this system: 1 1 1 x ˙ ( t ) = x ( t ) + u 0 2 0 with the same control approach A cl = A BK = 0 1 2 1 0 1 k 1 k 2 = 1 0 k 1 1 2 k 2 so that det( sI A cl ) = ( s 1 + k 1 )( s 2) = 0 So the feedback control can modify the pole at s = 1 , but it cannot move the pole at s = 2 . System cannot be stabilized with full-state feedback. Problem caused by a lack of controllability of the e 2 t mode. October 17, 2010
Fall 2010 16.30/31 11–4 Consider the basic controllability test: 1 1 1 1 M c = B AB = 0 0 2 0 So that rank M c = 1 < 2 .

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