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MIT16_30F10_lec16

# MIT16_30F10_lec16 - Topic#16 16.30/31 Feedback Control...

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Topic #16 16.30/31 Feedback Control Systems Add reference inputs for the DOFB case Reading: FPE 7.8, 7.9

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Fall 2010 16.30/31 16–2 Reference Input - II On page 15- 6 , compensator implemented with reference command by changing to feedback on e ( t ) = r ( t ) y ( t ) rather than y ( t ) r G c ( s ) G p ( s ) e u y So u ( t ) = G c ( s ) e ( t ) = G c ( s )( r ( t ) y ( t )) Intuitively appealing because same approach used for classical control, but it turns out not to be the best. Can improve the implementation by using a more general form: ˙ x c ( t ) = A c x c ( t ) + B c y ( t ) + G r ( t ) u ( t ) = C c x c ( t ) + N r ( t ) Now explicitly have two inputs to controller ( y ( t ) and r ( t ) ) N performs the same role that we used it for previously. Introduce G as an extra degree of freedom in the problem. Turns out that setting G = BN is a particularly good choice. Following presents some observations on the impact of G November 2, 2010
Fall 2010 16.30/31 16–3 First: this generalization does not change the closed-loop poles of the system, regardless of the selection of G and N , since x ˙ ( t ) = A x ( t ) + B u ( t ) , y ( t ) = C x ( t ) x ˙ c ( t ) = A c x c ( t ) + B c y ( t ) + G r ( t ) u ( t ) = C c x c ( t ) + N r ( t ) x ˙ ( t ) A BC c x ( t ) BN x ˙ c ( t ) = B c C A c x c ( t ) + G r ( t ) x ( t ) y ( t ) = C 0 x c ( t ) So the closed-loop poles are the eigenvalues of A BC c B c C A c (same as 15– 7 except “–” in a different place, gives same closed- loop eigenvalues) regardless of the choice of G and N G and N impact the forward path, not the feedback path Second: if N = 0 and G = L = B c , then we recover the original implementation, on 15– 6 since the controller reduces to: x ˙ c ( t ) = A c x c ( t ) + B c ( y ( t ) r ( t )) = A c x c ( t ) + B c ( e ( t )) u ( t ) = C c x c ( t ) With G c ( s ) = C c ( sI A c ) 1 B c , then this compensator can be written as u ( t ) = G c ( s ) e ( t ) as before (since the negative signs cancel). November 2, 2010

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Fall 2010 16.30/31 16–4 Third: Given this extra freedom, what is best way to use it? One good objective is to select G and N so that the state esti- mation error is independent of r ( t ) . With this choice, changes in r ( t ) do not tend to cause such large transients in x ˜( t ) For this analysis, take x ˜( t ) = x ( t ) x c ( t ) since x c ( t ) x ˆ( t ) x ˜ ˙ ( t ) = x ˙ ( t ) x ˙ c ( t ) = A x ( t ) + B u ( t ) ( A c x c ( t ) + B c y ( t ) + G r ( t )) = A x ( t ) + B ( C c x c ( t ) + N r ( t )) ( { A BC c B c C } x c ( t ) + B c C x ( t ) + G r ( t )) So x ˜ ˙ ( t ) = A x ( t ) + B ( N r ( t )) ( { A B c C } x c ( t ) + B c C x ( t ) + G r ( t )) = ( A B c C ) x ( t ) + BN r ( t ) ( { A B c C } x c ( t ) + G r ( t )) = ( A B c C ) x ˜( t ) + BN r ( t ) G r ( t ) = ( A B c C ) x ˜( t ) + ( BN G ) r ( t ) Thus we can eliminate the effect of r ( t ) on x ˜( t ) by setting G BN November 2, 2010
Fall 2010 16.30/31 16–5 With this choice, the controller: x ˙ c ( t ) = ( A BK LC ) x c ( t ) + L y ( t ) + BN r ( t ) u ( t ) = K x c ( t ) + N r ( t ) can be rewritten as: x ˙ c ( t ) = ( A LC ) x c ( t ) + L y ( t ) + B u ( t ) u ( t ) = K x c ( t ) + N r ( t ) So the control is computed using the reference before it is applied, and that control is applied to both system and estimator.

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MIT16_30F10_lec16 - Topic#16 16.30/31 Feedback Control...

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