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MIT16_30F10_lec18

# MIT16_30F10_lec18 - Topic#18 16.31 Feedback Control Systems...

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Topic #18 16.31 Feedback Control Systems Deterministic LQR Optimal control and the Riccati equation Weight Selection

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Fall 2010 16.30/31 18–2 Linear Quadratic Regulator (LQR) Have seen the solutions to the LQR problem, which results in linear full-state feedback control. Would like to get some more insight on where this came from. Deterministic Linear Quadratic Regulator Plant: x ˙ = A x + B u u , x ( t 0 ) = x 0 z = C z x Cost: 1 t f 1 J LQR = z T R zz z + u T R uu u dt + x T ( t f ) P ( t f ) x ( t f ) 2 0 2 Where R zz > 0 and R uu > 0 Define R xx = C z T R zz C z 0 Problem Statement: Find input u t [ t 0 , t f ] to min J LQR This is not necessarily specified to be a feedback controller. Control design problem is a constrained optimization, with the con- straints being the dynamics of the system. November 5, 2010
Fall 2010 16.30/31 18–3 Constrained Optimization The standard way of handling the constraints in an optimization is to add them to the cost using a Lagrange multiplier Results in an unconstrained optimization. Example: min f ( x, y ) = x 2 + y 2 subject to the constraint that c ( x, y ) = x + y + 2 = 0 2 1.5 1 0.5 0 0.5 1 1.5 2 2 1.5 1 0.5 0 0.5 1 1.5 2 x y Fig. 1: Optimization results Clearly the unconstrained minimum is at x = y = 0 November 5, 2010

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Fall 2010 16.30/31 18–4 To find the constrained minimum, form augmented cost function L f ( x, y ) + λc ( x, y ) = x 2 + y 2 + λ ( x + y + 2) Where λ is the Lagrange multiplier Note that if the constraint is satisfied, then L f The solution approach without constraints is to find the stationary point of f ( x, y ) ( ∂f/∂x = ∂f/∂y = 0 ) With constraints we find the stationary points of L ∂L ∂L ∂L = = = 0 ∂x ∂y ∂λ which gives ∂L = 2 x + λ = 0 ∂x ∂L = 2 y + λ = 0 ∂y ∂L = x + y + 2 = 0 ∂λ This gives 3 equations in 3 unknowns, solve to find x = y = 1 Key point here is that due to the constraint, the selection of x and y during the minimization
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