40 - lqr40 filled

# 40 - lqr40 filled - ME 575 Handouts Linear Quadratic...

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ME 575 Handouts 1 ME575 Session 40: Linear Quadratic Regulator School of Mechanical Engineering Purdue University Slide 1 Linear Quadratic Regulator Design Linear Quadratic Regulator Design Given: Given: Find: Find: u o to minimize to minimize xA x B u x(0) known = + ± TT 1 2 0 Vx Q x u R u d t =+ ME575 Session 40: Linear Quadratic Regulator School of Mechanical Engineering Purdue University Slide 2 Step 1: Step 1: Step 2: Step 2: Step 3: Step 3: [ ] T T 11 22 Hf x Q x u R u A x B u λ = + + λ + L T H 0R u B 0 u =⇒+ λ = 1 T T 1 T o Hx Q x B R B A x B R B −− λ λ + λ −λ LQR Solution LQR Solution

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ME 575 Handouts 2 ME575 Session 40: Linear Quadratic Regulator School of Mechanical Engineering Purdue University Slide 3 Step 4: Step 4: Step 5: Step 5: u o = T o H Qx A x λ=− =− −λ ± 1T o H xA x B R B == λ ∂λ ± LQR Solution LQR Solution ME575 Session 40: Linear Quadratic Regulator School of Mechanical Engineering Purdue University Slide 4 Algebraic Algebraic Riccati
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## This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue University.

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40 - lqr40 filled - ME 575 Handouts Linear Quadratic...

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