39 - opt_cont39 filled

39 - opt_cont39 filled - ME 575 Handouts Optimal Control Up...

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Unformatted text preview: ME 575 Handouts Optimal Control Up to now, we have designed controllers that primarily place closed-loop poles to establish stability and performance. We would like to change our focus to design the “best” controller to achieve some quantifiable objective. This requires specifying a performance index (or objective function). We call this optimal control. School of Mechanical Engineering Purdue University ME575 Session 39 – Optimal Control Slide 1 Performance Indices Typically involve an integral expression. Some examples: V=∫ tf V=∫ tf ti ti tf 2 ( y d − y ) dt T ( xd − x ) Q ( xd − x ) dt V = ∫ u T R u dt ti ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 2 1 ME 575 Handouts Calculus of Variations Given: V = ∫ L ( x, x, t ) dt tf ti Focus on scalar case for now (with fixed endpoints): x(t) x(t) = x o (t) + δx(t) x o (t) δx(t), variation t ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 3 Necessary Condition for a Minimum ∂V ΔV = V(x) − V(x o ) = ∂x 1 ∂2V δx + 2! ∂x 2 xo ( δx ) 2 + xo t f ⎡ ∂L ∂L ⎤ δV = ∫ ⎢ δx + δx dt ti ∂x ⎥ ⎣ ∂x ⎦ ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 4 2 ME 575 Handouts Euler-Lagrange Equations tf V = ∫ L(x, x, t)dt ti ⇒ ∂L d ⎛ ∂L ⎞ − =0 ∂x dt ⎜ ∂x ⎟ ⎝⎠ Vector Case: x T = [ x1 x 2 … xn ] ⇒ ∂L d ⎛ ∂L ⎞ −⎜ ⎟=0 ∂xi dt ⎝ ∂x i ⎠ ME575 Session 39 – Optimal Control i = 1, 2, School of Mechanical Engineering Purdue University ,n Slide 5 Optimal Control Plant is described by: x = f(x,u,t) Performance Index: tf V = ∫ L(x,u,t)dt ti We seek the optimal control u(t) or control law u(x) that will take x from x(ti) to x(tf) and minimize V. ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 6 3 ME 575 Handouts Optimal Control We could solve the plant equation for u in terms of x and x , then substitute into V and solve as before. But this is cumbersome. Instead, use Lagrange multipliers (λi): gi (x,x,u,t) = fi (x,u,t) − xi = 0 n n j=1 j=1 L* (x,x,u,t) = L(x,u,t) + ∑ λ jg j = L + ∑ λ j (fj − x j ) ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 7 Euler Equations ∂L* d ⎡ ∂L* ⎤ − ⎥=0 ∂xi dt ⎢ ∂xi ⎦ ⎣ ∂L* d ⎡ ∂L* ⎤ − ⎥=0 ∂ui dt ⎢ ∂ui ⎦ ⎣ ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 8 4 ME 575 Handouts Euler Equations (with respect to x) ∂ ∂xi n ⎡ ⎤ L(x,u,t) + ∑ λ j [f j (x,u,t) − x j ]⎥ ⎢ j=1 ⎣ ⎦ n ⎤ d∂⎡ − L(x,u,t) + ∑ λ j [f j (x,u,t) − x j ]⎥ = 0 ⎢ dt ∂xi ⎣ j=1 ⎦ ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 9 Euler Equations (with respect to u) ∂ ∂ui n ⎡ ⎤ L(x,u,t) + ∑ λ j [f j (x,u,t) − x j ]⎥ ⎢ j=1 ⎣ ⎦ n ⎤ d∂⎡ − L(x,u,t) + ∑ λ j [f j (x,u,t) − x j ]⎥ = 0 ⎢ dt ∂ui ⎣ j=1 ⎦ ME575 Session 39 – Optimal Control School of Mechanical Engineering Purdue University Slide 10 5 ...
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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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