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Unformatted text preview: Question 1 Part i: Show that u t = B e A t W 1 c (0 ,T )¯ x i transfers the state x = ¯ x i to x T = 0. Note that the solution to the ODE, ˙ x t = Ax t + Bu t , is, as was shown repeatedly and in various contexts, x ( t ) = e At x + e At R t e Aτ Bu ( τ ) dτ . To prove the result, it suffices to substitute u ( τ ) for the given policy and to evaluate x ( t ) for t = T . First, substitute u ( · ). We have x ( t ) = e At ¯ x i + e At Z t e Aτ B ( B ) e A τ W 1 c (0 ,T )¯ x i dτ. Note that W c (0 ,T ) is a fixed matrix and not a function of τ . Hence, x ( t ) = e At ¯ x i e At Z t e Aτ BB e A τ dτ W 1 c (0 ,T )¯ x i . Now, evaluating this expression for t = T yields, x ( T ) = e AT ¯ x i e AT Z T e Aτ BB e A τ dτ W 1 c (0 ,T )¯ x i , = e AT ¯ x i e AT W c (0 ,T ) W 1 c (0 ,T )¯ x i , = 0 , which completes the proof. Part ii: We again proceed via substitution. Fix x = 0 and show x T = ¯ x f . x ( t ) = e At x + e At Z t e Aτ Bu ( τ ) dτ, = e At Z t e Aτ BB e A τ W 1 c (0 ,T ) e AT ¯ x f dτ, = e At Z t e Aτ BB e A τ dτ W 1 c (0 ,T ) e AT ¯ x f , ⇒ x ( T ) = e AT Z T e...
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This note was uploaded on 05/27/2010 for the course MAE MAE143B taught by Professor Prof.bitmead during the Spring '10 term at UCSD.
 Spring '10
 Prof.Bitmead

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