# sol6 - h is varied This is intuitive since h measures error...

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EE 221 : Linear Systems HW #6 Solutions — Sam Burden — Nov 5, 2010 1 Note these solutions are overly terse and leave out some details; in your solutions, you should strive for greater clarity and rigor. Exercise 1. Given ˙ x = Ax + Bu = 0 1 0 0 x + 0 1 u, set z = x - x = x - ( 4 0 ) ; since x ∈ N ( A ), ˙ z = Az + Bu . Thus we may solve the LQR problem for z with cost J = R 0 z T Qz + ru 2 dt to obtain u = - Kz = - K ( x - x ). Exercise 2. (a) Given ˙ x = Ax + Bu = 0 a 0 b x + 1 0 u, J = Z T 0 | Cx | 2 + | ru | 2 dt = Z T 0 1 0 0 h x + | u | 2 dt, we know from lecture that the optimal control is given by u = - B T Px where P = P T > 0 satisfies the Ricatti differential equation - ˙ P = A T P + PA - PBr - 1 B T P + C T C . But - B T Px = - p 1 x - p 2 x , so we need only find the DE for the first column of P : - ˙ p 1 = 1 - p 1 , p 1 ( T ) = 0 , - ˙ p 2 = ap 1 + bp 2 - p 1 p 2 , p 2 ( T ) = 0 . (b) The parameter h does not appear in the expression for the optimal control, hence the control does not change as h is varied. This is intuitive, since
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Unformatted text preview: h is varied. This is intuitive, since h measures error in x 2 , but the control cannot aﬀect x 2 . Exercise 3. Since det BC = det B det C , χ A ( s ) = det( sI-A ) = det P det( sI-PAP-1 ) det P-1 = det( sI-PAP-1 ) = χ PAP-1 ( s ) , whence A and ¯ A = PAP-1 have the same eigenvalues. Exercise 4. A = n X j =1 λ j w j v T j = ⇒ A k = n X j =1 λ k j w j v T j = ⇒ e At = ∞ X k =0 t k k ! A k = ∞ X k =0 n X j =1 ( λ j t ) k k ! w j v T j = n X j =1 e λ j t w j v T j . Exercise 5. A ( A j-1 b ) = A j b , which is either contained in the given list of vectors or, by Cayley-Hamilton, linearly dependent on them....
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