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sol9 - k = n-1 C k Π k x = C k x-C k C k C k C k C k x = 0...

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EE 221 : Linear Systems HW #9 Solutions — Sam Burden — Dec 2, 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. (a) rank O = rank C CA = rank 1 2 0 1 = 2. (b) u = - Fx = A c‘ = - 2 - f 1 1 - f 2 1 0 = ⇒ O = 1 2 - f 1 1 - f 2 (c) f 1 = 1 , f 2 = 3 = rank O = 1 (d) C ( sI - A ) - 1 B = s +2 s 2 +2 s - 1 , C ( sI - A c‘ ) - 1 B = 1 s +1 , so loss of observability is due to a pole-zero cancellation. Exercise 2. (a) a k = a k - 1 , y k = C k a k , C k := ( m k - 1 m k - 2 · · · m k - N ) (b,c) Since m 0 6 = 0, can determine a particular a k that yields the correct output as a k = M - 1 k y 1: k := m 0 0 · · · 0 m 1 m 0 · · · 0 . . . . . . . . . . . . m k m k - 1 · · · m 0 - 1 y 1 y 2 . . . y k . The (affine) space of solutions is then S k = { a k + α : α ∈ N ( C k ) } , so to find the estimate ˆ a k that satisfies y k = C k ˆ a k while minimizing | ˆ a k - ˆ a k - 1 | , we project ˆ a k - 1 orthogonally into S k . Let Π k := I - C * k C k C k C * k ; then rank Π
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Unformatted text preview: k = n-1, C k Π k x = C k x-C k C * k C k C * k C k x = 0, and Π k x-x = C k x C k C * k C * k ∈ R ( C * k ) = ⇒ (Π k x-x ) ⊥ N ( C k ), so Π k projects orthogonally into the nullspace of C k . Therefore ˆ a k = Π k (ˆ a k-1-a k ) + a k is the desired estimate. Exercise 3. Note that Φ( t,τ ) is always invertible. To steer the system from x to y on [ τ ,t ] ⊃ [ τ,t ], choose an input that steers the system from Φ( τ,τ ) x to Φ( t ,t )-1 y on [ τ,t ]. As a counterexample when [ τ ,t ] ⊂ ( τ,t ), consider a controllable system for which B ( σ ) ≡ for σ ∈ [ τ ,t ]....
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