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2. Consider a system: ⎧ ⎪⎨ ⎪ ⎩ ẋ(t) = [ 0 2 1 −1 ] x(t), y(t) = [ 1 1 ] x(t). For each of the following specified locations of the observer poles: • eig(A − LC) = −2, −3. • eig(A − LC) = −3, −3. • eig(A − LC) = −1 ± j. is it possible to design an observer? If the answer is “yes”, design an observer. For the system given above, what is the condition for given observer pole locations to be achievable? (10pt) By the direct method, det(λI − (A − LC)) = · · · = (λ + 2)(λ + l 1 + l 2 − 1). Therefore, for whatever choice of L, the eigenvalue of s = −2 is impossible to move. Thus, the first observer poles are possible to achieve with L such that l 1 + l 2 = 4, but other two observer poles are not possible to achieve. The condition for given observer pole locations to be achievable is that they include s = −2. 3. Consider a system { ẋ(t) = αx(t) + u(t), y(t) = x(t). (a) For α = −1, design a state feedback with an integrator such that the closed-loop poles are s = −1, −1. (10pt) det(λI−(A aug −B aug [ K Ka ] ) = · · · = λ 2 +(K−α)λ−K a = λ 2 +2λ+1 When α = −1, K = 1 and K a = −1. (b) Suppose that our modeling is inaccurate and that the actual α is not −1. For the designed controller in (a), what is the range of the parameter α that results in zero steady-state tracking error for any step reference input? (10pt) For the designed controller parameters K = 1 and K a = −1, α must be less than one to keep the closed-loop stability. 2
4. Solve the following continuous-time infinite-horizon LQR problem: (10pt) min u(·) ∫ ∞ 0 subject to ( x 2 1 (t) + u 2 (t) ) dt, [ ẋ1 (t) ẋ 2 (t) ] = [ 0 1 0 −1 ] [ x1 (t) x 2 (t) ] [ 0 + 1 ] u(t). Verify the stability of the closed-loop system. Algebraic Riccati equation: [ √ ] 3 1 A T P + P A − P BR −1 B T P + Q = 0 ⇒ P = √ > 0 1 3 − 1 u(t) = −R −1 B T P x(t) = − [ 1 √ 3 − 1 ] x(t) The closed-loop system is stable since A cl := A − BR −1 B T P = [ 0 1 −1 − √ 3 ] 3
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solutions - UBC Mechanical Engineering

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