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sol8 - also showed through extension of equalities that the...

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EE 221 : Linear Systems HW #8 Solutions — Sam Burden — Nov 23, 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) A = TJT - 1 = 1 1 - 1 1 - 1 / 2 0 0 - 1 / 10 1 / 2 - 1 / 2 1 / 2 - 1 / 2 , so x = Tz = ˙ z = Jz + T - 1 Bu , T - 1 B = 1 20 1 - 1 1 1 , and y = Tx . (b) y ( t ) = x ( t ) = e At x 0 = Te Jt T - 1 x 0 . (c) The system is exponentially stable, whence BIBO. Exercise 2. cf. the last page of Lecture Notes 15. Exercise 3. (a) We showed in recitation that, if A is semi-simple (diagonalizable) with eigenvalues { λ i } n i =1 and left eigenvectors { v i } n i =1 , then L is semi-simple with eigenvalues { λ i + λ j } i,j and eigenvectors v T i v j i,j . In particular, L has n 2 linearly independent eigenvectors and none of its eigenvalues are zero, so it is bijective. Also, since each of L ’s eigenvectors are symmetric, L is a bijection when restricted to the set of symmetric matrices. We
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Unformatted text preview: also showed, through extension of equalities, that the same conclusion holds even if A is not semi-simple. (b) By (a), there is a unique symmetric solution to the equation. Let P = R ∞ e A T t Qe At dt . Then P > 0 since x T Px = R ∞ ( e At x ) T Q ( e At x ) dt and the integrand is > 0 whenever x 6 = 0 since e At is nonsingular and Q > 0. Further, P solves the equation since A T P + PA = Z ∞ d dt µ e A T t Qe At ¶ dt =-Q. Exercise 4. (a) χ ( s ) = s 2 + (4 + 2 f 2 + f 1 ) s + (6 f 1 + 7 f 2-7). (b) χ ( s ) = s 2 + (7 k + 4) s + (27 k-7). Exercise 5. k = ( a 1-α 1 a 2-α 2 a 3-α 3 ) yields the desired characteristic equation....
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