problemset3_Soln

# problemset3_Soln - The University of Texas at Austin...

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Unformatted text preview: The University of Texas at Austin Department of Electrical and Computer Engineering EE362K: Introduction to Automatic Control—Fall 2009 Solutions to Problem Set Three C. Caramanis October 5, 2009. 1. • We use the notation | A | to denote the determinant of a square matrix A . The rule for determinant of the product of matrices gives | T- 1 AT | = | T || A || T- 1 | . Also TT- 1 = I gives | T- 1 | = | T |- 1 , therefore, | T- 1 AT | = | T || A || T- 1 | = | T || A || T |- 1 = | A | . Denoting the characteristic polynomials of A and T- 1 AT by p A ( λ ) and p T- 1 AT ( λ ) gives p A ( λ ) = | λI- A | = | T |- 1 | λI- A || T | = | T- 1 || λI- A || T | = | T- 1 ( λI- A ) T | = | λT- 1 T- T- 1 AT | = | λI- T- 1 AT | = p T- 1 AT ( λ ) Since the characteristic polynomials of A and T- 1 AT are same, they have the same eigenvalues as eigenvalues are roots of the characteristic polynomial. • For a square matrix A , its eigenvalue λ and corresponding eigenvector v satisfy Av = λv . Since TT- 1 = I , we have ATT- 1 v = λv . Multiplying both sides by T- 1 from the left gives T- 1 ATT- 1 v = λ ( T- 1 v ) ⇒ ( T- 1 AT )( T- 1 v ) = λ ( T- 1 v ) ⇒ ( T- 1 AT ) w = λw , where w , T- 1 v . Hence, w = T- 1 v is the eigenvector corresponding to the eigenvalue λ for the matrix T- 1 AT . 2. (a) We assume α 6 = 0 or else the utility of feedback would be eliminated. Substituting u =- αx in the differential equation and solving for ˙ x = 0 gives 1 1 + x- 1- αx = 0 ⇒ x x + 1 + 1 α = 0 ⇒ x = 0 ,- 1- 1 α Hence, the equilibrium points are x e = 0 ,- 1- 1 α . (b) V ( x ) = 1 2 x 2 is positive definite about x e = 0 as V (0) = 0 and V ( x ) > 0 for x 6 = 0. ˙ V ( x ) = x ˙ x =- αx 2 x + 1 + 1 α x + 1 ! For stability about x e = 0, ˙ V ( x ) must be locally negative semi-definite about x e = 0. As ˙ V (0) = 0, we only require ˙ V ( x ) ≤ 0 for x 6 = 0. Since x 2 > 0 for x 6 = 0, we have ˙ V ( x ) =- αx 2 x + 1 + 1 α x + 1 ! ≤ ⇒ x + 1 + 1 α x + 1 ≥ 0 if α > ⇒ x + 1 + 1 α x + 1 ≤ 0 if α < 1 Consider the case α > 0; the range of x satisfying the inequality can be found to be x ≥ - 1 or x ≤ - 1- 1 α i.e. (-∞ ,- 1- 1 α ] ∪ [- 1 , ∞ ). This range always includes a neighborhood around...
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## This note was uploaded on 10/26/2009 for the course EE 362K taught by Professor Friedrich during the Fall '08 term at University of Texas.

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problemset3_Soln - The University of Texas at Austin...

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