problemset3_Soln

problemset3_Soln - The University of Texas at Austin...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 5

problemset3_Soln - The University of Texas at Austin...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online