20075ee141_1_hw5_solutions

20075ee141_1_hw5_solutions - EE 141 Fall 2007 Homework 5 1....

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Unformatted text preview: EE 141 Fall 2007 Homework 5 1. Calculate the characteristic polynomial of the matrix A = 1 1 a b a b For what values of a is the matrix A stable? (10 points) Solution: | I- A | = - 1 - 1- a - b- a - b = 2 ( - b ) 2- a ( - b )- a ( ( - b )- a ) = 2 ( - b ) 2- 2 a ( - b ) + a 2 = ( ( - b )- a ) 2 = f ( ) The real part of the eigenvalues of A has to be negative to have A stable. We could compute the roots of f ( ), but it is much faster to use Routh-Hurwitz (yes, here is useful as well), and see that we just have a seconde order equation. To have roots with negative real part we must have all coefficients greater than 0: ( - b )- a = 2- b- a = 0 b > Unstable regardless the value of a b < a < 0 to have a stable system 2. Let A be a matrix such that A 2 = Identity. Show that e At = I cosh t + A sinh t regardless of dimension. (20 points) Solution: e t = X n =0 t n n sinh( t ) = e t- e- t 2 =...
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This note was uploaded on 03/18/2008 for the course EE 141 taught by Professor Balakrishnan during the Fall '07 term at UCLA.

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20075ee141_1_hw5_solutions - EE 141 Fall 2007 Homework 5 1....

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