Exam 6 Solutions - Math 524 Exam 6 Solutions 1\/3 1\/6 1\/3 5\/6 The rst three problems all concern A = = 1\/2 0 0 2\/3 1 1 1 2 2 1 1 1 1 Solve the

Exam 6 Solutions - Math 524 Exam 6 Solutions 1/3 1/6 1/3...

This preview shows page 1 out of 1 page.

Math 524 Exam 6 Solutions The first three problems all concern A = - 1 / 3 - 1 / 6 1 / 3 - 5 / 6 = 1 1 1 2 - 1 / 2 0 0 - 2 / 3 2 - 1 - 1 1 1. Solve the discrete-time system given by x ( n ) = Ax ( n - 1), with initial condition x (0) = ( 0 1 ). A basis of eigenvectors is B = { b 1 , b 2 } , for b 1 = (1 , 1) T , b 2 = (1 , 2) T . We have [ x (0)] B = ( - 1 1 ) . We have x ( n ) = A n x (0) = ( PDP - 1 ) n x (0) = PD n P - 1 x (0). Be- cause P - 1 x = [ x ] B , we have [ x ( n )] B = D n [ x (0)] B = D n ( - 1 1 ) = - ( - 1 / 2 ) n ( - 2 / 3 ) n . Hence x ( n ) = P - ( - 1 / 2 ) n ( - 2 / 3 ) n = - ( - 1 / 2 ) n +( - 2 / 3 ) n - ( - 1 / 2 ) n +2( - 2 / 3 ) n , or x 1 ( n ) = ( - 2 / 3 ) n - ( - 2) - n , x 2 ( n ) = 2( - 2 / 3 ) n - ( - 2) - n . One may check this with some values such as n = 0 , 1 , 2; or, one may check that this satisfies the difference equation and initial condition. 2. Solve the first-order system given by d dt x = Ax , with initial condition given by x (0) = ( 0 1 ). We calculate a basis of eigenvectors as before, and note that x (0) = - b 1 + b 2 . This time x ( t ) = e At x (0) = e At ( - b 1 + b 2 ). We have e At b 1 = e - t/ 2 b 1 and e At b 2 = e - 2 t/ 3 b 2 , since they are eigenvectors of A . Hence x ( t ) = - e - t/ 2 b 1 + e - 2 t/ 3 b 2 = - e - t/ 2 + e - 2 t/ 3 - e - t/ 2 +2 e - 2 t/ 3 . One may check that this satisfies the DE and initial condition.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture