hw6_solution

hw6_solution - MAE107 Homework#6 Solution Prof MCloskey...

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

View Full Document Right Arrow Icon
MAE107 Homework #6 Solution Prof. M’Closkey Problem 1 1. The eigenvalues of A 1 are distinct so A 1 can be “diagonalized.” The eigenvalues and associated eigenvectors are λ 1 , t 1 = ± 1 0 ² , λ 2 , t 2 = ± 1 λ 2 - λ 1 ² . Note that t 1 and t 2 are linearly independent when λ 1 6 = λ 2 . Form the transformation matrix T = ³ t 1 t 2 ´ = ± 1 1 0 λ 2 - λ 1 ² . You can confirm Λ = T - 1 A 1 T = ± λ 1 0 0 λ 2 ² . The matrix exponential is e A 1 t = Te Λ t T - 1 = ± 1 1 0 λ 2 - λ 1 ²± e λ 1 t 0 0 e λ 2 t ²± λ 2 - λ 1 - 1 0 1 ² 1 λ 2 - λ 1 = ± e λ 1 t 1 λ 2 - λ 1 ( e λ 2 t - e λ 1 t ) 0 e λ 2 t ² . Turning to A 2 , we note that A 2 cannot be diagonalized, however, we can use the hints in the problem statement in order to compute e A 2 t . Confirm that the product of M 1 and M 2 commutes so we can use the expression e A 2 t = e ( M 1 + M 2 ) t = e M 1 t e M 2 t . Note that e M 1 t is easily computed to be e M 1 t = ± e λt 0 0 e λt ² . In order to compute e M 2 t , note that M k 2 = 0, for k = 2 , 3 , 4 ,... , so the matrix exponential power series has only a few non-zero terms, e M 2 t = I + 1 1! M 2 t + 1 2! M 2 2 t 2 + 1 3! M 3 2 t 3 + 1 4! M 4 2 t 4 + ··· = I + 1 1! M 2 t = ± 1 t 0 1 ² . Thus, e A 2 t = e M 1 t e M 2 t = ± e λt 0 0 e λt ²± 1 t 0 1 ² = ± e λt te λt 0 e λt ² . 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
2. The Laplace transform of a matrix is the matrix formed by the Laplace transforms of the individual elements, L - ( e A 1 t ) = " L - ( e λ 1 t ) L - ± 1 λ 2 - λ 1 ( e λ 2 t - e λ 1 t ) ² L - (0) L - ( e λ 2 t ) # = " 1 s - λ 1 1 λ 2 - λ 1 ± 1 s - λ 2 - 1 s - λ 1 ² 0 1 s - λ 2 # = " 1 s - λ 1 1 ( s - λ 1 )( s - λ 2 ) 0 1 s - λ 2 # . On the other hand, ( sI - A 1 ) - 1 = ³ s - λ 1 - 1 0 s - λ 2 ´ - 1 = 1 ( s - λ 1 )( s - λ 2 ) ³ s - λ 2 1 0 s - λ 1 ´ = " 1 s - λ 1 1 ( s - λ 1 )( s - λ 2 ) 0 1 s - λ 2 # . For A 2 , L - ( e A 2 t ) = ³ L - ( e λt ) L - ( te λt ) L - (0) L - ( e λt ) ´ = " 1 s - λ 1 ( s - λ ) 2 0 1 s - λ # . Computation of ( sI - A 2 ) - 1 yields, ( sI - A 2 ) - 1 = ³ s - λ - 1 0 s - λ ´ - 1 = " 1 s - λ 1 ( s - λ ) 2 0 1 s - λ # . 3. The transfer function are H 1 ( s ) = C ( sI - A 1 ) - 1 B + d = µ 1 0 " 1 s - λ 1 1 ( s - λ 1 )( s - λ 2 ) 0 1 s - λ 2 # ³ 1 1 ´ = s - λ 2 + 1 ( s - λ 1 )( s - λ 2 ) H 2 ( s ) = C ( sI - A 2 ) - 1 B + d = µ 1 0 " 1 s - λ 1 ( s - λ ) 2 0 1 s - λ # ³ 1 1 ´ = s - λ + 1 ( s - λ ) 2 . 4. The tables in the text can be used to determine the time functions associated with the transfer functions. The relation at the bottom of page 733 is s + A ( s + σ a )( s + σ b ) ←→ A - σ a σ b - σ a e - σ a t + σ b - A σ b - σ a e - σ b t , t 0
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/02/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.

Page1 / 9

hw6_solution - MAE107 Homework#6 Solution Prof MCloskey...

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