{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw6 - MAE107 Homework#6 Prof MCloskey Due Date The homework...

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

View Full Document Right Arrow Icon
MAE107 Homework #6 Prof. M’Closkey Due Date The homework is due at 5PM on Thursday, June 3, 2010, to David Shatto (38-138 foyer, Engineer- ing 4). Problem 1 Consider the following 2 × 2 matrices, A 1 = λ 1 1 0 λ 2 , λ 1 6 = λ 2 , A 2 = λ 1 0 λ . Answer the following: 1. Compute the matrix exponential for each matrix. Note that A 1 can be diagonalized but that A 2 cannot. One way to compute e A 2 t is to recognize that A 2 = λ 0 0 λ | {z } M 1 + 0 1 0 0 | {z } M 2 , and since M 1 M 2 = M 2 M 1 (confirm this), then e A 2 t = e ( M 1 + M 2 ) t = e M 1 t e M 2 t . 2. Show by direct calculation that L - ( e A k t ) = ( sI - A k ) - 1 , k = 1 , 2 . 3. Now let C = 1 0 , B = 1 1 , d = 0 . Compute the transfer functions H k ( s ) = C ( sI - A k ) - 1 B + d, k = 1 , 2 , 4. Use the unilateral Laplace transform tables in the back of the text to find the impulse re- sponses corresponding to H 1 and H 2 . 5. Compute the impulse responses using the matrix exponential functions you computed above and the formula h k ( t ) = Ce A k t ( t ) + ( t ) , k = 1 , 2 . Show that these expressions match what you derived using the unilateral Laplace transform tables (which implicitly assumes the impulse response is zero for t < 0). 1
Image of page 1

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

View Full Document Right Arrow Icon
Problem 2 Consider the block diagram shown below where G 1 = 1 s + 1 , G 2 = G 3 = 1 s , G 4 = - 1 , G 5 = 1 , represent transfer functions of the individual blocks, - r j - G 1 - j - G 2 - j - G 3 G 4 - ?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern