MMAE501_09_08_10

# MMAE501_09_08_10 - 2010 Illinois Institute of Technology...

This preview shows pages 1–4. Sign up to view the full content.

09/08/2010 2010 Illinois Institute of Technology ©

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

View Full Document
CHAPTER 2. THE EIGENPROBLEM AND ITS APPLICATIONS 53 or in matrix form L 0 CR 1 C ( R 1 + R 2 ) I 1 ( t ) I 2 ( t ) = R 1 R 1 0 1 I 1 ( t ) I 2 ( t ) , A 1 A 2 or A 1 x ( t ) = A 2 x ( t ) . To obtain the usual form ( x = Ax ) x ( t ) = A 1 1 A 2 x ( t ) . Therefore, A = A 1 1 A 2 . Diagonalize A to solve for I 1 ( t ) and I 2 ( t ). See “MathematicaDemo.nb” for sample solution. Note: No voltage source current decays with time. Example: Jeffrey 6.11 # 8 (example with f ( t ) = 0) Solve the system of equations: x 1 ( t ) = 2 x 2 + sin t x 2 ( t ) = 2 x 1 t or x ( t ) = Ax ( t ) + f ( t ) , where A = 0 2 2 0 , f ( t ) = sin t t . Eigenvalues and eigenvectors of A : λ 1 = 2 , u 1 = 1 1 , λ 2 = 2 , u 2 = 1 1 . Note: A = A T with λ 1 = λ 2 u 1 , u 2 = 0 . Forming the modal matrix and taking its inverse we have Q = 1 1 1 1 Q 1 = 1 2 1 1 1 1 . CHAPTER 2. THE EIGENPROBLEM AND ITS APPLICATIONS 54 We transform to the new variable y ( t ) using x ( t ) = Qy ( t ) ( x = Qy ) , with respect to which our system of equations becomes y ( t ) = Q 1 AQy ( t ) + Q 1 f ( t ) . Recalling that D = Q 1 AQ has the eigenvalues along its main diagonal and evaluating Q 1 f ( t ), this becomes y 1 ( t ) y 2 ( t ) = 2 0 0 2 y 1 ( t ) y 2 ( t ) + 1 2 sin t + t sin t t ; therefore, we have the two uncoupled equations y 1 ( t ) = 2 y 1 ( t ) + 1 2 sin t + 1 2 t y 2 ( t ) = 2 y 2 ( t ) + 1 2 sin t 1 2 t . Because the equations are nonhomogeneous, the solutions are of the form y 1 ( t ) = y 1c ( t ) + y 1p ( t ) y 2 ( t ) = y 2c ( t ) + y 2p ( t ) , where c represents the complementary (homogeneous) solution, and p represents the particular solution. The complementary solutions are y 1c ( t ) = c 1 e λ 1 t = c 1 e 2 t y 2c ( t ) = c 2 e λ 2 t = c 2 e 2 t . We determine the particular solutions using the method of undetermined coeﬃ- cients (see Jeffrey § 6.4, Table 6.2, or Kreyszig § 2.9). Consider the equation for y 1 ( t ) y 1p ( t ) = A sin t + B cos t + Ct + D, y 1p ( t ) = A cos t B sin t + C. Note: Be sure y p ( t ) isn’t part of y c ( t ), e.g. y p ( t ) = Ae 2 t + . . . . If it is, include t factor, e.g. y p = Ate 2 t + . . . (see Jeffrey, example 6.13). Substituting A cos t B sin t + C = 2( A sin t + B cos t + Ct + D ) + 1 2 sin t + 1 2 t. Equating like terms leads to four equations for the four unknown constants as follows cos t : A = 2 B A = 1 5 sin t : B = 2 A + 1 2 B = 4 B + 1 2 B = 1 10 t : 0 = 2 C + 1 2 C = 1 4 const : C = 2 D D = 1 8 .
CHAPTER 2. THE EIGENPROBLEM AND ITS APPLICATIONS 55 Thus, y 1p ( t ) = 1 5 sin t 1 10 cos t 1 4 t 1 8 , and y 1 ( t ) = y 1 c ( t ) + y 1 p ( t ) = c 1 e 2 t 1 5 sin t 1 10 cos t 1 4 t 1 8 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern