topic9 - Topic#9 16.31 Feedback Control Systems State-Space...

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Topic #9 16.31 Feedback Control Systems State-Space Systems What are the basic properties of a state-space model, and how do we analyze these? Time Domain Interpretations System Modes Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Fall 2007 16.31 9–1 SS: Forced Solution Forced Solution Consider a scalar case: ˙ x = ax + bu, x (0) given x ( t ) = e at x (0) + t 0 e a ( t τ ) bu ( τ ) where did this come from? 1. ˙ x ax = bu 2. e at [ ˙ x ax ] = d dt ( e at x ( t )) = e at bu ( t ) 3. t 0 d e x ( τ ) = e at x ( t ) x (0) = t 0 e bu ( τ ) Forced Solution – Matrix case: ˙ x = A x + B u where x is an n -vector and u is a m -vector Just follow the same steps as above to get x ( t ) = e At x (0) + t 0 e A ( t τ ) B u ( τ ) and if y = C x + D u , then y ( t ) = Ce At x (0) + t 0 Ce A ( t τ ) B u ( τ ) + D u ( t ) Ce At x (0) is the initial response Ce A ( t ) B is the impulse response of the system. September 22, 2007 Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Fall 2007 16.31 9–2 Have seen the key role of e At in the solution for x ( t ) Determines the system time response But would like to get more insight! Consider what happens if the matrix A is diagonalizable, i.e. there exists a T such that T 1 AT = Λ which is diagonal Λ = λ 1 . . . λ n Then e At = Te Λ t T 1 where e Λ t = e λ 1 t . . . e λ n t Follows since e At = I + At + 1 2! ( At ) 2 + . . . and that A = T Λ T 1 , so we can show that e At = I + At + 1 2! ( At ) 2 + . . . = I + T Λ T 1 t + 1 2! ( T Λ T 1 t ) 2 + . . . = Te Λ t T 1 This is a simpler way to get the matrix exponential, but how find T and λ ? Eigenvalues and Eigenvectors September 22, 2007 Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Fall 2007 16.31 9–3 Eigenvalues and Eigenvectors Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page 8–2) λ is an eigenvalue of A if det( λI A ) = 0 which is true iff there exists a nonzero v ( eigenvector ) for which ( λI A ) v = 0 Av = λv Repeat the process to find all of the eigenvectors. Assuming that the n
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