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# topic10 - Topic#10 16.31 Feedback Control Systems...

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Unformatted text preview: Topic #10 16.31 Feedback Control Systems State-Space Systems • System Zeros • Transfer Function Matrices for MIMO systems 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]. Fall 2007 16.31 10–1 Zeros in State Space Models • Roots of transfer function numerator called the system zeros . – Need to develop a similar way of defining/computing them using a state space model. • Zero: generalized frequency s for which the system can have a non-zero input u ( t ) = u e s t , but exactly zero output y ( t ) ≡ ∀ t – Note that there is a specific initial condition associated with this response x , so the state response is of the form x ( t ) = x e s t u ( t ) = u e s t ⇒ x ( t ) = x e s t ⇒ y ( t ) ≡ • Given ˙ x = A x + B u , substitute the above to get: x s e s t = A x e s t + B u e s t ⇒ s I − A − B x u = 0 • Also have that y = C x + D u = 0 which gives: C x e s t + D u e s t = 0 → C D x u = 0 • So we must find the s that solves: s I − A − B C D x u = 0 – Is a generalized eigenvalue problem that can be solved in MATLAB R using eig.m or tzero.m 4 4 MATLAB R is a trademark of the Mathworks Inc. October 1, 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]. Fall 2007 16.31 10–2 • There is a zero at the frequency s if there exists a non-trivial solution of det s I − A − B C D = 0 – Compare with equation on page 8–1 • Key Point: Zeros have both direction x u and frequency s – Just as we would associate a direction (eigenvector) with each pole (frequency λ i ) • Example: G ( s ) = s +2 s 2 +7 s +12 A = − 7 − 12 1 B = 1 C = 1 2 D = 0 det s I − A − B C D = det ⎡ ⎣ s + 7 12 − 1 − 1 s 1 2 ⎤ ⎦ = ( s + 7)(0) + 1(2) + 1( s ) = s + 2 = 0 so there is clearly a zero at s = − 2 , as we expected. For the directions, solve: ⎡ ⎣ s + 7 12 − 1 − 1 s 1 2 ⎤ ⎦ s = − 2 ⎡ ⎣ x 01 x 02 u ⎤ ⎦ = ⎡ ⎣ 5 12 − 1 − 1 − 2 1 2 ⎤ ⎦ ⎡ ⎣ x 01 x 02 u ⎤ ⎦ = 0? gives x 01 = − 2 x 02 and u = 2 x 02 so that with x 02 = 1 x = − 2 1 and u = 2 e − 2 t October 1, 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]. Fall 2007...
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## This note was uploaded on 11/07/2011 for the course AERO 16.31 taught by Professor Jonathanhow during the Fall '07 term at MIT.

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topic10 - Topic#10 16.31 Feedback Control Systems...

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