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lecture_10 - 16.333 Lecture 10 State Space Control Basic...

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16.333 Lecture # 10 State Space Control Basic state space control approaches
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Fall 2004 16.333 9–1 State Space Basics State space models are of the form x ˙( t ) = Ax ( t ) + Bu ( t ) y ( t ) = Cx ( t ) + Du ( t ) with associated transfer function G ( s ) = C ( sI A ) 1 B + D Note: must form symbolic inverse of matrix ( sI A ) , which is hard. Time response: Homogeneous part x ˙ = Ax, x (0) known Take Laplace transform X ( s ) = ( sI A ) 1 x (0) x ( t ) = L 1 ( sI A ) 1 x (0) I A A 2 But can show ( sI A ) 1 = + s 2 + s 3 + . . . s 1 so L 1 ( sI A ) 1 = I + At + 2! ( At ) 2 + . . . = e At Gives x ( t ) = e At x (0) where e At is Matrix Exponential 1 3 Calculate in MATLAB R using expm.m and not exp.m Time response: Forced Solution Matrix case x ˙ = Ax + Bu where x is an n -vector and u is a m -vector. Cam show t x ( t ) = e At x (0) + e A ( t τ ) Bu ( τ ) 0 t y ( t ) = Ce At x (0) + Ce A ( t τ ) Bu ( τ ) + Du ( t ) 0 Ce At x (0) is the initial response Ce A ( t ) B is the impulse response of the system. 1 MATLAB R is a trademark of the Mathworks Inc.
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Fall 2004 16.333 9–2 Dynamic Interpretation Since A = T Λ T 1 , then ⎤ ⎡ ⎤ ⎡ T e At | | λ 1 t . . . w . . 1 e = T e Λ t T 1 = v 1 ⎦ ⎣ ⎦ ⎣ . · · · v n | | λ n t T e w n where we have written T w . 1 . T 1 = . T w n which is a column of rows. Multiply this expression out and we get that n At λ i t T e = e v i w i i =1 Assume A diagonalizable, then x ˙ = Ax , x (0) given, has solution x ( t ) = e At x (0) = T e Λ t T 1 x (0) n = e λ i t v i { w i T x (0) } i =1 n λ i t = e v i β i i =1 State solution is a linear combination of the system modes v i e λ i e λ i t Determines the nature of the time response v i Determines extent to which each state contributes to that mode β i Determines extent to which the initial condition excites the mode
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Fall 2004 16.333 9–3 Note that the v i give the relative sizing of the response of each part of the state vector to the response. 1 v 1 ( t ) = e t mode 1 0 0 . 5 v 2 ( t ) = e 3 t mode 2 0 . 5 Clearly e λ i t gives the time modulation λ i real growing/decaying exponential response λ i complex growing/decaying exponential damped sinusoidal Bottom line: The locations of the eigenvalues determine the pole locations for the system, thus: They determine the stability and/or performance & transient be- havior of the system. It is their locations that we will want to modify with the controllers.
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Fall 2004 16.333 9–4 Full-state Feedback Controller Assume that the single-input system dynamics are given by x ˙ = Ax + Bu y = Cx so that D = 0 . The multi-actuator case is quite a bit more complicated as we would have many extra degrees of freedom. Recall that the system poles are given by the eigenvalues of A .
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