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

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Unformatted text preview: 16.333 Lecture # 10 State Space Control • Basic state space control approaches 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 ) dτ t y ( t ) = Ce At x (0) + Ce A ( t − τ ) Bu ( τ ) dτ + Du ( t ) – 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. 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 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 . 5 v 2 ( t ) = e − 3 t mode 2 . 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. 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 ....
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lecture_10 - 16.333 Lecture 10 State Space Control •...

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