Lecture 10 Notes State Space Control

# Lecture 10 Notes State Space Control - Lecture 10 Notes...

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Unformatted text preview: Lecture 10 Notes: State Space Control • Basic state space control approaches 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) – But can show (sI − A) − 1 = I + A + A 2 + . . . 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 3 Calculate in MATLAB R using expm . m and not exp. m 1 • 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 y(t) = Ce At x(0) + t 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. Dynamic Interpretati o n • Since A = TΛT − 1 , then ⎡ | e At = Te Λt T − 1 = ⎣ v 1 ··· | | ⎤⎡ e λ 1 t v n ⎦⎣ | .. . ⎤⎡ − ⎦⎣ e λ n t − w 1 . w n − ⎤ ⎦ − where we have written − T − 1 = ⎣ − which is a column of rows. w 1 . w n − ⎤ ⎦ − • Multiply this expression out and we get that e At = n e λ i t v i w T i= 1 • Assume A diagonalizable, then x = Ax, x(0) given, has solution x(t) = e At x(0) = Te Λt T − 1 x(0) n = e λ i t v i { w T x(0)} i= 1 = n e λ i t 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 excit es the mode • Note that the v...
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