MEM255Su06-07.8.sSpaceResponse.studs

MEM255Su06-07.8.sSpaceResponse.studs - MEM 255: Intro to...

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Unformatted text preview: MEM 255: Intro to Control ** State Response Dr.Ajmal Yousuff Dept. Mech. Engg. & Mechanics Drexel University Inverted Pendulum pendulum carriage A force (t) can be applied on carriage, in horizontal direction. Task: Obtain a math model Yousuff MEM 255 2 Schematics & Notations V s mg H Fs L m : pendulum mass M : carriage mass J : moment of inertia of pendulum w.r.t. its c.g H,V: reaction forces @ pivot (on pendulum) F : friction coefficient. (Coulomb) Yousuff MEM 255 3 Definition of States Assume small perturbations and define the states as: 1 0 0 -F / M ; A = 0 x3 = s + , pendulum displ. 0 , pend. vel. x4 = s + - 0 g / x1 = s, carriage displ. x2 = s, carriage vel. 0 0 0 0 1 g / 0 0 : a normalized length A.Yousuff MEM 255 Control 4 Scalar case (n=1) x = a (t ) x(t ); x(t0 ) = x0 Soln: x(t ) = (t , t0 ) x0 (t , t0 ) = exp{ a ( )d } t0 t A.Yousuff MEM 255 Control 5 Homogeneous,Time Invariant System A - constant Homogeneous System: u (t ) M0 x(t ) = Ax(t ); x(0) = x0 x(t ) = e At x0 e At 1 1 2 = I + At + ( At ) + ( At )3 + ...... 2! 3! A.Yousuff MEM 255 Control 6 Non-homogeneous x(t ) = Ax(t ) + Bu (t ); x(0) = x0 x(t ) = e At x0 + @( t - ) Bu ( )d eA 0 t eA(t- ) : TRANSITION MATRIX x(t ) = e A(t - ) x( ) + e A( t - ) Bu ( )d t A.Yousuff MEM 255 Control 7 Properties of State Transition Matrix e A ( t1 + t2 ) At -1 =e e At1 - At t At2 [e ] = e At At -1 e = Me M ; M = modal matrix e = L {( sI - A) } A.Yousuff MEM 255 Control 8 -1 -1 Similarity Transformation x = Ax + Bu; y = Cx + Du Transform (redefine) the states xx: x = T -1 x x = Tx Similarity transformation x = Ax + Bu; y = Cx + Du A = T -1 AT , B = T -1 B, C = CT , D = D A.Yousuff MEM 255 Control 9 Solution using eigenvectors Let T = M, the modal matrix, i.e., M-1AM=, a diagonal matrix. M = [ 1, 2, ... n], with A i = i i ; I = 1, 2, ..., n ith eigenvector ith eigenvalue Then, in the transformed state x, one gets = i i Assume distinct i A = T AT = M AM = -1 -1 x = x , x0 = T x0 A.Yousuff MEM 255 Control -1 scalar decoupled eqns. real s 10 Modal coordinates Solution to scalar equations: x 1 Recover the solution in original states x: x 2 x(t ) = Tx (t ) = Mx (t ) = [ 1 2 ... n ] M n x n i t xi (t ) = ei t xi 0 ; i = 1, 2,..., n x(t ) = @ i e xi 0 i =1 scalar, const. scalar, time-dependent vector, const. 11 A.Yousuff MEM 255 Control Initial conditions x0 The initial conditions are in terms of new states. x0 = T -1 x0 = M -1 x0 Express in terms of original states. Denote the rows of M-1 by ', so that 1' ' -1 2 x = ' x ; i = 1, 2,.., n x=M x= x i0 i 0 M ' n x(t ) = @ i ei ti' x0 i =1 A.Yousuff MEM 255 Control 12 n Modal response The term i ei ti' x0 is called the "modal response," since it is associated with the ith mode " i." Property of the product of and ': ' 1 ' -1 2 L ] = ' M M = In = [ n j i 1 2 M ' n { } 1 ,i = j i j = ij @ @ 0 @ , i @j ' A.Yousuff MEM 255 Control 13 Exciting ith mode Hence, if we choose x0 = i, for some i: x(t ) = e x0 0 i t The response remains along x with its magnitude changing with time. Thus, we can excite a specific mode with an appropriate choice of initial condition. A.Yousuff MEM 255 Control 14 Pendulum example Define the states to be 1 0 0 -F / M ; A = 0 x3 = s + , pendulum displ. 0 , pend. vel. x4 = s + - 0 g / x1 = s, carriage displ. x2 = s, carriage vel. 0 0 0 1 g / 0 0 0 ( A) = {0, - F g g , ,- } M A.Yousuff MEM 255 Control 15 Pendulum motion Associated eigenvectors: 1 1 0 F / M 0 - 0 , , , 1 1 0 - F / M g / - marg. stable unstable stable 0 1 g / 0 stable x1 = s x =x 1 2 3 = s + x x4 = x3 g/ @ g / - F2 / M 2 16 A.Yousuff MEM 255 Control ...
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This note was uploaded on 06/03/2008 for the course MEM 255 taught by Professor Yousuff during the Spring '08 term at Drexel.

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