lect4 - EL 625 Lecture 4 1 EL 625 Lecture 4 Solution of the...

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Unformatted text preview: EL 625 Lecture 4 1 EL 625 Lecture 4 Solution of the dynamic state equations x ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) y ( t ) = C ( t ) x ( t ) + D ( t ) u ( t ) (1) Homogeneous equation (Unforced system) : x = A ( t ) x (2) Consider the matrix differential equation, Q ( t ) = A ( t ) Q ( t ). If this can be solved, the solution to the homogeneous equation is x ( t ) = Q ( t ) Q- 1 ( t ) x ( t ) (3) x ( t ) = Q ( t ) Q- 1 ( t ) x ( t ) = A ( t ) Q ( t ) Q- 1 ( t ) x ( t ) = A ( t ) x ( t ) (4) Also, evaluating the left side of (3) at t = t , we have Q ( t ) Q- 1 ( t ) x ( t ) = x ( t ) (5) So, the solution satisfies the initial condition. EL 625 Lecture 4 2 Transition matrix , ( t, t ) = 4 Q ( t ) Q- 1 ( t ) (6) x ( t ) = ( t, t ) x ( t ) (7) The transition matrix characterizes the flow of the differential e- quation. Properties of the transition matrix : 1. x ( t ) = ( t , t ) x ( t ) ( t , t ) = I (8) 2. ( t 2 , t ) x ( t ) = x ( t 2 ) = ( t 2 , t 1 ) x ( t 1 ) = ( t 2 , t 1 ) ( t 1 , t ) x ( t ) ( t 2 , t 1 ) ( t 1 , t ) = ( t 2 , t ) (9) 3. x ( t 2 ) = ( t 2 , t 1 ) x ( t 1 ) = ( t 2 , t 1 ) ( t 1 , t 2 ) x ( t 2 ) EL 625 Lecture 4 3 ( t 2 , t 1 ) ( t 1 , t 2 ) = I (10) ( t 1 , t 2 ) = - 1 ( t 2 , t 1 ) (11) Given a transition matrix, ( t, t ), A ( t ) can be evaluated as follows. x ( t ) = ( t, t ) x ( t ) Also, x ( t ) = A ( t ) x ( t ) = A ( t ) ( t, t ) x ( t ) ( t, t ) = A ( t ) ( t, t ) (12) ( t, t ) | t = t = A ( t ) (13) We can also determine an expression for - 1 ( t, t ). ( t, t ) - 1 ( t, t ) = I (14) d dt ( t, t ) - 1 ( t, t ) + ( t, t ) d dt - 1 ( t, t ) = 0 (15) d dt ( t, t ) - 1 ( t, t ) =- ( t, t ) d dt - 1 ( t, t ) d dt - 1 ( t, t ) =- - 1 ( t, t ) d dt ( t, t ) - 1 ( t, t ) =- - 1 ( t, t ) A ( t ) ( t, t ) - 1 ( t, t ) =- - 1 ( t, t ) A ( t ) (16) EL 625 Lecture 4 4 Solution of the forced system equations : Assume : x ( t ) = ( t, t ) f ( t ) (17) = x ( t ) = ( t, t ) f ( t ) + ( t, t ) f ( t ) = A ( t ) ( t, t ) f ( t ) + ( t, t ) f ( t ) = A ( t ) x ( t ) + ( t, t ) f ( t ) = ( t, t ) f ( t ) = B ( t ) u ( t ) (18) Thus, f ( t ) = f ( t ) + Z t t ( t , ) B ( ) u ( ) d (19) x ( t ) = ( t, t ) f ( t ) + Z t t ( t, t ) ( t , ) B ( ) u ( ) d (20) x ( t ) = ( t, t ) x ( t ) + Z t t ( t, ) B ( ) u ( ) d (21) EL 625 Lecture 4 5 y ( t ) = C ( t ) ( t, t ) x ( t ) | {z } Zero-input response + Z t t C ( t ) ( t, ) B ( ) u ( ) d + D ( t ) u ( t ) | {z } Zero-state response (22) Using u ( t ) = Z - u ( ) ( t- ) d (23) and Z t t C ( t ) ( t, ) B ( ) u ( ) d = Z t C ( t ) ( t, ) B ( ) u ( )1( t- ) d (24) With t =- and x ( t ) = 0, Z...
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lect4 - EL 625 Lecture 4 1 EL 625 Lecture 4 Solution of the...

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