WUSTL ESE 543 - Lecture 3 and 4 eig cont obs (66 pages)

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Lecture 3 and 4 eig cont obs



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Lecture 3 and 4 eig cont obs

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Pages:
66
School:
Washington University in St. Louis
Course:
Ese 543 - Control Systems Design by State Space Methods

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Lecture 3 The State Transition Matrix Chapter 9 Time Varying Systems x A t x B t u y C t x D t u Eq 1 x R nx u R nu y R ny x A t x Theorem There exists uniquely for every x 0 a x t that satisfies Eq 1 Theorem Solutions to Eq 1 form a nx dimensional vector space over R Def U nx nx Fundamental matrix if its columns consist of nx linearly independent solutions to Eq 1 U nx nx U1 U nx 2 K A Wise Fundamental Matrix Example 0 0 x x t 0 x1 t x 1 t0 X 10 x 1 t 0 x 2 t tx1 t x2 t 1 2 t 2 X 10 X 20 Since the dimension of the vector space is 2 2 linearly independent initial vectors can be found and each one defines an unique solution X10 0 x t0 X 20 1 0 U1 t 1 X10 2 x t0 0 X 20 2 U 2 t 2 t 0 2 U t 2 1 t 3 K A Wise The State Transition Matrix The solution to the state equation describes the transition of the initial state x t0 state to the state x t at time t x t0 x t Define t t0 U t U 1 t0 as the state transition matrix Let U1 t and U 2 t be two fundamental matrices where U1 U 2 P 1 t t0 U1 t U 1 t0 1 t0 U 2 t P P 1U 2 U 2 t U 2 1 t0 t t0 exhibits the following properties 1 t t I 3 t2 t0 t2 t1 t1 t0 t0 t1 t2 2 1 t t0 t0 t 4 t t0 satisfies x A t x t t0 U t U 1 t0 A t U t U 1 t0 A t t t0 A t U t K A Wise 4 The State Transition Matrix From the previous example 0 2 0 2 1 U t U t0 U t0 2 2 1 t 1 t 0 Now consider the general solution to x Theorem The solution to x 1 2 t02 1 2 1 t t0 0 1 2 2 1 2 t t0 0 1 A t x B t u A t x B t u is t x t t t0 x t0 t B u d t0 Zero Input Response Convolution Integral 5 K A Wise Solution to the State Equation Proof Differentiate the solution to show it satisfies the differential equation t x t t0 x0 t B u d t t 0 t t B u d A t t t0 x0 t t B t u t t0 A t t t A t t t0 x0 t B u d B t u t t 0 x t A t x B t u The output is



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