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

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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 0 ) x ( t 0 ) (3) ˙ x ( t ) = ˙ Q ( t ) Q - 1 ( t 0 ) x ( t 0 ) = A ( t ) Q ( t ) Q - 1 ( t 0 ) x ( t 0 ) = A ( t ) x ( t ) (4) Also, evaluating the left side of (3) at t = t 0 , we have Q ( t 0 ) Q - 1 ( t 0 ) x ( t 0 ) = x ( t 0 ) (5) So, the solution satisfies the initial condition.
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EL 625 Lecture 4 2 ‘Transition matrix’ , φ ( t, t 0 ) = 4 Q ( t ) Q - 1 ( t 0 ) (6) x ( t ) = φ ( t, t 0 ) x ( t 0 ) (7) The transition matrix characterizes the ‘flow’ of the differential e- quation. Properties of the transition matrix : 1. x ( t 0 ) = φ ( t 0 , t 0 ) x ( t 0 ) φ ( t 0 , t 0 ) = I (8) 2. φ ( t 2 , t 0 ) x ( t 0 ) = x ( t 2 ) = φ ( t 2 , t 1 ) x ( t 1 ) = φ ( t 2 , t 1 ) φ ( t 1 , t 0 ) x ( t 0 ) φ ( t 2 , t 1 ) φ ( t 1 , t 0 ) = φ ( t 2 , t 0 ) (9) 3. x ( t 2 ) = φ ( t 2 , t 1 ) x ( t 1 ) = φ ( t 2 , t 1 ) φ ( t 1 , t 2 ) x ( t 2 )
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EL 625 Lecture 4 3 φ ( t 2 , t 1 ) φ ( t 1 , t 2 ) = I (10) φ ( t 1 , t 2 ) = φ - 1 ( t 2 , t 1 ) (11) Given a transition matrix, φ ( t, t 0 ), A ( t ) can be evaluated as follows. ˙ x ( t ) = ˙ φ ( t, t 0 ) x ( t 0 ) Also, ˙ x ( t ) = A ( t ) x ( t ) = A ( t ) φ ( t, t 0 ) x ( t 0 ) ˙ φ ( t, t 0 ) = A ( t ) φ ( t, t 0 ) (12) ˙ φ ( t, t 0 ) | t 0 = t = A ( t ) (13) We can also determine an expression for φ - 1 ( t, t 0 ). φ ( t, t 0 ) φ - 1 ( t, t 0 ) = I (14) d dt φ ( t, t 0 ) φ - 1 ( t, t 0 ) + φ ( t, t 0 ) d dt φ - 1 ( t, t 0 ) = 0 (15) d dt φ ( t, t 0 ) φ - 1 ( t, t 0 ) = - φ ( t, t 0 ) d dt φ - 1 ( t, t 0 ) d dt φ - 1 ( t, t 0 ) = - φ - 1 ( t, t 0 ) d dt φ ( t, t 0 ) φ - 1 ( t, t 0 ) = - φ - 1 ( t, t 0 ) A ( t ) φ ( t, t 0 ) φ - 1 ( t, t 0 ) = - φ - 1 ( t, t 0 ) A ( t ) (16)
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EL 625 Lecture 4 4 Solution of the forced system equations : Assume : x ( t ) = φ ( t, t 0 ) f ( t ) (17) = ˙ x ( t ) = ˙ φ ( t, t 0 ) f ( t ) + φ ( t, t 0 ) ˙ f ( t ) = A ( t ) φ ( t, t 0 ) f ( t ) + φ ( t, t 0 ) ˙ f ( t ) = A ( t ) x ( t ) + φ ( t, t 0 ) ˙ f ( t ) = φ ( t, t 0 ) ˙ f ( t ) = B ( t ) u ( t ) (18) Thus, f ( t ) = f ( t 0 ) + Z t t 0 φ ( t 0 , λ ) B ( λ ) u ( λ ) (19) x ( t ) = φ ( t, t 0 ) f ( t 0 ) + Z t t 0 φ ( t, t 0 ) φ ( t 0 , λ ) B ( λ ) u ( λ ) (20) x ( t ) = φ ( t, t 0 ) x ( t 0 ) + Z t t 0 φ ( t, λ ) B ( λ ) u ( λ ) (21)
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EL 625 Lecture 4 5 y ( t ) = C ( t ) φ ( t, t 0 ) x ( t 0 ) | {z } Zero-input response + Z t t 0 C ( t ) φ ( t, λ ) B ( λ ) u ( λ ) + D ( t ) u ( t ) | {z } Zero-state response (22) Using u ( t ) = Z -∞ u ( λ ) δ ( t - λ ) (23) and Z t t 0 C ( t ) φ ( t, λ ) B ( λ ) u ( λ ) = Z t 0 C ( t ) φ ( t, λ ) B ( λ ) u ( λ )1( t - λ ) (24) With t 0 = -∞ and x ( t 0 ) = 0, Z t 0 C ( t ) φ ( t, λ ) B ( λ ) u ( λ )1( t - λ ) = Z -∞ C ( t ) φ ( t, λ ) B ( λ ) u ( λ )1( t - λ ) (25) Hence, Zero-state response = Z -∞ [ C ( t ) φ ( t, λ ) B ( λ )1( t - λ ) + D ( t ) δ ( t - λ )] u ( λ ) = Z -∞ H ( t, λ ) u ( λ ) (26) where,
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EL 625 Lecture 4 6 H ( t, λ ) = C ( t ) φ ( t, λ ) B ( λ )1( t - λ ) + D ( t ) δ ( t - λ ) For fixed systems, A, B, C and D are constant matrices H ( t, λ ) = H ( t - λ ) (27) H ( t ) = ( t ) B 1( t ) + ( t ) (28) A different approach to the derivation of the forced response ˙ x ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) φ ( t 0 , t ) ˙ x ( t ) = φ ( t 0 , t ) A ( t ) x ( t ) + φ ( t 0 , t ) B ( t ) u ( t ) φ ( t 0 , t ) ˙ x ( t ) - φ ( t 0 , t ) A ( t ) x ( t ) = φ ( t 0 , t ) B ( t ) u ( t ) φ ( t 0 , t ) ˙ x ( t ) + ˙ φ ( t 0 , t ) x ( t ) = φ ( t 0 , t ) B ( t ) u ( t ) (29) d dt ( φ ( t 0 , t ) x ( t )) = φ ( t 0 , t ) B ( t ) u ( t ) (30) φ ( t 0 , τ ) x ( t ) | t τ = t 0 = Z t t 0 φ ( t 0 , τ ) B ( τ ) u ( τ ) (31) φ ( t 0 , t ) x ( t ) - x ( t 0 ) = Z t t 0 φ ( t 0 , τ ) B ( τ ) u ( τ ) (32) φ ( t 0 , t ) x ( t ) = x ( t 0 ) + Z t t 0 φ ( t 0 , τ ) B ( τ ) u ( τ ) (33) x ( t ) = φ ( t, t 0 ) x ( t 0 ) + φ ( t, t 0 ) Z t t 0 φ ( t 0 , τ ) B ( τ ) u ( τ ) (34)
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EL 625 Lecture 4 7 = x ( t ) = φ ( t, t 0 ) x ( t 0 ) + Z t t 0 φ ( t, τ ) B ( τ ) u ( τ ) (35) Computing the transition matrix The physical meaning of φ ( t, t 0 ) :
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