lect6 - EL 625 Lecture 6 1 EL 625 Lecture 6 Discrete-time...

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Unformatted text preview: EL 625 Lecture 6 1 EL 625 Lecture 6 Discrete-time systems : x ( k + 1) = A ( k ) x ( k ) + B ( k ) u ( k ) y ( k ) = C ( k ) x ( k ) + D ( k ) u ( k ) Initial condition, x ( k ) = x x ( k + 1) = A ( k ) x + B ( k ) u ( k ) x ( k + 2) = A ( k + 1) x ( k + 1) + B ( k + 1) u ( k + 1) = A ( k + 1) A ( k ) x + A ( k + 1) B ( k ) u ( k ) + B ( k + 1) u ( k + 1) x ( k + 3) = A ( k + 2) x ( k + 2) + B ( k + 2) u ( k + 2) = A ( k + 2) A ( k + 1) A ( k ) x + A ( k + 2) A ( k + 1) B ( k ) u ( k ) + A ( k + 2) B ( k + 1) u ( k + 1) + B ( k + 2) u ( k + 2) . . . . . . Transition matrix, ( k, i ) = 4 A ( k- 1) A ( k- 2) . . . A ( i ) for k i + 1 EL 625 Lecture 6 2 x ( k ) = ( k, k ) x + k- 1 X i = k ( k, i + 1) B ( i ) u ( i ) for k k Solution of the unforced system : x ( k ) = ( k, k ) x ( k ) y ( k ) = C ( k ) ( k, k ) x | {z } Zero-input response + k- 1 X i = k C ( k ) ( k, i + 1) B ( i ) u ( i ) + D ( k ) u ( k ) | {z } Zero-state response for k k Zero-state response =...
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lect6 - EL 625 Lecture 6 1 EL 625 Lecture 6 Discrete-time...

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