Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part21

Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part21

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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 5 23 Copyright © Orchard Publications Circuit Analysis with State Variables Solution: For this example, and Substitution of given values and rearranging, yields or (5.85) Next, we define the state variables and . Then, (5.86) and Also, and thus, or (5.87) Therefore, from (5.85), (5.86), and (5.87), we obtain the state equations and in matrix form, (5.88) ii L = Ri L L di L dt ------- v C ++ u 0 t () = 1 4 -- L 1 i L v C 1 + = L 4 i L 4v C 4 + = x 1 i L = x 2 v C = x · 1 L = x · 2 dv C -------- = i L C C = x 1 i L C C Cx · 2 4 3 --x · 2 == = = x · 2 3 4 1 = x · 1 4x 1 2 4 + = x · 2 3 4 -- x 1 = x · 1 x · 2 4 4 34 0 x 1 x 2 4 0 u 0 t + =
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Chapter 5 State Variables and State Equations 5 24 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications We will compute the solution of (5.88) using (5.89) where (5.90) First, we compute the state transition matrix . We find the eigenvalues from Then, Therefore, The next step is to find the coefficients . Since is a matrix, we only need the first two terms of the state transition matrix, that is, (5.91) The constants and are found from and with , we obtain (5.92) Simultaneous solution of (5.92) yields (5.93) We now substitute these values into (5.91), and we obtain xt () e At t 0 x 0 e At e A τ bu τ ()τ d t 0 t + = A 4 4 34 0 = x 0 i L 0 v C 0 0 12 == b 4 0 = e det A λ I [] 0 = λ I det 4 λ 4 ⁄λ 0 λ 4 λ 3 + 0 = λ 2 4 λ 3 ++ 0 = λ 1 1 and λ 2 3 = = a i A2 2 × e a 0 Ia 1 A + = a 0 a 1 a 0 a 1 λ 1 + e λ 1 t = a 0 a 1 λ 2 + e λ 2 t = λ 1 λ 2 3 = = a 0 a 1 e t = a 0 3a 1 e 3t = a 0 1.5e t 0.5e = a 1 0.5e t 0.5e =
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 5 25 Copyright © Orchard Publications Circuit Analysis with State Variables or The initial conditions vector is the second vector in (5.90); then, the first term of (5.89) becomes or (5.94) We also need to evaluate the integral on the right side of (5.89). From (5.90)
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Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part21

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