141_1_FinalSols

141_1_FinalSols - G(s) H(s) +- Figure 1: Feedback...

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Unformatted text preview: G(s) H(s) +- Figure 1: Feedback interconnection for Problem 1. Consider the following linear differential equation: d dt x 1 =- x 1 + 2 x 2 (1) d dt x 2 = x 2 + u (2) 1. Assuming that x 1 (0) = 0 and x 2 (0) = 0, what is the evolution of x 1 , in the time domain, when a step is applied as input at t = 0? We first apply the Laplace transform to the differential equations (1) and (2), using the fact that x 1 (0) = 0 and x 2 (0) = 0, to obtain: sX 1 ( s ) =- X 1 ( s ) + 2 X 2 ( s ) (3) sX 2 ( s ) = X 2 ( s ) + U ( s ) (4) (5) Eliminating X 2 in the above equations leads to: X 1 ( s ) = 2 ( s- 1)( s + 1) U ( s ) If the input is a step applied at t = 0 we have U ( s ) = 1 s so that: X 1 ( s ) = 2 ( s- 1)( s + 1) s = a s + b s- 1 + c s + 1 The coefficients in the partial fraction expansion are given by: a = X 1 ( s ) s | s =0 =- 2 b = X 1 ( s )( s- 1) | s =1 = 1 c = X 1 ( s )( s + 1) | s =- 1 = 1 Taking the inverse Laplace transform of X 1 ( s ) we finally obtain: L- 1 { X 1 ( s ) } =- 2 L- 1...
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This note was uploaded on 02/10/2011 for the course ELEC ENGR 141 taught by Professor Roychowdhury during the Fall '11 term at UCLA.

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141_1_FinalSols - G(s) H(s) +- Figure 1: Feedback...

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