This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
 Spring '09
 Tabuada

Click to edit the document details