s09_ps09_sol

s09_ps09_sol - Unified Engineering II Spring 2004 E1- + +-...

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Unformatted text preview: Unified Engineering II Spring 2004 E1- + +- y ( t ) R u ( t ) C C (a) We can use impedance methods to solve for Y ( s ) in terms of U ( s ). Label ground and E 1 as shown. Then KCL at E 1 yields 1 Cs ( E 1 0) + Cs + ( E 1 U ) = 0 (1) R Simplifying, we have 1 1 2 Cs + R E 1 = Cs + R U Since we are finding the step response, 1 U ( s ) = , Re[ s ] > 0 s Plugging in numbers, we have 1 (0 . 5 s + 0 . 5) E 1 ( s ) = (0 . 25 s + 0 . 5) s Solving for E 1 , we have . 25 s + 0 . 5 . 5 s + 1 E 1 ( s ) = = (0 . 5 s + 0 . 5) s s ( s + 1) The region of convergence must be Re[ s ] > 0, since the step response is causal, and the pole at s = 0 is the rightmost pole. Using partial fraction expansions, 1 . 5 E 1 ( s ) = s s + 1 Therefore, g s ( t ) = y ( t ) = e 1 ( t ) is the inverse transform of E 1 ( t ), so y ( t ) = 1 . 5 e t ( t ) The step response is plotted below: Normal differential equation methods are dicult to apply, because we cannot apply the normal initial condition...
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s09_ps09_sol - Unified Engineering II Spring 2004 E1- + +-...

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