Solving Electrical Circuits using Impedance Methods: Step Response Analysis, Exercises of Signals and Systems Theory

Download Solving Electrical Circuits using Impedance Methods: Step Response Analysis and more Exercises Signals and Systems Theory in PDF only on Docsity! � � � � � � � � 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 E1 as shown. Then KCL at E1 yields 1 Cs(E1 − 0) + Cs + (E1 − U) = 0 (1) R Simplifying, we have 1 1 2Cs + R E1 = Cs + R U Since we are finding the step response, 1 U(s) = , Re[s] > 0 s Plugging in numbers, we have 1 (0.5s + 0.5)E1(s) = (0.25s + 0.5) s Solving for E1, we have 0.25s + 0.5 0.5s + 1 E1(s) = = (0.5s + 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 0.5 E1(s) = s − s + 1 Therefore, gs(t) = y(t) = e1(t) is the inverse transform of E1(t), so y(t) = 1 − 0.5e−t σ(t) The step response is plotted below: docsity.com Normal differential equation methods are difficult to apply, because we cannot apply the normal initial condition that e1(0) = 0. This is because the chain of capacitors running from the voltage source to ground causes there to be an impulse of current at time t = 0, and the voltages across the capacitors change instantaneously at t = 0. It is possible to use differential equation methods, we just have to be more careful about the initial conditions. However, Laplace methods are easier. docsity.com