Question

Problem #2 letter a. Please!!!

University of Louisville Electrical and Computer Engineering Department Dr. Aly Farag Summer 2018 ECE 320: Hw 3 Due Tuesday 615/2018 Problem 1: For the circuit below, Derive the equation for the steady state voltage vo). Evaluate the state voltage when R1-R2 0.5 Ohms and L-1 Henry. itt) cos Problem 2: For the given circuit, and using the superposition property, evaluate the voltage voG) a) The Differential Equations Method. b) The Phasors Method. 42 10 cos 2r V Problem 3: Consider a continuous function x(E), defined for t 2 0. The Laplace Transform (LT) for x(t) is defined as: X(s) J x(t)e-st dt. Derive the following properties a) LT(6(t)) = 1, the ?(t) is the Dirac-delta function b) LT(u(t)) = 1/s, where u(t) is the unit-step function c) LT(sin(at))-0Ks'ta?) d) LT(x(t-t)u(t-t)) = e-STX(s), ? > 0. e) LT(tx()X(S) Note: Staple your papers and make your solutions neat and readable.

Answer 1

a) as it is not a transient problem we have to use steady state A C analysis when 10 cos 2t is considered: omega = 2 deactivate other sources i (t) = 10 cos 2t/2.123 + j 2.203 i (t) = 3.26 cos 2 (t - 23.02) v_0 (t) = (ln) [3.26 cos (2t - 46.04)] v_0_1 (t) = 3.26 cos (2t - 46.04) When current source 2 sin 5t alone considered: i = 5 sin 5t times j_10/(1.8 + j 8.4) = 5 sin 5t [1.138 + j 0.243]i = 5.818 sin (5t - 12.05) v_0 = 1 ohm [5.818 sin (5t - 12.05) v_0_2 = 5.818 sin (5t - 12.05) when 5v source alone: under steady state inductor acts as short capacitor as open i = 5/5 = 1 A V_0_3 = 1 ohm [- 1] = - 1v using super position Theorem v_0 = v_0_1 + v_0_2 + v_0_3