![]() ![]() (D) Form v_H by summing up the two terms. (C) Solve the characteristic equation for the roots s (A) Assume a solution of the form vH is Ae raised to st, where A and s are unknowns. So there's a step 2A, there's a step 2B, there's a step 2C, and finally a step 2D. OK, so let's write down what those four steps are. V_H is the solution to the homogeneous equation which is the original equation with the drive set to zero.Īnd as you recall, I've been using a four-step method to solve the homogeneous equation. If we try try v_p equals v_i, we will find that v_p equals v_i is a particular solution. And similarly, i(t) will also be equal to 0. Let's go ahead and solve it for the ZSR, OK, so initial conditions will be these. Let's pick the following initial conditions. And this input can be expressed as V_I, u(t). Input will step from 0 to capital V_I at time t equal to 0. The third step of the solution is to find the total solution as the sum of the particular and homogeneous solutions. So step two of the solution involves finding the homogeneous solution, which itself has four steps. And we will do this in a four-step process. The second step is to find the homogeneous solution.Īnd, as you recall from our solution of the LC circuit, the solution to the homogeneous equation will To solve the differential equations, there are three steps.The first step is to find the particular solution, vP. The node method always works but there is another cute little way of getting the equation. ![]() By equating, differentiating and arranging the equations, we will be able to get the second order equation. With node voltage analysis at nodes V and V_A, we will be able to come up with some equations. Video and lecture notes:- edX MITx: 6.002.2x Circuits and Electronics 3: week 1Īn inductor here, a resistor and a capacitor is connected in series.The voltage across the capacitor is V(t), and the current in the series circuit is i(t). ![]()
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