LC natural response derivation 4

So now we're going to use the initial conditions to figure out our values, our two constant values A1 and A2 that is in our proposed solution for current for the LC circuit.

So one thing we need to do, because this is a second order equation, we need to have two initial conditions for the variable that we're studying here. So we're studying I right now. We have one initial condition for I, and because we have a second order equation, that means we need two initial conditions for I. So we have one initial condition right here, and what we'd like to know is what is di/dt at time equals zero. So the other piece of information we have is this v_kn at time equals zero.

Let's use that and we'll just plug that straight into the inductor equation. So the inductor equation at T equals 0, the voltage across the inductor is V_kn, and that equals L * di/dt. All right, and that means that di/dt equals V_kn over L. So now I have two initial conditions in terms of I. There's one and there's one there, and we can use these now to go after A1 and A2.

First off, let's plug in I for time equals zero and then see if we can work out something over here. So that means at time equals zero, the current is zero, and that equals A1 * cos(ω_kn * 0) + A2 * sin(ω_kn * 0). And what does this evaluate to? Okay, this is sin(0) and sin(0) is 0, and cosine of 0 is 1. So that comes up with 0 equal A1.

Okay, and A1 equals 0 means that this entire term of our solution just dropped out. All right, let me rewrite what we end up with. I equals A2 * sin(ω_kn * t). This whole term here just dropped out of the solution.

So here's our proposed solution down here. Now we need to go after A2. Let's do that. As you might suspect, we're going to use our second initial condition to do that. So to use our initial condition, we need di/dt. So let's take d/dt of this.

We're going to take d/dt of this whole equation, and on the left side, we'll get di/dt, and on the other side, we'll get d/dt of A2 * sin(ω_kn * t). Okay, so far so good? Let's roll it down again. So let's take that derivative. We get di/dt equals A2 comes out of the derivative, and the derivative of sin(ω_kn * t) with respect to t is ω_kn * cos(ω_kn * t).

We apply our initial condition. Let's go to t equals 0 and we know that di/dt was V/L equals A2 * ω_kn * cos(ω_kn * 0). And cosine of 0 goes to one, and so we can solve for A2. A2 equals V_kn over L * ω_kn.

So now we've solved for our second adjustable parameter, and we can write I. I was A2 * sin(ω_kn * t). So let's fill it in for A2. I equals A2, which is V_kn over L * ω_kn * sin(ω_kn * t).

And I want to go back now. I want to write this a little bit differently. I want to go back and plug in our value for ω_kn. So if we remember, we said ω₀ equals 1/sqrt(LC).

So now L * ω_kn equals 1/√(LC) * L, and that equals √(L/C). Lastly, I'll write 1/(L * ω_kn) equals √(C/L), just the reciprocal.

And now we can write I equals √(C/L) * V_kn * sin(ω_kn * t). And that is the solution for the natural response of an LC circuit. It's in the form of a sine wave, and the frequency is determined by ω_kn, which is the two component values, and the amplitude is determined by the energy we started with, which is represented here by V_kn and the ratio of the two components again.

So this is why I said at the beginning that this is where sine waves are born.