LC natural response derivation 1

In this video, we're going to begin the derivation of the LC natural response, the response of an inductor capacitor circuit. This is a difficult derivation, but it really pays off in the end. There's a real fun surprise at the end, and that is this is where sine waves are born. We're going to end up with sine waves at the end of this, and that's a really nice result because these are everywhere in electronics and in the natural world.

We're going to start out by saying we're going to put some charge on this capacitor here, and we're going to have the current in this circuit, I. We're going to have that start at 0. So I zero is zero, but this is the current through the circuit, and that means that current exists over here as well. Let's put a switch in this circuit, and we're going to close that switch at t equals 0.

So we have one variable; one variable is I, and the other variable we're going to use is V. V is this voltage here after the switch closes. So we want to find I and V, and for this we're going to focus on finding I. Once we find I, it's straightforward to find V. So this is going to be our independent variable, the current.

Let's begin the analysis at the moment t equals 0. The charge that exists here is going to start flowing in the circuit, and both voltage and current are going to start to change. So let's write some expressions. Let's start out by writing some things we know that are true about our two components, and we'll start with the capacitor.

So we'll start by writing an expression for our capacitor, and there's a little sign flip that happens here. So we got to be a little bit careful. If I have V on a capacitor and I in a capacitor, I would say that I capacitor equals C dv/dt. Now, if we look here, Vc is the same as V here, positive sign at the top, positive sign at the top, and negative sign at the bottom.

So Vc and V are the same, but we have to be careful; I is in the opposite direction of the current that I picked for our independent variable, so that's upside down. So there's a negative sign we get to flip here. It equals negative C dv/dt, so that's the IV equation for our particular circuit. We have that little extra negative sign, and now I want to write the integral form of the capacitor IV equation which is V equals 1 over C times the integral of I dt. Remember our minus sign; this is the important minus sign to bring along.

Okay, so let's do the voltage across the inductor. The inductor voltage is the same variable; V is L di/dt, so there's no extra minus sign here. This is assigned according to the sign convention for passive components. So now we have an expression for V on the capacitor, and we have an expression for V on the inductor, and we know that that's the same voltage. So let's set these two things together.

We can say that L di/dt equals minus 1 over C times the integral of I dt. All we did was set the two identical voltages equal to each other. Now, I'm going to manipulate this a bit. It says that L di/dt plus 1 over C the integral of I dt equals 0. We just gathered the terms on one side, and the other side became zero.

Now, since I'm not used to having integrals inside equations, I'd rather have derivatives since I have some experience with differential equations. If we take a derivative of this whole equation, here's what we're going to do. We're going to take a derivative with respect to time of this whole thing here, and that gives us the second derivative of the first term.

So we get L d²I/dt² plus 1 over C, and then the derivative of the integral of I dt turns into just I, and the derivative of zero on this side is zero. So this is now the differential equation for the LC circuit, and it's called—it has a name. It's called a second-order homogeneous ordinary differential equation.

It's a differential equation because it has derivatives in it. It's homogeneous because it only has derivatives of I with respect to t and nothing else; the indicator is that this side is equal to zero. There's no forcing term over here on this side, so when you can write the equation this way, we say it's homogeneous. It's called a second-order equation because it has this second derivative right here.

So now we've set up our second-order homogeneous ODE, and in the next video, we'll go about solving this. We're going to go through it step by step.