LC natural response derivation 3

In the last video, we took a guess at what the solution was for our differential equation, and we came up with an exponential as our guess. As we did the analysis, we developed a characteristic equation. We ended up with a complex answer for one of the adjustable parameters: the natural frequency of our circuit.

So here's the form of our proposed solution. Now, this is getting kind of complicated, but I promise you, we're going to simplify this pretty soon. We have two solutions for s, s1 and s2, that we plugged in as plus or minus j omega naught, and we have more adjustable amplitude parameters that we have to figure out.

So now, in this video, we're going to continue on solving our differential equation. This is our proposed solution; this is a pretty complicated looking expression. What we're going to call on here now is something really important in electronics and in general, and it's called Euler's identity. We're going to use this identity to figure out what to do with these complex exponential terms.

If you search on this term on Khan Academy, you'll find an explanation of where these identities come from, but I'm just going to state them right here. What it says is that Euler's identity is e to the j x anything up there equals cosine x plus j times sine x. That's one of the identities, and the other identity is e to the minus j x equals cosine of x minus j sine x.

So these are useful because we have this exponential function with the complex unit inside of it. When we go over to this side, this is a sort of a normal complex number. Cosine x is some number between plus and minus one, sine x is some number between plus and minus one, and it's just a normal complex number. So this may help us simplify our life here as we move forward.

All right, so that's a really important identity that we get to use to solve our LC circuit. Now, I'm going to go back and rewrite these two exponentials using Euler's identity here, and this is going to get big, but it'll collapse down pretty soon.

Okay, let me move over here: i equals k1 times e to the plus j omega t. Let's use this one here, so that would be cosine x is omega naught t plus j sine x is omega naught t, and we use the first one. Okay, now the second term is plus k2 times. Now we have the negative up in the exponent, so we use this one: cosine omega naught t minus j sine omega naught t.

Now we have our solution spread all out across the screen, and let's see if we can tidy things up here. So what I'm going to do is I'm going to gather all the cosine terms together, this one and this one, and then I'll gather the sine terms together.

Okay, so i equals cosine omega naught t, and cosine is multiplied by k1 and plus k2. All right, now let's add to that; we're gonna have j sine and j sine, so I'll write sine omega naught t over here, and j times what? j times k1, and this time we have this minus sign that makes it minus k2.

All right, so now current is some number times cosine plus j times some number times sine. Now these are two arbitrary constants, and I'm going to just make up another one. I'm just going to call this one a1, and we'll call this one a2, and we'll call that; I'll use the j k1 minus k2.

And this now I can rewrite this as i equals a1 cos omega naught t plus a2 sine omega naught t. Good! So from now on, we're going to work with these a's, and if I ever want to know what the original k's were, I would just come back to these equations here. Once I figure out a, I can figure out both the k's.

So let's keep pressing on. How do we figure out a1 and a2? To do that, we're going to use the initial conditions (the ICs). And if we think back, we remember that in our original schematic we had some q here, which means we had a plus or minus v zero, v naught, and we had—we said the current through here started at zero.

So that's our initial conditions: v of time equals zero equals v naught, and current of time equals zero equals zero. Let's use these two values to help us figure out what a1 and a2 are, and we'll do that in the next video as we continue the derivation of the natural response.