LC natural response derivation 2

- [Voiceover] In the last video, we set up this differential equation that described an LC circuit, and now we're gonna go about solving this second order circuit. The technique that works here is the same that worked with first order ordinary differential equations. We're looking for a function of I that makes this whole equation true, and we're gonna do that by guessing at an I, put it back into the equations, and see if it works. If it works, we win. If it doesn't work, we have to think of something else, and we keep doing that until we solve it. So, knowing what I know about derivatives, and knowing and looking at this equation, can I guess at a plausible solution? So what we have here is two things. We have two terms that have to add up to zero for all time, and so that means that whatever function I pick for I, and there's some scaling factors here. Lemme write this equation like this. So what I have here is a scaling factor times I has to somehow fully subtract from the second derivative of I. So these two terms have to somehow look alike for all time. There's one function I know where its derivatives sort of look like what we started with, and that's the exponential function. So I'm actually gonna make a guess. I'm gonna guess that I of T is something of the form some constant times the exponential of time with some other scale factor. Now, K is an adjustable parameter, and that's an amplitude. So it tells us how big the signal is, and what's S? S is up in the exponent along with T, and we know that by the time we take an exponent of anything that whatever's up here has to have no units, so that means that ST, S times T, has no units, and that means that S has units of one over time so S is a frequency of some sort. And in particular, it's gonna be a radian frequency. It's gonna be in radians per second. So, S is gonna be called a natural frequency. So let's keep working on this. We're gonna plug I back into our equation, and see if it works. So we can plug I straight into here, and we need the second derivative of I. Let's first take the first derivative D, DT of I equals D, DT of K E to the ST, and we can take that derivative, and that equals K times S times E to the ST. So it's also an exponential, and now we need the second derivative, so we wanna take the derivative of this guy, so second derivative of I with respect to time equals the first derivative of S, K, E to the ST, and that equals-- Another S comes down, so it's S squared, K, E to the ST. Good, now we have our second derivative. We can plug that in here, so let's do that. So the equation becomes S squared, K, E to the ST, plus one over LC, times K, E to the ST, equals zero. Let's do a little factoring. There's a common term here. There's a common term. K, E to the ST, K to the E, to the ST, so lemme factor that out. K, E to the ST times, what's left? S squared plus one over LC equals zero. Okay, we have how many adjustable parameters here? We have K, S. Those are the two. L and C are constants that are values of our circuit. So, we need to find some values of K and S that make this equation zero. Okay, I can make K equal to zero, and that would mean that zero equals zero, so our amplitude would be zero. So if we put nothing in the circuit, we get nothing out. Okay, that's totally boring. So that's not so interesting of a solution. Now, does E to the ST ever become zero? E to something, does it ever become zero? It never does. If I let T go to plus infinity, and S is negative, then E to the ST would become zero, but plus infinity time from now is pretty far in the future, and I don't wanna wait that long. So the interesting solution becomes can we make S squared plus one over LC equal to zero? This equation is referred to as the characteristic equation. So let's see what happens when we try to solve this. Well, the first step of this is I'm gonna get S squared equals minus one over LC, or S equals square root of minus one over LC. Uh oh, look what happens here. We're taking the square root of a negative number. So what's gonna happen here? We're gonna get an imaginary number for our answer. I can write this as square root of minus one times square root of one over LC, and that gives me two answers, and I'm gonna call the first answer S one, and that's gonna be equal to J, which is the imaginary number. That's the square root of minus one for electrical engineers, times what? Times square root of one over LC, and the other, S two, is the negative of that, minus J. So these are two possible solutions to our differential equation. Now, I'm gonna give a nickname to this expression here because I don't wanna write it so much. I'm gonna call this omega knot. So this is a lower case omega Greek letter. This is the capital omega that we use for Ohms, but little omega is often used as this variable here. So I can say that S one equals plus J, omega knot, and S two equals minus J, omega knot. So we're gonna use this notation for a little bit, but just remember I made that simple substitution. We have two different roots that can cause our differential equation to go to zero. So when we combine these into a solution for I, we're gonna use a combination of these two. We don't know which one it is. It could be both. So they could be superimposed on each other, and this is what superposition is for. So we're gonna now have, we need to have K one. We'll have two constants, E to the S one T plus K two, E to the S two T, and I could also write this as-- Let's fill in some of the numbers here, K one, E to the plus J, omega knot, T, plus K two, E to the minus, J omega knot, T. So that's my proposed solution, and what we need to do now, we found S. We found two values of S. Now we have these two constants. We gotta work those out. So, in the next video, we'll use the initial conditions to figure out what K one and K two are.