RC step response solve (2 of 3)

- [Voiceover] In the last video on step response, we set up the differential equation that describes our circuit, and we found that it was a non-homogeneous equation, and now we're gonna follow through on the strategy of solving it with a forced response plus a natural response. So here's two copies of our circuit, and we're gonna, on the top one here, we're gonna solve for the natural response, and on the bottom one, we'll use that one for the forced response. Alright, let's go to work on the natural. Now, to do a natural response, what we do is we set the initial conditions to whatever they were for the original circuit, so that involves some q and a v zero, a v naught on the capacitor. So let's just draw in quick, let's draw in our q. There's some q here, that'll be a plus or minus v naught on the capacitor. Now the other thing we do, is for the inputs, we set the inputs to zero to do the natural response. And how you do that, our input here is a voltage source, and we're gonna suppress the voltage source, that's how we do it with super position. When you suppress a voltage source, what you do is you set the voltage to zero. And that means that's the same as creating a short circuit for this. So, for the natural response we're gonna take out our voltage source and replace it with a short circuit, like this. So, this circuit now is identical to the circuit that we used for the natural response. So, if you haven't seen that video on how to solve for the natural response, this would be a good time to go look at that. What I'm gonna do now is just write down the answer from that, this will be plus or minus, we'll call this v natural, and that equals some k natural times e to the minus t over r c. This is the natural response of an r c circuit, so I'll put a square around that so we can remember it. And I've left the constant in here, and we'll work out what this constant is a little bit later. Okay, so now let's move on and do the forced response for this circuit. And for the forced response, we remember the initial conditions are set to zero. So that means that q equals zero here, and that means that initial v is zero. I'll just write in zero. So that's what it means to set the initial conditions to zero. And the inputs are equal to, we use to inputs this time, the input is equal to v s. And in particular, the input is equal to v s, we're gonna solve the forced response after time is equal to zero, so that means that the input is v s, capital v s. And as a reminder, what we're trying to solve here is the differential equation from the previous screen, and that is c times d v d t, I'll put a forced in here, plus one over r, times v forced equals one over r times v s, and I can plug in capital v s here because we're trying to solve this for this initial condition. So this is our differential equation for the forced response, and we're gonna take a strategy here that's like we did with the natural response, is we're gonna guess at an answer for v f, and then plug it into this differential equation, this non-homogeneous differential equation, and we're gonna see if it works. And a good guess here is to guess something that looks like the input. So a good guess is gonna be some function that looks like v s. And v s looks like what? V s looks like, over here, v s looks like a constant. So we're gonna make a guess here that the forced response looks like some constant, we'll call that k f. The way we test our guess is to plug it back into here. Plug it back into the differential equation. So I'm gonna do that, alright so we get c times the derivative of k f, with respect to time, plus one over r, k sub f, equals one over r times v s. Alright, now here's the interesting thing that happens next. What's the derivative of a constant with respect to time? That's That goes to zero. So this first leading term here of our differential equation goes to zero, and now I'm left with k f times one over r equals v s times one over r. So that makes k f equal to v s. So I'll tuck this in here, our forced response, v f equals v s. So now I have our natural response right here, and in this square we have our forced response, the forced response is just a constant v s. So now we're ready to come up with our total response, and we'll call that v capital t, the total response is equal to the natural response plus the forced response. So now we're using our principal of super position. Okay, v t equals k some constant times e to the minus t over r c. And let's add the forced response, and the forced response is right here, it's v s, and we're getting close, the only thing we have left is to figure out this value now, we have to figure out the gain factor in front of the exponential term of the natural response. Now the way we go after that, is we would know this if we knew what v s is, and we do, it's called v s. If we knew v t at some time, we could plug in two values here. We could plug in a v t and a t, and one of the most convenient times to know this, if we set t equal to zero, and we know that the total at time equals zero is what? It's basically, let me roll it up here a little bit, let's go backwards a little bit. Let me go back here and use this diagram again, this was the forced response, so we're not gonna use that right now. We're gonna use, this is now the total circuit all assembled together, and we have to figure out what is the actual initial charge on here, and if we recall, the initial charge on here was v naught, it was this value here, just before time equals zero, this was the value of the voltage on that capacitor. So I can fill that in here, v naught. Alright, let's go back to our total solution and plug in these two values, time is zero and v t equals v naught. So that looks like v naught equals natural constant times e to the minus zero over r c plus v s. And let's solve for k n k n equals, this term is e to the zero is one, so this term goes to one, that says that it's k n plus v s, is this side, and I can write down here v naught minus v s. So now I've solved for k, and we can finish our total response, we can say the total response is v t equals k sub n, which is v naught minus v s, e to the minus t over r c and plus v s. And that is the total response of our circuit, and we saw that in two steps, first we did the natural and then we did the forced and we added them together and worked out the last constant, whatever the constant was, and there's our answer. So in the next video we'll do an explicit example with values for r and c in the step and we'll see what it actually looks like.