RC step response 2 of 3 solve

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. Now we're going to follow through on the strategy of solving it with a forced response plus a natural response.

So here's two copies of our circuit. On the top one, we're going to solve for the natural response, and on the bottom one, we'll use that one for the forced response. All right, let's go to work on the natural now.

To do a natural response, what we do is set the initial conditions to whatever they were for the original circuit. So that involves some Q and A, A V0, A V KN on the capacitor. Let's just draw in quick, let's draw in our Q. There's some Q here, and it'll be a plus or minus V on the capacitor.

Now, the other thing we do is for the inputs, we set the inputs to zero to do the natural response. How do you do that? Our input here is a voltage source, and we're going to suppress the voltage source. That's how we do it with superposition. 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 going to take out our voltage source and replace it with a short circuit like this. This circuit now is identical to the circuit that we used for the natural response. 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 going to 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 K natural * e to the minus t over R C. This is the natural response of an RC circuit, so I'll put a square around that so we can remember it. I've left a 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. For the forced response, we remember the initial conditions are set to zero, so that means Q equals 0 here, and that means that the initial V is zero. I'll just write in zero. That's what it means to set the initial conditions to zero, and the inputs are equal to we use the inputs this time.

The input is equal to V s, and in particular, the input is equal to V SS. We're going to solve the forced response after time equals zero, so that means that the input is V s, Capital V s. As a reminder, what we're trying to solve here is the differential equation from the previous screen, and that is C * DV DT. I'll put a forced in here, plus 1 / R * V forced equals 1 / R times V s.

I could 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. We're going to take a strategy here that's like we did with the natural response; we're going to guess at an answer for V F and then plug it into this differential equation, this non-homogeneous differential equation, and we're going to see if it works.

A good guess here is to guess something that looks like the input. So a good guess is going to be some function that looks like V s. V S S looks like what? V S S looks like a constant. So we're going to make a guess here that the forced response looks like some constant, and we'll call that K F.

The way we test our guess is to plug it back into here, back into the differential equation. So I'm going to do that right. We'll get C times the derivative of K F with respect to time plus 1 / R K sub F equals 1 / R * V s.

All right, now here's the interesting thing that happens next. What's the derivative of a constant with respect to time? 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 1 / R equals V s times 1 / 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. 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. 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 principle of superposition.

Okay, V T equals K some constant times e to the minus t over R C, and let's add the forced response. The forced response is right here; it's V s. We're getting close. The only thing we have left is we have 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 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. One of the most convenient times to know this is if we set t equal to zero, and we know that V 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 going to use that right now. We're going to use this as now the total circuit all assembled together, and we have to figure out what the actual initial charge on here is.

If we recall, the initial charge on here was V Kn; that 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 K n. All right, let's go back to our total solution and plug in these two values. Time is zero, and V T equals V.

So that looks like V Kn equals natural constant times e to the minus 0 over R C plus V s. Let's solve for K n. K n equals this term; e to the 0 is 1. So this term goes to one, so that says that it's K n plus V s on this side, and I can write down here V Kn 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 minus V s e to the minus t over R C plus V s s. That is the total response of our circuit, and we solved 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.