RC step response 1 of 3 setup

In the last video, we looked at this RC circuit, and we gave it a step input with this step source. A step from V naught up to V s, with a sharp change right here at t equals zero. We sort of took an intuitive guess at what this voltage looks like—here's what V of t is. In this video, we're going to develop a precise expression for what happens to this voltage across the capacitor.

As we said, this is a very, very common circuit. There are millions and millions of these types of circuits inside every computer, so it really pays to understand this one quite well. Because this is a function of time, we have a function of time here; this is going to end up needing techniques from calculus to really figure out properly. So, this is going to be a place where our calculus skills are really necessary.

I'm also going to assume that you've listened to the video on RC natural response, where we worked out the expression for the natural response when there was no voltage here when this was zero. The natural response looks like V equals V naught e to the minus t over RC. There's a video on Khan Academy about natural response, and there's also an article you can review too.

So, let's get back here and analyze this formally. One of the things we can do is—well, the first thing we always do is we always label the quantities that we want in here. I'll put, there's the current, and here we'll call this voltage of R. If I've used a lowercase letter 'v' or 'i', that means the quantity is changing with time. If I use a big capital letter 'V' here, that means that those values don't change with time, so that's my notation.

When we analyze the circuit, a good thing to do is just to look at what the component equations for this thing are. Let's talk about Ohm's law for this resistor. Here, we know that VR equals I times R. Another way we can write that is I equals VR over R. So, the other expression I can write for the capacitor is the capacitor IV equation, which is I equals C dV/dt. This means the slope of the voltage, or the rate of change of the voltage, times C is equal to the current.

All right, so far so good. I have one 'I'; there's actually only one 'I' in this circuit, so this 'I' here is equal to this one here. That's a good relationship for us to know. We also have two different 'V' variables: we have the regular 'V' that's across the capacitor, and we have V sub R. Let me see if I can write V sub R in terms of V.

One thing I can tell from this circuit: this point right here is V s, and this point right here is V c. So, I can write VR equals VS minus VC, right? That's going to be useful.

Now, what I'm going to do is I'm going to set these two 'I's equal to each other, and at the same time, I'll plug in this expression VS minus VC for VR up here. It'll go in right here, so we can write C dV/dt equals 1/R times VR. VR is VS minus VC. I shouldn't write VC; I'll just write V. We'll take that out there because we don't need that subscript, and multiply through by 1/R, and on we get C dV/dt equals 1/R VS minus 1/R V.

We'll get C dV/dt minus 1/R times V equals V s. All right, so now we have our differential equation. It's a differential equation because it has derivatives in it, and this is, in particular, called a non-homogeneous ordinary differential equation. It's ordinary because there's the first derivative—the first derivative has a power of one here; that's the ordinary. It's a differential equation because it has a derivative, and it's called non-homogeneous because this side over here, this is not V or a derivative of V.

So, this equation is sort of mixed up; it's non-homogeneous. When we did the natural response analysis, this term right here was zero in that equation, and so we were able to solve this rapidly. We're going to develop a new strategy here for solving this kind of equation, this non-homogeneous ODE.

This equation is a little tricky to solve because this is not zero over here. We have a mixture of things going on here. This side of the equation represents the forcing function; this is what the circuit is being driven to by that step response input voltage source. Then, this side here, we're going to get tied up with the initial conditions that that capacitor has.

So, there are sort of two things going on here that make this a little more complicated than usual to solve. Our strategy is to break the problem down into two pieces. This is always what engineers do when you're faced with something that is a little complicated: you try to break it into bits and pieces.

So, what we're going to do is our strategy is to figure out what we call the natural response. We're going to figure out something called the forced response, and then we're going to add those together with a plus sign. That's going to be equal to what we call the total response. So, this is what the circuit really does.

We've got two new words here: one's the forced response, and one's the total response. If we remember, before, the natural response was what the circuit did when nothing was forcing it. We just put in some initial energy, some charge, and let the circuit sort of sag toward wherever it was going—that's the natural response.

The forced response is going to be what the input source forces the circuit to do. When we take those two things and add them together, we'll get what the circuit is going to do all together, and this is another good application of superposition. We're going to take two things that the circuit does, analyze them separately, and then add them together to superimpose them to get the total response.

The way we do this—the strategy for the natural response—basically we take for the initial conditions, that's the charge on the capacitor. We'll take whatever we get; we'll take the initial Q and the initial V W—the V naught. When we look at the inputs, we set the inputs to zero. That's how we analyze; that's the conditions for the natural response.

For the forced response, we do the opposite. What we do is we set the initial conditions to zero, simplify the circuit down, and use the inputs that are actually applied—in this case, it was V of S. Then, to get the total response, we add those together with a plus sign.

For the total response, you get the initial conditions, which are Q and V naught, and you get V s all together. So, that's our strategy. By turning it into two simpler problems, basically, we make this differential equation a little bit simpler to solve two times than trying to solve it all together at once.

We'll stop here now, and in the next video, we'll follow through on the strategy of combining the forced response with the natural response.