Now, this is your chance to use all of the AC circuit analysis techniques that we've learned so far in the course. To figure out, in great detail, exactly how this circuit works. So I want to take you through this analysis. I and I really want to encourage you to sit with a paper and pencil. Draw the circuit diagrams. label all of the components, and write down the equations. And try to reproduce every step that we take here. The only way to do this, to learn this, is to do it yourself. So, I'll, I'll take you through it. But you really need to sit down, and figure this out on your own, at the same time. So, here's the model of the pickup that we talked about a couple of weeks ago. When I introduced electric guitar pickups. So you have the, the pickup inductance from all of the coils of wire. That wire has some resistance that is in series, then, with the inductor. And then you have a voltage that's generated by the motion of the string in Faraday's Law. And, but unfortunately you also have small turn to turn capacitance, a parasitic capacitance in the pickup that gives you a shunting capacitance. That goes across this, this whole thing. So this was the model of the pickup and if you look back at those old videos to see how the frequency responds, the pickup was a function of the R, L, and C values here. Now, to analyze the circuit in more detail, the first thing I want to do is transform this model. Now, I'm not going to drag you through all of the detail of, of how to do this, but you can figure out a way to represent this circuit as the circuit. And actually, this is a more advanced topic then we're going to go into in this course but you can use Thevenin's and Norton's Theorems. They're really, they're just techniques for finding equivalent circuits for. If I have a complicated circuit, I can reduce it to a simpler equa, a thevenin or Norton equivalent circuit. Now if you want to look up those techniques, and learn a little bit more about that. Then, wonderful. But we just don't have time to go over that in any detail here. But we can figure out what I want to calculate here, is, if I look at this terminal. And ask, what is the open circuit voltage at that terminal? And what is the apparent. Impedance, looking from this terminal to ground. Then I can construct an equivalent circuit that has the same open circuit impedance and the same impedance to ground. I'm sorry. It has the same open circuit voltage and the same impedance from that point to ground. And this is the answer. The way you figure this out, and I encourage you to do this, is assume there's a current circulating in this loop, call it I. Then you can write Kirkov's Voltage Law going around this loop. And so you have the sum of the voltage drops across the capacitor, the resistor, the inductor, and then the voltage of the source that all has to sum up to 0. So then you can find the current, and then the open circuit voltage is just going to be that current times the impedence of this capacitor. And if you work that out, you'll find the open circuit voltage comes out to be this expression. So it's v, the original output voltage. But then it's divided by this factor that depends upon the inductance, capacitance, and resistance of the pickup. And it's frequency dependent. So you write down, J omega L for this impedance, one over J omega C for that impedance, write out the algebraic equation, solve for I, multiply by one over J omega C and you'll find this. So, now this circuit here. If I look into this terminal, and if I don't put a load on it. But I just look in the, ask what's the open circuit voltage. Well, this voltage is going to be what I see. because there's no current through, through these. So there's no voltage drop. And so the open circuit voltage here is V1, and then if I assume that the voltage, the source was zero and I look into this terminal, this is the impedance that I see R and L in series. That's in parallel with C. It's the same thing here. If I replace the voltage source by a short. In both cases, then these two circuits look the same. So this circuit is an equivalent representation of that circuit, the original circuit there. Now what I want to do just to simplify this is I'll just, this RLC combination, I'm just going to. write that as an impedance, a general impedance, Z, and that Z is the parallel combination of the top and the bottom. While the impedance in the top up here is R plus J omega LP. So there's the impedance of the top branch. And then the parallel combination of two impedances is their product. And so it's the top branch impedance times the impedance of that capacitor. So it's the product divided by their sum. And so, here's the first term. And there's the impedance of the capacitor. So then I just multiply the J omega C through the denominator here and you end up with this factor over this, okay? So work that out. Now So now I've got the, the pickup reduced to this relatively simple single voltage source with that voltage. And a one impedance with this value. Now, the next step I want to take is take my representation of the pickup. And then connect it to my volume control circuit and figure out what's going to happen then. Now you have to sit and stare at this a little bit, a little bit to realize that this is really a fairly simple circuit. What you have here is you have a voltage source, there's a pickup impedence. And then I just have another impedance to ground. So, this thing is really just a voltage divider. So it's the pickup impedance and then Zt here for the tone control. That's just a series combination of Rt and Ct. So this circuit here I can just represent with a block zt, with these impedence. It's a series impedence of Rt plus the p to this capacitor, which is over one over j omega ct. Now, the middle point of this voltage divider. So now I've got Zp up here, Zt down here. The midpoint of that voltage divider goes to this resistor, R-out. And R-out has a wiper that I can move up and down.
