Okay so, that’s the basics of how the guitar pickup Now, what we want to do next is make an electrical model for that guitar pickup. So, here is the coils of wire comprising the pickup. And we’re not going to worry about the permanent magnet in this circuit model, it’s not important. but each turn of wire has a certain amount of inductance, and a little bit of series resistance. Because after all this wire is very thin and if you actually by the time you're done wrapping a five thousand turn pickup, you have several thousand ohms of resistance just from the series resistance of the wire. So, every little turn has some inductance and resistance and there's a whole bunch of those turns on the the pick-up. And from one turn to the next, there is a little bit of capacitance. And so, every turn the wire is, the metal wire is insulated of course and so, you have the metal wire and then there's a coating of insulation and then the neighboring wire also has its insulation. So, you have these two metal wires with some insulation between them, and that forms a capacitor. And so, every turn in the pickup is lying on other turns and there's a little bit of capacitance from every turn to every other turn. And so, this is the, the model of one loop and so every loop is, adds some series inductance and series resistance and there's what we would call a shunting capacitance from one loop to the next. And so, if we take and combine all of these inductors and resistors in series, we can turn this into one big resistor and one big inductor. And all of these shunt capacitors, that are shunting the inductor and resistors can all be put together over here and so, we have this kind of R, L and C in the equivalent circuit for the guitar pickup. Now, the other thing that we're putting in the equivalent circuit here is the signal voltage. Now, this is the voltage that is produced at the terminals of the output of the inductor, or the output terminals of the inductor. It's the voltage produced by the motion of the string, near this inductor and the change of magnetic flux linking the inductor. So, this is the signal voltage produced when the guitar string, which has been magnetized, vibrates in the proximity of the pickup. Now, a typical pickup say 6000 turns on the sort of form that we just showed you in the last slide. Has a really big inductance of a couple of Henrys, it's an awful lot of turns and you know that the inductance is proportional to the number of turns squared. I showed you that formula a couple of lectures ago and you probably thought you'd never see it again, but I just wanted to show it to you so, you would appreciate how the inductance goes with the number of turns squared. So, when we have 6000 turns and you square that, that becomes a very large number. So, there's a lot of inductance in a typical pickup, the amount of capacitance in total is not that much really. It's only maybe 100 to 200 picofarads and the series resistance, the DC resistance might be something like 5,000 ohms. Now, and as I said, the voltage vs is just the voltage signal from the induced voltage in the coil from the spring motion. Now, there's one thing that's quite important here. The actual effective resistance at audio frequencies is much higher than this DC resistance. The DC resistance of the of the wire itself is only, is a few kilohms. But the permanent magnets also have losses in them. And so, the effective total resistance might be about ten times this or even, even greater. So, it depends upon the material that you've used for the magnets, and primarily that. So, there's quite a bit of art in designing electric guitar pickups, and choosing the right kind of magnetic materials to use. And to determine the amount of resistance that you want and, and there's just a lot of parameters that can be varied. Now, let's analyze this circuit. So, what we want to do, is here is the circuit, I just turned it on its side so we can fit it on the page better. And so, I've got this signal voltage being produced by the motion of the string. And what I want to do is compute the output voltage here, which is what you measure. This is what, this is the voltage that's going to go into the your electric guitar amplifier. So, instead of going and doing, solving just a general RLC circuit, I thought [INAUDIBLE] we could achieve the same results just by looking at this specific result. or this specific example of a, of a guitar pickup. Now, to analyze this, we're going to use Kirchhoff's voltage law and so, there's going to be a [SOUND] a current circulating around this loop. And we'll assume the output terminal here is not connected to a low impedance, it's connected to a very high impedance and so, no current will actually go through this, this line. So, the current coming through the R and L, will go through the capacitor. In practice the a little bit of that current is going to go down this line too/g, into the input of your amplifier but for our purposes we can ignore that. So, applying Kirchhoff's voltage law here, I'll start at the ground and I have a voltage drop which is i times R, and another voltage drop that is j omega l, times i. Just the impedance of that inductor. Then I pick up a voltage vs, the signal voltage and then going around the loop the rest of the way, there's another voltage drop that's the impedance of the capacitor times the current. And then, after that, you're back here at the ground. And so, these voltages all have to sum up to zero and be careful with the sign. Now, I'll solve that for i. So, this is a good example that you should really take your time here and work through these steps with paper and pencil yourself. Because I, I want to skip a few steps and this is a way for you to really practice your ability to manipulate complex numbers. So, I've just put all of the terms with i on one side and then factor out an i and so I have the R, j omega l and 1 over j omega C. And factor and then I have vs on the other side and I just divide by that factor. So, this is the current is the voltage over this total impedance, which is the sum of the resistance, the impedance of that inductor, and the the impedance of that capacitor. So, it's just as simple as this voltage is driving a current through the series combination of the R, L, and C, and this is just iv over that total series impedance. Now, what we're looking for is the output voltage, which is the current going through the capacitor times the impedance of the capacitor. So, the voltage here is just the voltage across the capacitor. So, it's from this side of the capacitor to that side, which is grounded. Now, I could also compute this using the top branch, but it's easier to do this here. So, it's just the current times that impedance, 1 over j omega C. So, here's the expression for the output voltage. Now, we want to simplify this so we're done this is the answer. But I want to take and put it in a simpler form.