We'll take SEPIC, it's a fourth-order circuit, right? There's four poles in whatever transfer function you are going to look at. And the goal that we have right here is going to be to, first of all, try to simplify the analysis of the transfer functions. This is an important converter, it has many practical applications. It has the ability to step up or step down voltage, and doesn't invert it. The output voltage is positive for a given positive input voltage. And people have found a number of cases where this is a very useful configuration to use. And then you say, well, all right, I'd like to make a voltage regulator out of a SPEIC, and then you're faced with this issue of, well, what is Gvd, right? The best we can do at this point here right now is really, as I said, sit down and do brute-force analysis, and hope that we find Gvd correctly. Or forget about it and just put it in a simulator and see what happens. Okay, but we want to do extra elements right now to simplify analysis of the transfer functions. We'll focus on Gvd, but we're going to actually make a good use of that analysis to figure out the nature of the transfer functions in this circuit. And you will see that it exhibits pretty unpleasant resonance. That resonance is going to have high peaking in the magnitude response. It makes it really, really difficult to put together a control system around that fourth-order transfer function. So then we'll actually find a way to deal with that by what is called damping. We'll figure out how to damp the SEPIC as a circuit and, by doing so, make it much easier to put a control system around. So there's going to be a very practical point to this exercise. In addition to working through a complicated case of extra element theorem, application will actually have a result that can be used in practice, where we really prefer to damp these nasty resonances in the SEPIC. Which element do we consider an extra element? Now, this is entirely up to us right now. There is nothing that is driving us to considering any of these particular elements as an extra element. But as we show on the next page, one choice is to consider C1 as an open. Because when you look at the complexity of this circuit here, it's really that it's coupled through two different paths. It's coupled through this transformer action right here with these d hat sources on both sides, and it also is separately coupled through C1 right here. This is what makes writing equations for this circuit here complicated, and requires quite a bit of algebra to work through. So a fairly natural choice for the extra element is to consider C1 as an open. If you take that out, the circuit sort of splits out and it's easier to solve, right? So we can do the G old transfer function in a simpler manner if you take out C1. So that's the strategy, we're going to take C1 out and consider it an open circuit. So when you look at the original equivalent circuit model of the SEPIC and you take C1 out, then you want to redraw this circuit a little bit so that you can visualize better how it looks. And you can see that what we have here on the output side is really a series combination of an L2 right here, and this primary side of the transformer. I'm sorry, the secondary side of the transformer, they're connected in series. Now I'd like to see this transformer as primary next to secondary, and so I can just swap L2 and the secondary side of the transformer, and redraw it in this form right here. Pay attention to the dot, right? So the dot was connected pointing to the plus side of the output. Here, dot is pointing to the plus side of the output. And all we did is swap the order of these two elements in the series connection on the output side of the circuit. The input side of the circuit, we haven't touched here at all, so we have the input voltage, we have L1. And in series with that, we have these V1 over DD primes, d hat source with the primary side of the D prime to the transformer. If you look at this circuit here, you can say, well, it's not necessarily trivial by inspection, but at least I have a working chance, right? I can solve this for the old transfer function relatively easily. I wouldn't say it's by inspection immediately, but you can actually work on it, right? So there are two independent sources as far as d hat is concerned. If you were to solve, for example, for Gvg, it would actually be fairly simple, right? You would have just a series combination of inductors and Vg hat driving an LC filter, pretty easy. With d hat, it's just a little bit more difficult, and that's what we're going to do. So we're going to solve this for Gvd old that we're going to call Gvdbb. And that's equal to Vc2 hat, that's output voltage, over d hat, with all other independence sources set to 0. That means Vg hat is equal to 0, this goes away. And we just have this source right here, this source right here, and the rest of the circuit. So this is the old transfer function, right, this is a simplified circuit that we obtained where we removed C out as an open circuit. Instead of Gvd, call it Gvdbb because the circuit that we obtain once we remove C1 can remind you of the equivalent circuit model for a Buck boost converter. And so this BB really, instead of old, is to remind us that really, once the C1 is out of the picture, SEPIC goes back into what looks like a Buck boost type equivalent circuit model. All right, so let's do that Gvdbb, so here's the setup for solving that circuit. This goes away, and now you can solve it in any way you like. I'll take the left hand side, the primary side of the transformer, and move that through the transformer, so we obtain an equivalent circuit that looks like this. So we'll have here V1 over DD prime times D over D prime, so that's going to be 1 over D prime squared times d hat. The L1 is going to become D over D prime squared times L1. Then unfortunately, we still have this other source present right here, C2 and R. And this is L2, so this guy right here is I2 over DD prime d hat. You can again start writing equations, but let me do a little bit more of circuit manipulation. So we'll take these two right here and present in an equivalent form that's going to look like this. So we have the original source right here, V1 D prime squared d hat. Then we have the product of this current times the impedance of the inductor, so that's going to be our 2 over DD prime times sL1 D over D prime squared. And then we have now this series combination of two inductors. That's going to be D over D prime squared L1 + L2. And the rest of the circuit is like this C2, R, and we're looking for Vc2 hat voltage. The combined effect of these two sources in series, this is d hat right here. The combined effect of that is going to be V1 over D prime squared minus I2 over DD prime times sL1 D over D prime squared, that times d hat. That is the source that we have a combination of these two in front. Take V1 over D prime square in front, and that's going to be 1 minus I2 over V1 DD prime sL D square. Let's see if we can sort this through, so this D and this D goes away, and now what is I2 and V1? Well, in the SEPIC, these are DC values, right? The V1 is the average voltage across the switch on the transistor switch. So we have V1 is equal to the input voltage. I2 is a DC value of the current that goes through the inductor L2, that must be equal to Iout. So this term right here is going to be equal to Iout over Vg D over D prime sL. And then Iout is equal to V over R, that's V output over R. Put that in here, and we have sL over R D over D prime squared. So this is just going through these little Steps so you can understand when you look up the results that we're going to show next, where they really are coming from. The numerator is going to contain this right half plane 0, just like we have a right half plane 0 in the buck boost converter, right? The rest is just an LCR circuit that I'm sure everyone knows how to solve, so we're going to go ahead and write down the result for Gvd-bb(S). So here is the result for Gvd-bb(S). Here is that right half plane 0 that I just worked through, and then we have a pair of poles. Notice the effect of this combined inductance L2 plus D over D prime squared L1. And we have that present of course in the F naught, the corner frequency of the pair of poles. The final result for the bb version of the transfer function with C1 gone looks like buck boost control to output transfer function. And here is the summary of the salient features of that transfer function. So if you look at this point right now, so we have, with a little bit of work. So again it's not completely trivial or by inspection. But with a little bit of work following fairly standard analysis without really doing much algebra, we obtain a transfer function for the this epic without C1. With C1 gone, considered an open circuit, and we find interestingly, that it looks like a buck boost control to output transfer function. Now at this point here you stop for a second, then you start thinking about it. If I could make the insertion of this C1, have relatively negligible correction factor, correction factor that's close to 1. Then maybe I can design my control loop around this epic in the same way I would design a control loop around the buck boost converter. So we're going here a little bit ahead of ourselves. We haven't got to that point yet. We still need to workout through what is correction factor is going to look like, find ZN and ZD, go through that exercise. But in the end, we're going to come back to it and say, well, can we actually make this C1 not have this resonance that is going to be unpleasant from the point of view of constructing a control loop around it. And we'll figure out how to do it through the technique that's called damping.