So here's the SEPIC, and as we have already discussed, we would like to find out the frequency responses of this converter. We realize it is a complicated converter that has four further transfer functions. And from a prodigal perspective, it is a converter that is generally considered more difficult to control. So the approach to analysis of transfer functions is based on the application of the extra element theorem. We decided to consider C1 as the extra element, that in what we would call an old circuit, would be considered open. All right, if you open C1, you obtain the old circuit that you refer to as being similar to a boost model. And we can solve for the transfer function in that circuit relatively easily, and we obtain that transfer function Gvd-bb. That's really the old transfer function in the context of the extra element here, and to have a right of plane 0 and a pair of poles. Now SEPIC does have this extra element. C1 is something that you cannot take out of the circuit. The circuit no longer works correctly if you take C1 out. And so we have to examine, what is the effect of C1 on the actual transfer functions? And we do that by finding the impedances ZN and ZD, seen. And the port where the extra element is connected. So we'll do ZN first, so let's look at the setup for ZN. So we have a test source inserted at the location where the extra element was connected to the port, or the extra element was connected. ZN is found as the ratio Vtest over Itest, under the condition that the output that you're interested in is null. So remember, we're looking for control to output transfer function. So when you're looking for ZN, the output methods, that's what is nulled. And of course, it's the output for something else, input current for example, then you would null the input current. So in the context of the application of the extra element here, it's important to always keep in mind what is really the output you're interested in? In a GVD, transfer function, the output is the output voltage. Therefore we compute or find ZN under the condition that the output we had is nulled. When you do the ZN calculations, the first thing to do is really just spend some time with the circuit diagram. So first of all, immediately take advantage of the nulling condition at the output. And you realize that if the output is equal to 0, not shorted but equal to 0. That means that the current going through the load resistance must be equal to 0. That means that the current going through the C2 resistence must be equal to 0. The sum of these two currents as the current is going through here, that current must be equal to 0. And we have as a result that this source that is proportional to D hat, that D hat has to be present for null double injection. That source is going to produce current that is going to flow through the secondary side of that transformer, representing the average switch model. So that current i2 over dd prime d hat, is going to be going out of the dot on the secondary side of the transformer. Which means that on the primary side of the transformer, we will have a scale version of that current, scaled by the trans ratio. So here we will have a current that is equal to D over D prime times that current, I2 over D D prime D hat. And that gives us, as you see right here, D hat times I2 over D prime square as the current known to go through that particular branch of the circuit. What else can we see? We also see that here, since this current must be equal to that current, and it must be equal to 0. The entire test current must be going through the inductance L2. And so, the voltage drop across the inductance L2 is equal to simply the product of the impedance, so if L2 and i test, so sl2 times Itest. That voltage is the voltage that is going to be inserted across the primary side and the secondary side of the transformer in the evidence switch model. So this is where we see sL2 I test hat. That voltage is going to be then visible on the primary side, as the voltage across the primary side of the transformer. And that voltage is going to be equal to D prime over D times SL2 I test. And so now, we have looked at the circuit, and we identified the important relationships that we see in currents and voltages. At this point, we have a little pause, because it does not appear that it is one line of work to find Vtest or Itest under the condition the output is null. So we have gained quite a bit of insight already, just by looking at the circuit and labeling through what we know about currents and voltages. Bu the solution for Vtest or Itest still does not just pop up immediately. Why is that the case? It is the case because we have these two sources, Itest and D hat, so we have two independent sources present at the same time, Itest hat and D hat. And their contributions are visible everywhere, they are visible in currents and they are visible in voltages. And so we do need to do some analysis of the circuit, then some algebra to eliminate one of the two. Of course we are going to eliminate D hat in the process, but we have to do that by writing down equations for the circuit that we have. So let's do that. All right, so now we are actually solving the circuit. When you look at this particular node right here, you have the current flowing down to the primary side of the transformer. You have also the Itest current going this way. So the current that is going right here is going to be the sum of Itest hat plus this I2 over D prime square times D hat current. Then the voltage that we see across that L1, that's going to be voltage we can call -VL1 hat according to how VL1 is labeled right here, is going to be just SL1 times that current. Right, so that's great, so we have that voltage right here and now we have another way of writing that voltage. When we look at this side, we look at this loop right here, from here to here, going through the transformer. So in that loop we're going to have a total voltage that is going to be the sum of V1 over DD-prime D hat, plus this voltage draw. Because the primary size of the transformer is really reflected from the secondary side based on the knowledge of the voltage we have across L2. So that's going to be plus D Prime over D as L2 times Itest. So you see we are writing that same voltage from here to here in two different manners. Okay, its really from here to here in two different manners. And we can just equate them, so on the left-hand side we can say, all right we have SL1 times Itest hat + I2 over D prime squared D hat. That's what we have through this loop right here. And then on the right hand side, we have the green loop that's equal to V1 over D, D prime D hat plus D prime over D SL2 Itest hat. All right, so now you say, what is the point of this, right? The point of this is that we have now the ability to eliminate D hat in favor of Itest. So at this point right here we are going to eliminate D hat, in other words solve for D hat in terms of I test, and the result for that is as follows. We have SL- D prime over D, and this is SL1 sorry. -D prime over D times SL2 over V1 over DD prime- SL1 I2 over D prime squared, and that's all times Itest hat. So now that we have D hat as a function of Itest, then we can use really any number of ways of writing voltage loop equations to solve for finally v test as a function of Itest. We can do that through this loop right here, we'll just label that as blue. Okay, so write equation for that loop. We are going to have Vtest hat over ITest hat. That's going to be equal to, we have SL2 + V1 over DD prime D hat. So it's going to be V1 over DD prime. But then D hat you plug in from here, and then finally we have the drop because the primary side of the transformer. And that drop is equal to D prime over D SL2, that's it. So this is the expression that we're looking for, and you do a little bit of algebra right here and you get the final result in the following form. This here is the final result for ZN. Let's stop here for a second. So first of all, earlier we talked about that ZN is usually easy to find. And in fact we have had examples where you can write down ZN by inspection. This is not one of those cases. And it is not one of those cases because there is a need to eliminate the other independent input. In this case here, there are always a new nullable injection. There are two independent inputs. If you are lucky, things go very easily if you don't need to worry about the second independent input, you just make use of the nulling condition. But in this case here you'll see that it is necessary to go through these steps right here to eliminate the other independent input. Find out how large that needs to be in order to null the output and that relationship then goes back in to circuit analysis. And with a little bit of algebra, you get the final value of ZN, so it's not too bad. This is all that's remaining is easier than solving the full SEPIC, right? But it's still not trivial, and so it is a good example of what can happen. There are cases where ZN does require solving that interaction between the two independent sources. And that usually requires some circuit analysis, and some algebra, to complete. The second comment on ZN, is the form of ZN. So when you look at the ZN, right here, you see that, not surprisingly, we have that it behaves inductively, right? Why is that not surprising? It's not surprising because once the output is null, capacitor is gone and all you are left with in terms of dynamic components are inductors. And so of course ZN does behaves inductively. But further more it actually has a 0 and a pole. And both of those do look unusual in the sense of having the minus sign in front in the numerator and the minus sign in the denominator. In fact, if you look at the ZN expression right here with the denominator, we have a right half plane pole. So the comment is that this is entirely possible. Having two independent sources acting upon the circuit at the same time to produce the nulling condition can result in right half plane zeros or right half plane poles in the expression for ZN. This does not mean the system by itself is unstable, ZN is not a system transfer function. And so pole of ZN could be half plane pole. It doesn't really imply anything in terms of stability or the behavior of the circuit, right? So again, first look at the circuit carefully, don't write equations right away. Look at the circuit and just assign what you can see on the circuit itself from the nulling condition. That in many cases is just enough and you just weed out what ZN is. In this case it's not, and then you have to make use of what you have labeled on the circuit. And go through the circuit solution to find out what the expression for ZN actually is.
