n-Extra Element Theorem is an extension of the extra element theorem that allows us to consider multiple inductors and capacitors as extras at the same time. And this particular approach, will give us an opportunity to evaluate complex transfer functions by splitting the task of solving a complicated circuit at the same time into many solutions of very simple circuits instead. And those solutions are going to be finding the DC gain in front and finding the individual coefficients that we have in the transfer function in the numerator and denominator. With n inductors and m capacitors, the transfer function of a system can be up to n+m order. So you may have up to n+m poles and up to n+m sinodes. Of course in many cases, you will find that some of these coefficients are going to be equal to zero and the the order of the polynomials may be less than n+m. To start with, we need to really define a state so these extra elements in the following manner. So when you look at the case where the transfer function has this form, so transfer function G(s) thats output over input is equal to DC gain times the ratio of the polynomials. In those cases, we would consider what we call a reference state of the elements to be the DC state of the inductor and the capacitor or inductors and capacitors. So for inductors, the DC state is short. For capacitors the DC state is open. That in more general terms, is what we referred to as being a reference state. So the reference state is what we start with in general, but in the case where the transfer function is of the form of Gdc times this form, then the reference states are going to be really the DC states of the inductors and capacitors. Very simple. Short for inductors, open from capacitors. The inverse state, the opposite if you wish, of that reference state, the inverse state is going to be the high frequency state for these components. And thats going to be an open for the inductors and short for capacitors. So this is just defining what we're going to be calling states of the reactive elements. So we will first do a discussion of the n-extra element theorem for this case right here. This is the form of the transfer function we assume up front and the beginning, which means that the reference stage for inductors and capacitors are DC states and inverse States for inductors and conductors are high frequency States. Well let's look at the first into how we actually determine Gdc. How do we find Gdc? That's going to be able to output over input under the condition that all other elements are in CD state. So all L's and C's in DC state. Okay. Short inductors, open capacitors, fine Gdc. It's difficult to find an example where this would not be by inspection. Right. So that's the Gdc. Now let's look at what happens with the evolution of the coefficients we have in the numerator and denominator. Those are the ones that they will have to spend, you know, pay attention and be careful how we actually do them. So here is the form of the transfer function. I will talk first about the coefficient with s. So that's the coefficient that is called a1 in the numerator or b1 denominator. This term here a1s must be such that a1 has to have dimensions of one over hertz, because s in frequency is going to have Dimensions of hertz(HZ). This must have the same dimensions as one, which means no units. And thereby a1 or b1 have to have dimensions of hertz(HZ) minus 1, 1 over hertz. How do you get one over hertz when you have the active components in the circuit? Well, you get that in two different ways depending on the nature of the reactive element. You either have a time constant associated with the capacitor, which is a product of sum resistance and capacitance or you have a time constant associated with an inductor, which is of the form of inductance over resistance. Okay. So the conclusion that we have for these terms a1 and b1, hand-waving conclusion but nevertheless pretty important as we have understanding of how these systems look like, Is that b1 or a1 are going to be of the form of sum of the time constants associated with capacitors and time constants associated with inductors. So let's look at b1 first. So there's going to be sum of all individual inductors and all individual capacitors, of the form of RiCi plus Lj over Rj. The question is whether these resistances with respect to the individual components Ci or Lj. The answer to that question is really what the n-extra element theorem is about. So we will find out that Ri does the resistance associated with a time constant for capacitor Ci is the resistance seen at the port where Ci is connected in the circuit, under the condition that all other elements are in DC state. So we will find Ri by using a test source attached to the port where the capacitor Ci is connected. All other elements are going to be rolled to their DC state and we will solve for the resistance seen by that source and the port where Ci is connected. Exactly the same holds for Rj except that's going to be the resistance seen at the port where the inductor Lj is connected. When that resistance is evaluated all other elements, all other C's and L's are in the DC state. Now, when you're looking for the resistances associated with the bi coefficient in the denominator, What else do we do in the circuit when we do that test? We're looking for resistance. When we put that source at the port, we find voltage over current of the test source. In the rest of the circuit, we must not have any other independent sources. And in particular, that means our input is equal to zero. So this resistance is seen by Ci or Rj are found under the condition that all other elements are in DC state and that the input is set to zero. How about the a1? Well a1 is the desk, exactly the same set of bullets hold for a1. It must have dimensions of hertz minus one, must be of the form time constant with the capacitor or time constant with an inductor. And for a1, we have similarly a conclusion that it must be a sum of these time constants over all capacitors and all inductors. But how are these resistances found in this case. For a1 resistances are found under the condition that all other elements are in the DC state. But, instead of setting the input zero we have no double injection in which the output is null to zero. Okay. So here's the statement, where Ri ( or Rj) is the resistance seen by Ci (or Lj) with all other elements in DC state and the output Yo. Now as we go through the rest of this n-extra element theorem, you will see a little bit of a difficulty with notation. There are so many different things that we're going to show up. The way we've decided to put this in a textbook is to try to simplify the notation as much as possible and be a little bit loose on that end. In particular here, we see b1 is the sum of RiCi plus Lj over Rj. And we have exactly the same thing right here. What is different here is this Ri here, is not the same as Ri here. All right. So, keep that in mind. So these two here are not the same and the same holds for these two. Right. Just to simplify notation and not put million subscript right here to make things even more messy, you will see an application doesn't require any special notation. We will go through that super simply. But for the definition of the terms, pay attention to what is written on the side here in addition to the expression itself.L Stop here for a second and see if there are questions about this. Yes. When you're finding the denominator coefficients, you set an input to zero, but then in the numerator, you don't. Is that primarily because all the feed forward effects are always captured in the numerator and not in the denominator? The question is, why do we have these two different conditions really with respect to the denominator coefficients versus the numerator coefficients? Denominator coefficients are going to be giving poles on the system. The denominator of the system is really poles of the system. Poles of the system is a property of the system. It's a property, no matter what transfer function you're looking at, you're always going to get the same denominator, same poles, same characteristic values for the system. It's a linear system. When we look for those resistances seen right here, certainly we have no other inputs involved. These are driving point impedances at certain ports in the network, no other independent inputs. Now, the zeros of a transfer function do depend on where the output is. You're going to have very different zeros if you're looking for Gvd versus Gvg versus zero towards any other transfer function. Zeros actually do depend on where is the output that you're looking at, but as the denominator, does not depend on what the inputs and outputs are, the numerator does depend on that. You can see intuitively how that is captured in the form of evaluating these time constants under null-double injection that keeps the input at the point for the input actually is, and nulls the output. Nulling the output is really, if you wish, equivalent to finding a zero of the transfer function. Nulling the output is zeroing the transfer function. That's why, intuitively you can understand that this nulling condition applied to a specific output is the conditions under which these resistances are going to be evaluated. By the way, we are going to skip the proof of the extra element theorem, it's pretty convoluted. We are not going to go there. There is a reference to a paper where this is actually shown. It's a long paper with really, really difficult notation. It's not difficult by itself, but the notation is really cumbersome. We're not going to go through that. We're just going to make these statements here, and rely on a somewhat intuitive knowledge of what we have gained from the extra element theorem that we have, in fact proved and derived. Any other questions about this? This is a starting point, theorem A1 and B1. I'm just wondering, can nEET give the n result, if EET. N extra element theorem, can you derive extra element theorem from n extra element theorem? You probably can. Yeah. I think, I'm pretty sure you can. Yes. Let's see the next coefficient. What do you think this is going to be, a_2 or b_2? It's going to be the products of time constant. The term that we have with s has this form right here. When you look at the term with s squared, now we'll have to have products of time constants. Now, you can see how things can get. The number of these terms that you're going to have with s squared is going to be higher. Because here we have each individual element with its own time constant. Here we have pairs of elements with products of their time constants. As to the cube, you're going to have products of three elements at a time. Then whwith s_4 you're going to have four elements at a time. In the end, it's going to be s_2 n + 10, when you have all elements. All right, so the numerator and denominator both have the same general form, where the increasing orders of s are going to be corresponding to the increasing number of time constants that we need to multiply. The question is going to be really what these time constants are. That's what I want to introduce first with the coefficient with s squared. Then we'll do the s cubed and then we'll conclude what the pattern is. Let's look at the s squared terms. We have L's and C's. You can see this is sum over all possible combinations of two elements at a time. We have L_L_j or R_iR_ j-i. I'll explain what this is in just a second. Then we have the time constants that we'are combining L's and Cs and then you have time constants that are combinations of two C's at a time. We have to consider all possible combinations of two elements at a time, but in those combinations the order does not matter. It turns out that this is another version of the reciprocity relationship is that the product of these two resistances regardless of the order which comes first the j or i, or i or j, the product of these two is the same. This term here can be written in two different forms. They are equivalent. Any one of these terms can be written in two different forms that are equivalent. We can pick any route we like to get to a particular product of time constants. Now the key point is the following. What is this resistance here? What is this extra resistance in the extra time constant that we combine with the first one? All right, well to start with, this is coming from the term with s. L_i or R_i or right here or right here. Those are something that we have already evaluated, already done. We've already done those time constants in the process of computing the coefficient with s. The question is, what is this time constant right here that is now multiplying the time constant that he already had found? This resistance right here, is this resistance right here. That's the resistance that's going to be seen by the jth element,w hen all other elements are in the DC state, except the element i. In this particular case, so let's say you have this L_i/R_I times C_jR_j - i. This one here is going to be seen by C_j, when all other are in DC state, except L_i in high frequency state. Moving forward from the time constants that you have with s, you're adding time constants with resistances that are found in the same manner as for s. Except that now the element you are combining this with is going to be in the inverse state, in this case the high frequency state cake. This is the most important item to note right here and that's going to be the pattern that's going to follow up through to the higher order of the coefficients. The difference between the numerator and denominator is again exactly the same as before. The denominator coefficients, the resistances that are seen when you compute denominator coefficients are done under the condition that input is set for zero, versus the numerator coefficients are found under the condition that the output y is nulled by null double injection of the port test source and the input, so y is not for the coefficients in the numerator. Now that you see how it goes with s squared, I am hoping that you can see what the pattern is already. Let's look at the coefficients with s cube. Now we have products of three terms, three time constants. Three time concentrate times was all possible combinations of three elements at a time. Again, the order in which you do these elements does not matter. But importantly now this extra turn, right in the extra time constant R_k-ij, is the resistance seen by the k element when all other elements are in the DC state except elements I and j and so on. All right, let's look at this carefully again and if see there are questions. Looking at which one do we want, this one for example. You look at this particular resistance, that's going to be resistance seen by L_k. All other elements are DC state except L_i and L_j in high frequency states. Once we do examples this becomes a little bit easier to grasp. Right now we're just telling you how to do it. But then once we do it I think you will realize this actually is maybe spending too much time on how to do it. It's pretty routine. All right, so the same pattern then continues to hire are the coefficients. As you're adding more and more time constants you go up further in the order of the coefficient, each time you add a new element the one that you're combining with are going to go to the inverse state, to the high frequency state.