Here are the equations that we derived previously for the buck–boost converter example. So these are the averaged AC equations of the converter. We call them large signal or non-linear equations for the reason that the right hand sides of these equations involve the multiplication of time varying quantities. For example here, on, in the inductor equation, we have the duty cycle d of t multiplied by vg of t. We have d prime of t multiplied by the average v of t. Whenever you multiply time varying quantities together you can generate harmonics. So for example if d of t and v of t are both sinusoids, then when we multiply we get terms like sine squared that has harmonics and the components are not simply at the original frequency. So, in fact, multiplication is a highly non-linear function. And so the, this is a set of non-linear differential equations. To apply Laplace transforms and talk about transfer functions we need to linearize them, and so that's what we're going to do in this lecture. Now, how do we linearize a non-linear equation? The way this is done classically in electronics is to construct a small signal model about a quiescent operating point. Another way to say that in mathematical terms is that we can construct a Taylor series expansion of the non-linear function. And take the linear terms, that gives us a linearization of the function about some some given operating point. So, here's how we do it. Let's consider that we drive the inputs to the converter, first with some DC or steady state values. So the duty cycle and vg are the independent inputs to the converter system. So let's consider first that we apply some steady state or DC value of duty cycle that we'll call capital D, so we let d of t equal capital D. Likewise, we apply a DC or a steady-state or quiescent value of vg that we'll call capital Vg. So you let the average vg equal capital Vg. Incidentally, quiescent means quiet, and when we talk about a quiescent operating point it's the same thing as saying we're at the steady state, or DC values of the converter. So, if we apply these DC inputs to the converter then the converter will go through some transient, like we talked about early in the semester. And eventually the transient will decay, and we'll be left with the converter operating at its steady state operating point. So in that case the average inductor current and average output voltage and average ig will go to their DC or quiescent values that we already know how to find. We, we solve the DC equivalent circuit models that we talked about in previous weeks, to solve for the DC values of these quantities. For the buck-boost converter without losses, here is the answer that we had. Okay, so this defines the quiescent values, or steady state values of the voltages and currents within the converter, in response to these applied constant inputs. Next, what we're going to do is to consider what happens if we add some small AC variation to the inputs. So now, suppose we let the average vg equal this quiescent value, capital Vg, plus a small AC variation about that quiescent value that we're going to call vg hat of t, like this. Likewise, we could let the duty cycle v of t be equal to the quiescent value capital D plus a perturbation, d hat of t, that is a small AC variation as well. So if we apply that to the inputs, what we would expect is that the voltages and currents in the converter would also have some corresponding small AC variations in them. So that we could write say, the average i, inductor current, as being equal to capital I, the quiescent value, plus i hat of t, the perturbation or var, AC variation in average i. Likewise, for v and ig. Okay, now, the small signal assumption that we will make, and what allows us to linearize the equations, is to assume that the perturbations on the left side of these inequalities are very small in magnitude compared to the quiescent values or DC values on the right-hand side. So, provided that all the perturbations are small compared to the quiescent values, then we will be able to linearize the non-linear equations and construct small signal linear AC models for the converter. Okay the the lecture notes have slides for this next part, but I'm going to actually just work them out by hand. So what we're going to do, is for each of these AC quantities, I'm going to plug in the, the perturbed expressions. So on the left hand side, we get L times the derivative of the average i, which we're going to let be equal to capital I plus i hat of t. And to keep the equation simple, I'm not going to write the of t part, but it's assumed to be here. Okay, on the right hand side of the equations, d of t, we're going to let be capital D plus D hat of T. And the average vg we're going to let v equal to capital Vg plus vg hat. Okay, then we have to add d prime, so what is d prime of t? Well if d equals capital D plus d hat, that means that d prime, which is 1 minus d, would be equal to 1 minus the quantity capital D plus d hat, which is then 1 minus capital D minus d hat, and we're going to write that as capital D prime minus d hat. So, capital D prime is equal to 1 minus capital D. Okay, everyone, people often forget the minus sign in front of the d hat, if d hat makes d bigger then it makes d prime smaller, so d prime becomes capital D prime minus d hat. Okay, so for this second term then, d prime we will write as capital D prime minus d hat, and the average v we will write as capital V plus v hat. All right, let's multiply this out. On the left hand side, what is the derivative of capital I? Well, capital I is defined to be DC, so its derivative is zero. So the left hand side becomes L times di hat dt. Okay, on the right hand side we have a lot of terms to multiply out, but what I will say is that we get three different kinds of term. The first kind of term involves the products of DC quantities. So here we have a capital D times capital Vg, and we also have a capital D prime times capital V, so these terms are what we call DC terms, they're all constants. The second time, kind of term that we have is the product of a hat quantity with a DC quantity. So, for example, we have d hat times vg, we have capital D times vg hat, there's a D prime, or capital D prime times v hat, and there's a minus d hat times capital V. So these involve the DC quantity times a small AC quantity, we call these first order linear AC terms. And then the third kind of term that we have, involves the products of two hat quantities. So we have a d hat vg hat term, and we have a minus d hat v hat term. So, these involve the products of two time varying AC quantities. The hat terms are small, or we're taking them to be small, so this is a small number times another small number. We call these second order terms, and they're non-linear because they involve the products of time varying quantities. Okay, what, what do we do with these different kinds of terms? Well, we treat each of them separately. First of all, for the DC terms, what do the DC terms equal in this equation. Well, if you solve the DC model of the converter by applying volt second balance to the inductor, what you get is that these DC terms are the average voltage applied to the inductor in steady state. And by volt second balance, they equal zero. So the DC terms in fact are equal to zero. You can also see that by looking, by equating them to the DC terms on the left hand side of the equation. The derivative of capital I was zero and we got zero on the left hand side of the equation, so the DC terms add up to zero. Now how about the second order terms. Well, we're going to make the small signal approximation now, which says that the product of two hat quantities is a very small number. This is a small number times a small number, which is a very small number. And the small signal approximation, or small signal assumption says that we will throw these terms out. So okay, they're not exactly zero, but they're second order or higher order terms, that we will neglect. And so the first order, or linear terms, are all larger than the second order terms because they involve a hat quantity times a large DC quantity. So for example, here we have d hat times vg hat, that is much smaller than d hat times capital Vg. So we will retain only the first order terms, and our equation then becomes L times di hat dt on the left-hand side, equals the first order terms on the right-hand side, which are capital D times vg hat plus capital D prime times v hat plus d hat times capital Vg minus v. And this is the desired result. This is the linearized small signal equation for the inductor. Okay, here, here is the capacitor equation and we can do a similar process for it. So we'll write C times the derivative of capital V plus v hat is equal to minus d prime which is capital D prime minus d times the average i, capital I plus i hat, minus the average v which is capital V plus v hat over R. We can simplify this. On the left hand side, the derivative of the DC, capital V is zero, so we get C times dv hat dt on the left, and on the right-hand side, we'll get DC terms that are minus D prime capital I minus capital V over R. So these steady state terms equal zero. Then we get first order terms. So they'll be d hat times capital I. And they'll be, let's see, minus D prime times i hat, and then there'll be a minus v hat over R, so these are the first order terms. And then we have, let's see, one second order term, that would be d hat times i hat. So we'll throw out the second order terms making the small signal approximation. The DC terms add to zero. And finally, we get C dv hat dt is equal to the first order terms. And this is our small signal equation for the capacitor. Finally, for, for ig, we need to linearize this equation as well. So the average current ig is capital Ig plus ig hat. This is equal to the duty cycle, capital D plus d hat times the average i, capital I plus i hat. We multiply this out. On the left side we have capital Ig plus ig hat. And on the right hand side we have DC terms, so it would be capital D times I, plus small signal AC terms, d hat capital I plus capital D times i hat. And we have second order terms which in this case would be just d hat times i hat. For this equation we don't have a derivative, and there are DC terms on both sides of the equation. So what we can write, in fact, is we can equate the DC terms on both sides, and write that capital Ig equals capital D times I. And this is, in fact, what we get out of our DC model for the converter. Then we'll throw out the second order terms, making the small ripple approximation. And the other equation we'll get is, equating the first order terms on both sides of the equation, so that ig hat equals d hat times I plus capital D times i hat. And this is the small signal, first order AC equation for the input port of the converter. In the small signal equations, such as this Ig hat equation, we can, we take the capital quantities, or the quiescent terms capital I and capital D for this example, to be givens. We solve the DC model at a quiescent operating point, to find these things like capital I and capital D. And then in the AC small signal model, to solve for the hat quantities, the DC quantities are just constants. All right, we treat them as constants, they're already known. And the variables in the small signal AC equations then are the hat quantities. Okay, so this process of pertur, perturbation and linearization allows us to construct small signal AC equations for the converter. The next step that we're going to do is to take these equations, and construct a small signal AC equivalent circuit model in the next lecture.