In this lecture, we will generalize the low-Q approximation to the case of a higher order polynomial. So what we're going to do is use numerical values to justify making approximations, but those approximations then can be expressed as analytical functions of the component values. And these analytical functions can be very simple. So we have an nth order polynomial. Here P of s with coefficients A1 through An. And what we would like to do is to factor it into roots, in this form, so the taos essentially are the roots. And under the case that the roots are real and well separated in value. And we're going to find that some simple ways to do this. We'll also look at some special cases, for example, when we have two roots that are nearby or near each other, and where we need to leave the roots in complex form. Here's how it works. What we do, is we take the the factored form and we multiply it out. And then equate like powers of s to the original polynomial. So if you do that, here's what we find. That the coefficient of s to the first power, or a1, is the sum of the taos. A2 is the sum of the taus that are products taken two at a time. A3 is this, involves the sums of products taken three at a time and so on. And an is the product of all the taos, tao1 through taon. Now so far we haven't really accomplished anything. We haven't solved the problem because we can't solve this in general. And in fact it's not known, in general how to, to factor exactly a higher order polynomial. However, this does give us a way to, to make approximations. And what we're going to do is first assume that the taos are written in order. So tao1 is bigger than tao2, tao2 is bigger than tao3, and so on. And we can make that assumption without loss of generality either. There is an order for the taos. Okay, and in that case, if you look at these expressions, the first expression that's written is in fact the largest. So tao1 is bigger than tao2, which is bigger than the rest of the taos. Then an approximation you can make, is to simply take the largest in the sum, and throw out the remaining, the remainder. So a1 is tao1. Likewise, for a2, tao1 times tao2 is the largest of all of these terms, that's bigger than tao2, tao3 and so on. So we'll take the tao1 tao2 term, and throw out the rest of those. And likewise we keep doing this. Then in the case where the taos are written in this order, wihch means they're real and well separated in value, then we can take the largest or the first time in, in each of these expressions. So that we're left with this system of equations, and this system is in fact easy to solve. Tao1 is a1, tao2 can be found by dividing the second equation by the first, which would give us that tao2 is a2 over a1. Tao3 can be found by take, dividing the third equation by the second, which means that tao3 is a3 over a2, and so on. So we have an approximate factorization. And then here is the answer. So these are the, this is tao1, this is tao2, tao3 and so on. And here is our factorization. So, we have a set of inequalities that has to be satisfied. We can easily check these inequalities if we know the element values, and see if they're satisfied, and if they are, then this is the factorization. It's also very easy to evaluate these. If we simply have the ratio of terms, it's pretty easy given expressions for the coefficients, to find their ratios. We'll do an example in a few minutes. But this form is one that lends itself well to getting simple analytical expressions for the roots. Another thing about this is we can plug in numerical values to check the first, this set of inequalities, but having plugged the numerical values in and found the inequalities are satisfied, we don't have to plug numerical values into this equation. We can instead plug in analytical values So, analytical expressions. So for example, a1 is something R C. A2 might be L over R, or else L times C, and so on. So we can plug or obtain analytical expressions for the roots in this second form. Okay, well, we just need to worry next about special cases. So, suppose some inequality is not satisfied, let's say, this inequality out here in the middle of, of this string of inequalities. In that case, what we have to do, is take the two roots that are, the [INAUDIBLE] are on each side of this inequality and leave them in a quadratic form, like this. So, this means we have two roots that are not well separated, and we can treat them in a quadratic form to get a more accurate expression for, for the corresponding terms. Here is, then, the way to do it. So this term here is ak over ak minus 1, which is what it would have been in the original approach, and the second term is ak plus 1 divided by ak minus 1. So this is a different term. One can multiply this out and show that this is the set of inequalities that are required for this approximation to be accurate. So this is in fact what happens when we have two roots that are close to each other. We have another special case where the first inequality is not satisfied. And in that case we need the first two roots in their original form, which is a quadratic form like this, and then we continue as usual for the remaining roots. One can multiply this out and find the inequalities that need to be satisfied and here's what they turn out to be. So the first two inequalities are different. Other cases, if we have several isolated inequalities that are violated, then we just have to leave the corresponding roots in quadratic form, so we may have more than one quadratic form in our factorization. we'll do some examples of that. And then if we have several inequalities in a row that are violated, then that means we have more than 2 roots that are close together. So then [INAUDIBLE] we have to use cubic or high order roots and then you're on your own. You know, that's a higher order case and we have to handle it with other means. Here is an example, and this is one that we actually the circuit that we build in our second power electronics course at Colorado it is a damped input filter. So we have a switching converter here like, for example, a buck converter, whose input current is a pulsating current. There are regulations typically that limit the amount of high frequency current harmonics, we're allowed to inject back into the power source, and so we have to build LC filters. And in the process of adding an LC filter what one finds is that this LC filter can disrupt the feedback weep of this converter and actually cause instabilities. So we have to damp the filter. And here's one way to do it. We have, we add an RL network in parallel with our inductor, which adds some damping to the resonant filter here, and makes it easier to stabilize and control the converter. So, this is an example of a third order input filter circuit. Now you can do the analysis of this circuit, and find that its transfer function from converter current to current going into ig is this expression. I'm, I'm not going to take the time to do the algebra. You'll have to take my word for it. But we get a third order denominator because there are two inductors and one capacitor in our circuit. So what we would like to do is to factor the denominator into a simpler form with factor grids. So, if we apply our approximate factorization method, we identify that this is a1. This here is a2, and this is a3, which is noted here. And with that we can now apply the factorization method. If the roots are real and well separated then here is what the factorization will be. We said it would be 1 plus a1s, times 1 plus a2 over a1s, times 1 plus a3 over a2s. This is the a1, this right here is a2 over a1, and this is a3 over a2. So simple enough to apply. Here are the inequalities that are required. Basically, the first root is much greater than the second root, which is much greater than the third root. Okay, we could leave it in this form, but, in fact, this result can be further simplified. And if you look at the, these inequalities for a minute, you'll see, first of all, how can it be that L1 plus L2 over R is much greater than L2 over R. One thing I would caution you is that you have to be careful with these much greater than symbols. We can't just do normal algebra like this, subtract L2 over R from each side of the equation. If we do, we'll get L1 over R is much greater than 0 which is not a very useful inequality. In fact, what this inequality means, the only way it can, can be obeyed, is if L1 is much bigger than L2, which then means that we can ignore the L2 here, we can ignore there also, and then we get L1 over L1 which is just 1. And so the whole thing simplifies then to this. So, this is our inequal-, or set of inequalities. And in fact, these are our simple order expressions for the three roots. So, we can actually take the factored form and simplify it in the same way to this. So this is a nice form, I think. It's a easy way to see how each of the three roots relates to one of the Ls or Cs. Okay, that was the real root case. Now let's suppose that the second inequality is violated. So in that case, we have to leave the second and third roots in quadratic form. And so here is the correct form from the second case. 1 plus a2 over a1s plus a3 over a1s squared. Here is our a1, now which is this term. Here is a2 over a1. [SOUND] And this is what a3 over a1 works out to be. Okay, we can go back and plug in for this case what inequalities are required. We still have three inequalities required, but they're different than before, and in particular the third one is different. Now if you look at this set of inequalities, how is it that the second term can be much greater than the third? Both terms have RC multiplying factors, but the first one has L1 over L1 plus L2. The second one has L1 in parallel with L2 over L1 plus L2. So basically it means that L1 is much greater than L1 in parallel with L2. And again, the only way that can happen is if L2 is much smaller than L1. We still have this inequality, which we had before. And having made that inequality we can now go and throw out the L2 here, throw out this L2, and this is just RC. And then this one gives us really L2 over L1 plus L2. So the second inequality is really saying, oh, 1 is much greater than L2. So what we get then from these is, really this, these two inequalities are required, but we no longer require this third inequality. So when this inequality is not satisfied we can leave the second and third roots in a quadratic form like this. And in fact, you can see that this, this expression then simplifies down to, to this. We have one more case where it's the first inequality that is not satisfied. So, in that case, we have to leave the first and second roots in quadratic form, like this. Basically, we leave them alone. And then the third root is, is a well separated root that is a3 over a2. Okay, so again, plugging in a1, a2 and a3 into that, here is the factorization that we get for this case. And by plugging into the, this third special case, here are the inequalities that are required here. In this case, we require L1 much greater than L2, and also we require RC much greater than L2 over R. But the, the in between or second inequality is no longer required. So it, with these, we can again simplify these expressions and write the polynomial like this. So this method of approximate factorization can lead to very simple expressions, there's simple functions of the element values. So this is a very nice design oriented way to, to conquer the algebra, and simplify the factorization process. And really get expressions that, that give a lot more insight into how the circuit works. One thing, other thing I would add is that, in the case where none of the inequalities are satisfied, then basically we're designing a system that has three roots all close to each other. So you have a cubic type system, and we can't approximately factor it. We're going to have to deal with all of the rot, roots being close together. Okay, we're going to generalize this even further in a couple of lectures, where we're going to talk about how to draw bode plots by inspection. And it is in fact a, one more small step beyond this approach, where we'll be able to just look at a circuit. Draw some asymptotes and get a very simple bode plots. But, that are equivalent to doing this kind of factorization.