[BLANK_AUDIO] In this continuing review lecture, we'll talk about combining the forms that we have discussed in the previous two reviews, to construct the Bodie plots of more complicated transfer functions. So the basic idea here is that we can combine terms such as poles and zeroes that we know how to draw the, the asymptotes for according to some simple rules. And the simple rules can be seen or derived really by writing the expression for a complex number in polar form. So suppose we have a transfer function G1, and another one G2 and G3, and we express each of these in term, in the polar form as some magnitude that's called R here, times e to the j times theta, the, the phase. So, if we call G3 is the product of G1 and G2, we can find the magnitude and phase of G3 by multiplying the terms for G1 and G2 together. So we'll get R1 times R2 times e to the j theta 1, times e to the j theta 2. And the the product then we can write as R1R2, in magnitude, which is the magnitude of G3, times e to the j, theta 1 plus theta 2. So the phase adds. Okay, when R1 and R2 are expressed as decibels and their product is found by adding the dB values. So R3 is the sum of the dB magnitudes R1 and R2, and theta3 the phase is the sum of the phases. So in general then to find the composite phase when we multiply terms, we simply add the individual phases. And define the composite magnitude in decibels we add the composite dB individual magnitudes. So here is an example, suppose we have a more complex transfer function that has a constant term, G0, we'll to take as a real positive number, and then it has two poles. So there are two denominator terms that look like the poles that we previously discussed. And here are some numbers plugged in, so G nought is 40. If you express that in decibels, 20 times the base 10 log of 40 turns out to be 32 dB. F1, the corner frequency omega 1 over 2 pi, let's suppose is a 100 hertz. Let's suppose F2 associated with omega 2 is 2 kilohertz. So here's a plot of that. And to, to construct the magnitude we construct the magnitudes of the individual terms and then add their dB together. So G nought is 40 dB. I'm sorry, it's 40 or 32 dB, which is up here. So that's the G nought. The first pole is 0 dB for frequencies below f1. F1 is a 100 Hertz right there, so here's f1. So it will be 0 dB until we get to f1. And after f1 we have an asymptote that has a slope of minus 20 dB per decade. That was supposed to be a straight line. Okay, then our second pole, at frequency F2 of 2 kilohertz, let's see there's F2. This one has zero dB asymptote until we get up to F2, and then it has a minus 20 dB per decade slope after that. So, to get the composite, we have to add the blue, the red, and the green lines. And at any one frequency, the composite magnitude will be the sum of the individual magnitudes. So for example, down here at low frequency, we have 32 dB plus 0 dB plus 0 dB. This is a total of 32 dB. And continue like that until we get to the first corner frequency at a 100 hertz. After a 100 hertz the first pole adds a minus 20 dB per decade slope. So the composite will decrease with that slope because the the first pole is decreasing. And so we'll follow this minus 20 dB per decade slope until we get to the second corner at F2. At that point, we have another minus 20 dB per decade slope, and the composite will go down with a minus 40 dB per decade slope. So the yellow asymptotes are the total asymptotes for G of s. Likewise, to construct the phase asymptotes, we construct the individual phases of the different terms and then we add them together. So the G nought term is a real positive number, its phase is 0. The first pole at f1 gives us 0 degrees phase until we get to one tenth of the corner frequency f1. So f1 is a 100 hertz. One tenth of that is 10 hertz. So here's f1 over 10 or 10 hertz. Beyond that corner, this first term has a, an asymptote with a slope of minus 45 degrees per decade. And we continue at that asymptote or that slope, until we get to the next break frequency which is at 10 times our corner. 10 f1 is a 1,000, so we're at a 1,000 Hertz. And at that point, we flatten out and have a constant minus 90 degree asymptote. Finally, our last hole has a similar shape, except its corner is at 2 kilohertz. So let's see, here's 2 kilohertz up here. one tenth of that is 200 Hertz. So we'll, follow a 0 degree asymptote until we get to 200 Hertz. And we'll go down at minus 45 degrees per decade until you get to 10 times 2 kilohertz or 20 kilohertz. And after that we have a minus 90 slope or a minus 90 degree flat asymptote after that. So a composite phase asymptote for this transfer function are found by just adding the individual phase asymptote. So at low frequency, we have 0 degrees. When we get past 10 Hertz the first pole makes our phase start to decrease. When we get to the second pole or at, when we get to 200 kilohertz, one tenth of the second pole, we get another break. And the second pole starts contributing negative phase as well. So the composite will have two phase asymptotes that are decreasing, and our total will go down with twice the slope or minus 90 degrees per decade. So that's supposed to be a straight line there. When we get to 10f1 or 1 kilohertz, the first asymptote flattens out, but the second keeps going. So we continue with a slope of minus 45 degrees per decade until we get to 10 times the second corner at 20 kilohertz. And after that all of our comp, individual asymptotes are flat, and so the composite as a flat asymptote also of minus 90 plus minus 90 which is minus a 180 degrees. So, there yellow is the composite phase asymptotes for this transfer function. Let's do a second example. Here is a set of asymptotes that we're given, and the object here is to write the transfer function. So this is a transfer function A of s. And we're supposed to write what is, what is the expression for A of s. Well, there are, there's more than one way to do this, but one way is to say okay, we're given what the, this low frequency asymptote is, it's A0. Here the asymptote is actually expressed in dB. We have to, rewrite the transfer function, we have to write A0 that's not in dB, but just as a regular number. at frequency f1, the asymptotes change and we have a plus 20 dB per decade slope. What kind of term does that? Well a 0 would do that. So we can have a 0 function term in the numerator with a corner frequency f1. Or in our transfer function we expressed this in the corner in radians per seconds, so omega1, where omega1 is 2 pi f1. We continue until we get to f2, and after f2, the the slope decreases by 20 dB per decade, and it goes back to a flat slope. Well, the kind of term that decreases the slope by 20 dB per decade is a pole. And so the transfer function A of s can be written in this way. Okay. we can also write expressions for the asymptotes. And I'm going to illustrate how to do that. here in order to write the equation of an asymptote, what we do is, wherever we have a sum, we take the term at its largest. And this is in fact how we derive the asymptotes in previous lectures. So for example, what is the equation of the mid frequency asymptote? Yeah, between f1 and f2? So, for f between f1 and f2, we can write A of s as follows. If we're at frequency higher than f1, so you can think of s as being j omega and what is the magnitude of the j omega over omega1 term relative to 1? Well if omega is greater than omega1, then this term has larger magnitude than the 1 and to write the equation of the asymptote we throw out the 1. Okay. Likewise for the denominator, what do we do? Well, over this frequency range we're at a frequency less than f2. So, let S equal j omega and omega is less than omega2. That means that this second denominator term will be smaller in magnitude than 1. So we throw that term out. And what we get then, is A of s is A0 times the numerator, S over omega1. And then for the denominator, we just have 1. [COUGH] And that's our function. Now if we want to take the magnitude, magnitude would be the magnitude of A0 which is a real number times the magnitude of this numerator with s equals j omega, gets us omega over omega1. And then the denominator is 1. Okay, or in terms of f, we could also write this as A0 times f over f1. So that's the equation of the mid-frequency asymptote. Okay, it's very useful to be able to write the equations of asymptotes. When, when we design our feedback loops, we are going to need to do this to calculate things like the crossover frequency. And other quantities that we need to know. Okay, and I'll write our A of s again. Let's write the equation of the high frequency asymptote. For f greater than f2, which is in turn greater than f1. What can we say about this high frequency asymptote? Well again we take, when we have a sum we take the largest term and throw out the smallest. So here S is larger or ed, omega is larger than both omega1 and omega2. So the terms on the right are the largest and we can throw out the 1 in both cases. So A of s, then we'll have an asymptote, or say we could say A of s is approximately equal to A nought times the S over omega1 in the numerator. And the S over omega2 in the denominator. Let's see, the S's cancel out, and we get A0 times omega2 over omega1, or in terms of f, A0 times f2 over f1. So that's the equation of this high frequency asymptote. Now, here the high frequency asymptote is labeled A infinity, that's just the value of the asymptote. So A infinity apparent, apparently is given by that expression. So we have a straight forward way to easily write the equations of the asymptotes, by simply taking the largest term wherever we have a sum of terms. And here this also gives us an easy way to find say the high frequency gain in terms of the low frequency gain. Okay, these slides 35 and 36 document what I just said. Now there's one other way to write the transfer function that, that can be very useful. Suppose we're not interested in A0 or it's not so important in our application. Suppose instead we're more interested in the high frequency gain. We could express A in terms of the high frequency gain instead of the low frequency gain. And what we do is we use inverted forms. So if we use inverted poles of 0s, each of those kind of terms has a high frequency gain of 1 or 0 dB. So, if we express our transfer function in terms of inverted forms only, then we know that each of these forms will be 1 at high frequency. And then any multiplying constant out in front is in fact the high frequency gain of the transfer function. So if we take that viewpoint, we start with a high frequency gain of A infinity, and if we decrease frequency below f2, the gain decreases. So that is in fact what a pole does, and this is an inverted form, so 1 plus omega2 over S, would, would cause our transfer function to decrease. And then when we get to f1, the slope increases again, as we decrease frequency. So this is in an inverted 0. So we can put an inverted form 0 in our expression as well. This happens at f1. So this is another completely equivalent way to, to write the transfer function. This way of writing the transfer function emphasizes the high frequency gain rather than the low frequency gain. And if that is what is important in our application then that's a good way to write the term. I'll leave it to you to verify by algebra, that this form is, in fact equivalent to the original form with non-inverted poles and zeros. One last application of inverted poles and zeroes. Suppose we're interested in a transfer function that is a band pass filter [NOISE]. So let's suppose we want, say, a transfer function like this, where this slope is plus 40 dB per decade. This is plus 20. This is a flat slope at this mid band gain. This slope is minus 20, and this slope is minus 40. How should we write the transfer function of this? Well, what I would suggest is the easiest way is to reference our transfer function to the mid band gain. And a band pass filter that's usually what we're interested in. So we'll write A of s as equal to the mid band gain A sub m, multiplied by some poles and zeros. And these poles in zero terms have to be chosen such that each of them gives a gain of 1 or 0 dB over this mid band range of frequencies. And then these terms can make the gain roll off at higher frequency or lower frequency. Well, so what, what we can do is first we can have regular poles at omega3 and omega4, like this. [NOISE] Okay. These poles give zero dB gains at frequencies below omega3 and omega4. But over the, the important mid band range of frequencies, their gains' asymptotes are 1. Then for the omega2 and omega1 corner frequencies, we use inverted forms. The inverted forms will cause the transfer function to roll off at low frequency. But they will have 0 dB asymptotes, over the mid band and higher frequencies. So we can have an inverted pole at omega2, and an inverted pole at omega1. And the numerator apparently is just 1. So this is a valid way to write this transfer function, it mixes both regular and inverted forms. But it's, it's written directly in terms of the things that are important. So express, it expresses the mid band gain directly here and then it gives each of the corner frequencies with an appropriate term. So this is another example of how to use inverted form terms to make the transfer function be meaningful and, and straightforward.