In the last several lectures, we've discussed how to use Graphical construction directly on the Bode plot, to construct the asymptotes of series, parallel, and more complex impedances. In this lecture, I want to extend these ideas now to talk about constructing transfer functions. And in an upcoming lecture, we're going to further extend this to talk about, constructing the closed loop transfer functions of feedback systems. So con, to construct a transfer function using these graphical techniques, what we do is to express the transfer function as a ratio of impedances. We then construct the impedances graphically and then divide to get the transfer function. So, here's a very common example where we have an RLC filter network, such as the one in our canonical model. And we want to find the transfer function H of s from the input voltage to the output voltage. So, we can do this using the familiar Voltage divider formula. If we call this element Z1 and this, these collection of elements here as Z2, then we can express the transfer function as Z2 divided by Z1 plusZ2 according to the voltage divider formula. Now, Z1 plus Z2 is in fact the input impedance looking in here from, from the input source. We'll have Z1 in series with Z2. So we can, also write this simply as a ratio of impedances, Z2 divided by that input impedance. So one way to construct the transfer function, is to construct the Bode plots of Z2 and of the input impedance and then divide. We'll talk in a minute about how to divide graphically. Another way to do it, is to express the transfer function in this way. You could, for example, multiply by Z1 and then divide by Z1, and write the transfer function like this. And you recognize that this is Z1, Z2 over Z1 plus Z2 as the parallel combination of the two impedances. Yet another way to look at that, is to take the input source and Z1 and construct a Norton equivalent circuit. So the input source and, and Z1 can be written as an effective current source in parallel with an impedance. And the Norton equivalent is found in which the current source is the short circuit current between these terminals, or the voltage divided by Z1. So V1 over Z1, and the Norton equivalent impedance here is in fact Z1. Or the impedance between the terminals when the voltage is set to zero. So, you we could apply this transformation, and we have to connect Z2 in parallel. And you can see then, that the output V2 is equal to the current V1 over Z1 times this parallel combination of impedances, Z1 in parallel with Z2. The parallel combination is, in fact, the output impedance of the network. If you short the input voltage source, and find the impedance scene from the output terminals, you get Z1 in parallel with Z2. So this alternate method then is to write the transfer function as Z out divided by Z1, according to this formula. And again, we've expressed the transfer function as a ratio of impedances. So you can use either way, either the first one or the second one and I would suggest that you use whichever one is easiest to construct. It turns out that the second one, for this particular circuit, I think is easier, although you could do it either way. In the, the second case, the output impedance is just the parallel combination of the three elements and we've already constructed that. And Z1 is simply the inductor impedance. So, let's construct then, this transfer function H of s as the ratio of the output impedance divided by Z1. Okay, I'm going to do an example of that here on some semi-log axes, although the next slide contains the results. So let's suppose that we've constructed and we have some values for r, l and c, I'm going to just choose some arbitrary numbers. So let's suppose the inductor has some omega L asymptote that looks like that. The capacitor has a one over omega C asymptote, perhaps looking like this. The resistor maybe is here, just for some arbitrary choice. And so to construct the output impedance or parallel combination, as usual, we take the lowest asymptote, so we get asymptotes like this, in red, for Zout. We have a corner frequency right here at f0, which is 1 over 2pi root LC, as we discussed previously. There's a characteristic impedance R naught that is omega naught L or one over omega naught C, or simple root l over c if you eliminate omega naught. And, we have a q factor where the inductor and capacitor cancel out at the corner frequency and, the, the output impedance rises up to be what's left, which is the load resistance. So, we get a q factor, that would be R divided by R naught. So that is our Zout asymptote, our set of asymptotes. We have to divide that by Z1, which is the inductor asymptote. So Z1, I'll draw as simply following omega L, and I'll draw that one in green. And finally, we'll construct the, the ratio. So to divide asymptotes, or to divide impedances, we simply divide the expressions for the asymptotes. So over some given range of frequencies, we express each impedance by its asymptote and divide by the asymptote of the transfer function. Once again, I'll remind us what H of s is. It's the Zout divided by Z1, so at low frequency, below f naught, Zout and Z1 are both omega L and they follow the omega L asymptote. So, the asymptote for H is omega L divided by omega L or one. My axes are labelled in dB ohms but I'm going to draw H and just DB on the same axes, so we'll draw H at 0dB all the way up to F knot. So I'm going to label the magnitude of h, which is the magnitude of Zout divided by the magnitude of Z1. And for these low frequencies below f naught, we get omega L divided by omega L, or 1. Okay, above f naught Z1 remains omega L but Zout becomes one omega C. So what we have for Zout divided by Z1 will be one over omega C divided by omega L. And the slope will have a Z out will have a slope of minus 20 db per decade. When we divide by Z1, we subtract its slope, which is 20 db per decade. So we get a composite slope of minus 40 dB per decade, so we need to draw a minus 40 dB per decade slope there. [SOUND] Like this. And the equation of this asymptote here then, H is again Zout over Z 1. But Zout as I mentioned, what now becomes one over omega C and then we divide by omega L and so the expression of this asymptote is one over omega squared LC. So we have two poles at f naught. Now, one last question is, what is the q factor. Well, to find the q factor for H, we find the exact value of H at the corner frequency. And, the exact value of H is the exact value of Zout divided by the exact value of Z1. So, what will that be? Well, Zout, the exact value is R, and Z1 has an exact value of R naught at f naught. So, the ratio will be R over R naught which is q, therefore H has the same q factor that Zout had. So if Q is that distance, then we'll get the same distance here and the actual function will have a resonance curve as well, so I'll label that distance. So Q is Zout of j omega naught divided by Z1 of j omega naught which is again R over R naught. So graphically, we can get everything we need simply by taking the ratios of the impedances. Okay. Again, all of those quantities are listed here on the next slide, let's apply this then to the canonical model. Our, our canonical model of the, of power converters contains exactly the filter that we just solved with a transfer function that we've been calling He of s. So we can construct He of s by the method I just described. The only slight complication is that we have an effective inductive cell E that might depend on duty cycle. However, we can handle that the, the asymptotes for the inductor impedance, again follow as usual. So, here's our output impedance as we just dis, discussed. Here's the graphical construction of the output impedance with our capacitor, resistor, and our Le asymptotes and we can identify the Q factor. So, we get a transfer function then with asymptotes like we just constructed. With an f naught corner frequency given by the point where the capacitor and Le asymptotes intersect and with a Q factor defined by R created by that characteristic impedance. In the boost and the buck boost converters, Le is given by L over D prime square, that's what we found in an earlier lecture. So, what happens to the inductor asymptote then, when we change the duty cycle? Well, let's see, if you increase the duty cycle that will make D prime decrease and so one over D prime squared will increase. So increasing the duty cycle makes L over D prime squared increase, and the asymptote then will go up like this as you increase the duty cycle. Well, you can see what that does to the corner frequency, if we have a, a larger effective inductor asymptote, then that will move the corner frequency down to a lower frequency. And you can also see what that does to the Q factor, when we increase our inductance, we increase R naught, so a new R naught will shift from here to here. And then the Q factor, which is this distance, will become smaller. So, we get a lower Q when we increase the duty cycle. So it's a fairly simple matter from constructing the asymptotes to see what happens when things like the operating point change. What does this do, then, to the corner frequency or the transfer function? So we previously had a transfer function that had a low frequency asymptote of 0dB and we get a corner frequency here at f naught with some Q factor. And when we increase the duty cycle, the low frequency 0 dB asymptote doesn't change, but the high frequency asymptote, corner frequency and Q factor all change. So this will move down to this frequency, with this smaller q, and then we'll have our minus 40 dB per decade sloped in. So, the graphical construction method then can give us everything we need, we can just construct all of the impedances. and transfer functions, asymptotes, corner frequencies, and Q factors directly, and in the process get a great deal of, I think, physical insight as to how changing things change the results of our converter circuit.
