So far we've talked just about series impedances. Let’s talk next about parallel impedances. So, here is an RLC example where we have a parallel network and we want to compute the composite Z of X for this. So we are going to start out in the same way by constructing the individual impedance asymptotes. And here [SOUND] I'd use the same values as in the last example. So R is 10 ohms. L is 1 millihenry, which we drew in asymptotes like this per L. And C is the endpoint of one micro-ferret which we drew with the asymptotes right here. With a parallel combination, which component dominates? Well, let's write the equation Z. So Z of S is found by the parallel impedance formula, which is inverse addition. So R in parallel with SL and parallel with 1 over SC. This notation means that we, add the impedance's using inverse addition, so this is 1 over, 1 over R plus 1 over SL plus 1 over the capacitor impedance 1 over SC. At any given frequency, which impedance will dominate the inverse sum. Well, the smallest impedance will have the largest inverse, so it will dominate the inverse sum, and then when we take 1 over that we get that impedance back. So at low frequency the inductor is smallest and we'll follow the inductor impedance. Here at mid frequency the resistor is smallest, so we follow it, until we get to high frequency where the capacitor impedance is smallest, and then those are the asymptotes of the composite as the of S. Looks like we again have two corner frequencies, back and F1 and an F2. Again F1 is where the inductor asymptote intersects the resistor asymptote. We've already done that one, we found that F1 was R over 2 pie L. Likewise, at F2, that's where the capacitor asymptote intersects the resistor asymptote. That happens at 1 over 2 pie R C. So we, looks like we have two real poles in this case. The asymptotes are changing from a plus 20 slope to a 0 slope to a minus 20 slope. Alright, let's change the value of R back to 1K. For this parallel resonant circuit. Let's see at 1K, R ends up being up here again. [SOUND] Let's draw our R, L and our C. And there's Omega L and Omega C. [SOUND] We'll go there. So who dominates the parallel combination now? Well at low frequency, the inductor is smallest so it will dominate. At high frequency the capacitor is smallest so it dominates. The resistor is never smallest so it doesn't determine the asymptotes anymore. We have a plus 20 [INAUDIBLE] per decade asymptote at low frequency and minus 20 at high frequency. Right here where the inductor asymptote and the capacitor asymptote are equal which is at F naught the inductor asymptote, inductor and capacitor asymptotes. become equal and we get a corner frequency. Again, since we change from plus 20 to minus 20 slope at this corner frequency, we should suspect that there's a resonance. This looks like there could be complex poles in this case, and so we again need to work out, what is the exact value of Z at this corner. Well, let's use the inverse impedance or inverse some formula z is 1 over, 1 over R plus 1 over j omega L plus j omega C. We'll let omega B omega 0. At omega 0, both the inductor and capacitor asymptotes have the same magnitude. Again, R 0 or the characteristic impedance. And so Z we can write as 1 over, 1 over R plus 1 over jR 0 for the inductor, plus, in this case, j over R 0 for the capacitor. If you multiply top and bottom of the inductor impedance by j, we ge, we get minus j over R0 for the inductor sss. And plus j over R0 for the capacitor. So the inductor and capacitor terms again cancel out, and Z goes exactly to R. So we peak up and touch the R asymptote right at the resonant frequency. So again we get a curve. With a resonance that goes up and touches R in the parallel combination the inductor and capacitor again cancel out in the inverse sum, just as they did in the sum. So with, with this parallel resonance circuit, then, the impedance then goes to the, what's left, which is the resistor. We again have a Q factor, then in this case Q would be this value and D B minus the characteristic impedance in D B, or if it's not in D B it would be R divided by R 0. So, we have an expression for the Q factor in this case as well. In the parallel-resonance circuit, Q is R over R0, whereas in the series resonant circuit it was R0 over R. So again we can, with the resonant circuit, we can work out the individual asymptotes, find the composite, and get everything that we need. Okay, I want to do one last example. Here's a somewhat more complex circuit. You can do the same tricks with more complex circuits. We just keep applying the series and parallel formulas taking the larger or smaller asymptotes to build up the composite asymptote at Z. The good way to start here is to break this down into smaller sections, so maybe we'll call this parallel R1 and C1. I'll call that Z1. So that's R1 and parallel with 1 over SC1. And we'll call this lower network a Z2 which would be R2 in parallel with 1 over SC2 in parallel with SL2. So what we'll do is construct those and then finally put them in series to get Z so that Z is Z1 plus Z2. So let's, let's start. Z1 is R1 which is 100 ohms or 40 dB ohms. So there's R1. C1 is 16 microfarads. We can work out the frequency where 1 over omega C1 is 1 ohm. And that turns out to be at 10 kilohertz. So here's the 1 over omega C1 asymptote. The parallel combination of R1 and C1 then will take the smallest. [SOUND] which is this and that's a Z1 that has a corner frequency right there, the frequency we will call F1. F1 is where R1 equals 1 over omega C 1 and that works out to be 100 Hertz. For Z2 lets construct those asymptotes. R2 is 10 ohms so that's 20 [UNKNOWN] and that's right there, C 2 is 1.6 micro ferrets. It goes through 1 ohm at a 100 kilohertz which is right there [SOUND]. So, here's our C 2 asymptote. And then L2 is 16 millihenrys. We can work out where omega L2 equals 1 ohm, and that turns out to be at 10 hertz, right here. So, we can draw a plus 20 dB per decade slope for L2. Like that. So let's put all the red asymptote in parallel R2, C2, and L2. So we'll follow L2 at low frequency. Then R2 at mid frequency. And then C2 at high frequency. And that would be the magnitude of Z2. We get a corner in it right here at 100 hertz and another corner here at 10 kilohertz. I'm going to call the 10 kilohertz corner F2 which is 1 over 2pie let's see, R2 times C2. [SOUND] To cocnstruct the composite S of tote L, we simply take the larger of Z1 and Z2, so take the larger of this blue cross hatched line with the red crosshatched line. You can see that Z1 dominates at low frequency, until we get right here at 1 kilohertz. We'll call that F3 where the C1 asymptote crosses the R2 asymptote. So F3 is at 1 over 2 pie times C1, times R2. After that, we follow the Z2 or red crossed hashed asymptote. We get to F2 and then we follow capacitor C2. And the green asymptotes then are the composite asymptotes for Z. So, we can build up the impedance of, a fairly complicated circuit and easily construct the approximate asymptotes of, of the composite impedance Z. One thing I'll say here is that unlike maybe the very first example, these are not exact asymptotes. They're approximate. And one way you can see that is for example at high frequency. Where the two capacitors dominate. At high frequency here, we're taking C2, as being the dominate impedance. If you look at the circuit, at high frequency really we have C2 in series with C1, which is, in fact, slightly less than C2 in value. We're making an approximation here, and it's a pretty good approximation, but it's an approximation nonetheless. You can't expect to be able to simply draw straight line asymptotes, and get exact factorizations of higher order polynomials. So this is, in fact, an approximate solution within approximate factorization. I would suggest that it is really the graphical equivalent of the analytical factorization method that we talked about last week. Another interesting thing that comes out of this, is that the L2 asymptote doesn't play at role in the final composite impedance asymptotes. Omega L2 is actually too large in value to come into play where it would have a significant effect. And so it, it may not be obvious from looking at the original circuit that, that’s the case but you can see from looking at the asymptotes here in the way we constructed them as to why that happens. It's actually just as important to be able to make approximations to see which components don't have an impact. As it is to see which ones do. And so, if in fact you simply threw out [UNKNOWN] 2, approximated it as an open circuit. And just drew the asymptotes of these components, the, the resistors and capacitors, you would get a good approximation of the final result. So, being able to simply the, the circuit and construct the asymptotes that depend only on the dominant impedences and nothing else. Is a very variable thing to be able to do. And this is a structured way that leads to the approximation such as that as well. In a radial frequency business, this kind of approximation is widely used, but generally in an earlier day than today. Something called Reactance paper, or also here called Impedance graph paper, was widely distributed among RF engineers. And this is basically a Bode plot, but it has inductor and capacitor impedance asymptotes already plotted for you. So this slide has, something like that in PDF form. So if we have a, say, combination of, maybe of, what a 1 millihenry inductor? In series with a, say a, 1 ohm, resistor, and then that's maybe in parallel with maybe 10 microfarads. Then we can easily plot the composite impedance by just following the 1 millihenry line until we get to 1 ohm. And that's the series combination of 1 millihenry and 1 ohm. And then, we put the 10 microfarads in parallel, which is this Impedance and there you go, there's your vote plot. So this a very easy and quick way to construct pretty complex impedance's right on the plot [SOUND].
