In this lecture, we're going to talk about Graphical construction of the bode plots of impedances. We will directly construct the asymptotes, of impedances on the bode plot, doing a minimum or no amount of algebra beforehand. Effectively, what we are going to do, is do the algebra on the plot itself, by combining asymptotes. This way, I think gives a great deal of insight into what is going on. And if you can construct the asymptotes by just looking at the circuit, then you have a lot more understanding of how the circuit works. This method also naturally give approximations it is in fact an approximate method. And it provides a framework for coming up with good approximations. Yet another advantage of it is that we can see which impedances don't matter. And so, the ones that don't matter we can leave out, and not, not bother with. And it begins a formal way to do that, as well, So, basically we're going to construct the asymptotes essentially by inspection and I'm going to illustrate this with several examples. So, we're going to start with a simple example of a series combination of an R and a C. So, here Z of S is the resistor impedance. In series or plus the capacitor impedance. And what we're going to do is construct the asymptotes of each of these individual elements first. And, then we'll combine them in the series combination. So, first, let's do the resistor. That one's pretty easy. It's at 10 ohms, and it's independent of frequency. In decibels, 10 ohms is 20 dBohms so, we're right here, and we'll have this impedance for all frequencies. That line is supposed to be at 10 ohms so, I'm going to label that R equals 10 ohms. Okay, the second impedance, the capacitor, we know the capacitor impedance is 1 over omega C, and this is a function that goes like frequency to the minus 1 power. So, we can draw an asymptote that has a minus 20 dB per decade slope. And the only other thing we need to do is to find one point on this curve to draw our line through. So, one way to do is to find the frequency where this impedance equals one ohm. Okay, so, one over omega C equals 1 ohm. At what? Omega equals 1 over 1 ohm times C or f, which is 2 pi omega, would be 1 over 2 pi times 1 times C, which is 1 microfarad or 10 to the minus 6th farads. So, we get 10 to the 6 over 2 pi. 1 over 2 pi is 0.159 so, this is 0.159 times 10 to the 6th hertz, or 159 kilohertz. Okay, let's see. 159 kilohertz is just about there, 1 ohm would put us right about there. So, we'll draw a line that goes through that point with a slope of minus 20 dB per decade. So, if I go down a decade in frequency, we would go up 20 dB so, we'll go through these points and I'll draw a line through those. So, that's 1 over omega C with a slope of minus 20 dB per decade. So, these are the asymptotes of the individual impedances. Now, next how do we combine them? We, we have two quantities that are in series or actually mathematically we add them. And what are the asymptotes of the sum? Well, let's look at what happens at high frequency and at low frequency. Let's take it maybe high frequency first. At high frequency, the capacitor impedance is tending towards 0, and if you go to infinite frequency, the capacitor becomes a short circuit. And the function then goes to just the resistor. In fact, the resistor asymptote is the asymptote of the function, the series combination, at high frequency. Likewise, at low frequency what happens is that the capacitor tends towards an open circuit. And a DC in the capacitor impedance goes to infinity. Do it dominates the series combination or the sum. So, at low frequency we follow the capacitor asymptote [SOUND] like that. In fact all we do then for a series combination is we take the larger asymptote and use that as the asymptote for the series combination or the sum. In this particular example, this actually gives the exact asymptotes of Z. In other more complicated cases, it won't be the exact asymptotes. This will give an approximation. But nonetheless, taking the largest asymptote is a good approximation for the sum of, of quantities. So, we have a minus 20dB per decade slope at low frequency in a constant asymptote at a high frequency. And we know in fact what they are, they are resistor or the capacitor impedances. We can also find the [INAUDIBLE] frequency right here. Let's call that f1. What is f1? Well, f1 is the frequency where the two asymptotes intersect. So, we can write at f1, that's where the resistor asymptote or R is equal to the capacitor asymptote, 1 over omega C with omega equal to omega 1. So, we can solve them for omega 1. Omega 1, would be what 1 over RC and you can plug in numbers if you like or you can just read the value off the plot. This turns out to be at 15.9 kilohertz. F1 is 1 over 2 pi RC. And that works out to be 15.9 kilohertz. So, we have everything we need. We have the asymptote. We have the corner frequency. Admittedly this was a simple case, but. This approach generalises to much more complicated cases. Let's do one that's slight harder. Here's a an RLC network. Three impedances in series. So, in case its Z of s is the resistor in impedances R plus the inductor impedances sL. Plus the capacitor impedance 1 over sC. So, again we will construct the individual asymptotes first. The resistor is 1K in this case. So, 1K ohm is 60 dB ohms, and we get this value right up here. The inductor is 1 milihenry. Let's find where 1 milihenry times omega equals 1 ohm. So, omega L equals 1 ohm at omega equals. Let's see, 1 ohm over L, and f will be omega over 2 pi so, that's 1 ohm over 2 pi times L with L as 1 millihenry, or 10 to the minus third, 3 henries. So, let's see, that's 10 to the 3 over 2 pi, which is 0.159 times 10 to the 3 or, 159 hertz. Well let's see, there's 159 hertz, where at 1 ohm is 0dB ohms for the inductor. So, we'll draw a line with a slope of 20 dB per decade through that. Next, let's do the capacitor impedance. So, we have a 0.1 micro [UNKNOWN] capacitor in this example. That impedance equals 1 ohm at what omega would be 1 ohm over C. So, f would be 1 ohm over 2 pi times C, which is 10 to the minus 7th. So, I'll get 10 over the 7th over 2 pi, which is 1.6 mega hertz. So, we'll go, we're right about there. You can draw some points for that. [SOUND]. Now to construct the series combination, just as in the last example we take the largest asymptote at any given frequency, and that is the asymptote of this series combination. So, you can see that at low frequency, the capacitor dominates and we'll follow the capacitor asymptote. At mid frequency, the resistor dominates, so we'll follow the resistor asymptote and then, at high frequency the inductor dominates, and we follow it. And so, those are the composite asymptotes of the magnitude of Z. It looks now like we have two corner frequencies, one here and one here. [SOUND]. And again, we can work out what those are, at f1 thatâ€™s where the resistor process the capacitor asymptote. So, we have R equals capacitor 1 over omega C where the omega equals omega 1. So, that omega 1 is 1 over RC. At f2, that's where the inductor asymptote equals the resistor asymptote. So, R equals omega L, with omega equal omega 2, so we can solve for omega 2, and get omega 2. Is R over L. Again, we have it's analytical expressions for the corner frequencies, and we know what each asymptote is, and so that's everything we need. Alright, let's do another example, let's suppose that we changed the value of R, so that it's now ten ohms instead of 1,000 ohms. We'll leave the values of L and C the same. So, R is now 10 ohms, which puts it down here at 20 dB ohms. The inductor asymptote is the same. It was 1 ohm at 159 hertz. [NOISE]. So, I'll draw it again in the same place, like that. And the capacitor asymptote is also the same. And I'll draw it in the same place as well. Like this, so using our procedure in the same way, we take the largest asymptote at any one frequency. At low frequency, that's the capacitor. Looks like this. At high frequency, that's the inductor, so we follow this one. And you can see that there's no mid-frequency resister asymptote for this case because the R is too small. So, we have a minus 20dB per decade slope, at low frequency and a plus 20dB per decade slope at high frequency. Right here at this corner, we'll call it omega naught, we switch from minus 20 to plus 20. Well, let's see. Let's first calculate omega naught. At omega-naught, that's where the inductor asymptote. Omega L equals the capacitor asymptote, 1 over omega C, with omega equals omega naught. So, you can solve for omega naught then. Omega naught squared is 1 over LC, so we get that omega naught is 1 over root LC. Okay? Again we get a simple analytical expression for the corner frequency right off the asymptotes. Now, one other thing about this, is that you can see that an omega naught, slope changes by 40 dB per decade. So, that sounds like we have two zeros. It may also be that the zeros are complex, and we should suspect that. So we need to do a little work here, beyond this to figure out what is the Q factor of Z. And, and do we have complex of roots here, really complex zeroes or not. Well, the way we found the Q factor, previous, in previous lectures, was we evaluated the exact function. And its exact value at the corner frequency. So, let's work out what the exact value of Z is at the corner frequency omega naught and see what it does. So, at omega equals omega naught, Z of s with s equals j omega. Well, that omega equal omega naught. This would be equal to the sum of the three impedances so, the resistor is R, the inductor is sL with s of j omega naught, so I get j omega naught L. And a capacitor is 1 over sC where s is j omega naught. We'll get 1 over j omega naught C. And we need to simplify this. Well, we can simplify it by looking at these omega naught L and omega naught C quantities. Now, if you look at omega naught, that is in fact, where omega naught l equals 1 over omega naught C. This is in fact, the value of the asymptotes at the corner frequency omega naught. That value we, we classically called the characteristic impedance, R naught. R naught is defined as the value of the asymptotes which we could write as omega naught L or 1 over omega naught C. [NOISE]. So, at the corner frequency, the inductor has impedance jR naught. Likewise, the capacitor has a magnitude of R naught, and then it's divided by j because it's a capacitor. So, we get plus R naught divided by j. Now you can simplify the 1 over j factor. Again, by multiplying top and bottom by j, the square root of minus 1, and the denominator you get j squared, which is minus 1. And so, the whole thing goes to R plus jR naught minus jR naught. So, the inductor and capacitor impedances cancel out. At the corner frequency omega naught, they have the same magnitude of R naught, but they have opposite phase. The inductor has a phase of plus 90 degrees and the capacitor impedance has a phase of minus 90 degrees. And when you add them together, they cancel. And so, the whole thing just goes to R. Now to me this is very interesting. That the quantities that depend or that determine the asymptotes are exactly canceling out at omega naught. And then the function goes to whatever's left. Which in this case is R. So, the actual function goes there and touches the R asymptote right at omega naught. And in the vicinity, we get this resonance curve where the function deviates from the asymptotes as necessary to go and, and become equal to R. Well, R could be anything. it could be absolutely anything, but at omega naught, the function goes to R. So, this is another way to explain a resonance that the function will deviate wildly from the asymptotes because the, the components that determine the asymptotes cancel out. And we go to whatever is left over. The Q factor we defined as the deviation of the actual curve from the asymptotes, add omega naught. So, this distance here is Q in decibels. if we, if it's not in decibels, it's the ratio. So, Q here is equal to R naught, this value divided by this value, which is R. So, we can get an expression for what the Q is also.