We have one more important standard form to consider in construction of Bode plots. And that is the second order system that may contain a resonance. Here's an example of a two-pole low pass filter that contains an inductor and capacitor, and these LC type filter circuits commonly appear in a converter models, and in fact we had seen them already in the canonical model last week. So, if we solve this circuit to find the transfer function from input to output. We're going to call it g of s in this lecture. We can do the analysis and find the g of s is given by this expression. Which has a quadratic form polynomial in the denominator. So this quadratic or second-order polynomial, we can write, in general, like this, where a1 and a2 are the coefficients of s to the first power and s to the second power. And by equating a1 and a2 to the coefficients l over R and LC in the original transfer function. We find that a1 is L over R and a2 is LC. So, in a case like this, how should we draw the bode diagram. Well, one, one approach, I think the most straight forward way is to factor the denominator. We already know how to draw the Bode plot of a single pole. And so if we could factor the denominator and the second order system into two single poles then we might, perhaps, apply the, the methods we already talked about in the preceding lectures. So, here is our polynomial in the denominator. We could factor it with two roots, and here I've denoted the roots as S1 and S2. And using the quadratic formula, here are the solutions for s1 and s2. So, as usual, it contains a radical, and the quantity inside the radical is the discriminant. And so we have the case where it's possible. For the discriminant to be negative, and we can get complex roots. So, this indeed can happen so in the case then where the discriminant is positive, and we have real roots then we can go ahead and factor it this way, and then construct the bode plot as two real poles. In the way that we have already been talking about. Although I am going to talk a little bit more about that in the next lecture and there are some, perhaps even easier ways to do it than that. But in the other case where the discriminant is negative, and we have complex roots, and we really can't use what we've already talked about in the case of the single pole. It doesn't apply. Here going back to the slide from the previous lectures for the single pole. What we did was we let s equal j omega, and found what G of j omega was for the, the single pole case. We got this expression. And then to write the equation for the magnitude, we, we found the square root of the sum of the squares of the real and imaginary parts. And while I didn't point it out at the time, there was an implicit assumption there that omega naught is a real number. And if omega naught is complex, which happens in the complex roots case, then we're not correctly finding the real and imaginary parts. So the single pole derivation from the earlier lecture assumes that we have real poles and it's not valid in the case where we have complex roots. So in the complex case, what we're going to do instead is define a new standard normalized form. For the second order type polynomial, and then derive the magnitude and phase asymptotes without assuming that we have real roots. There are two forms that are in common use in the control business for the second order case. The first one uses a damping factor called zeta here. This quantity. And the second uses a q-factor here. And you can use either form, in fact. Just from equating the coefficient of s to the first power, apparently two zeta is the same as one over cube. In the case of the q-factor, this is in fact not the most general way to define q. And in the RF business, the Q is, is defined in a more general way. so we can talk about the Q factor of an, a single component such as an inductor. And the Q factor is, a measure of how ideal the inductor is and how large the loss is. So Q actually comes from quality factor, and it's most generally defined in this way, as 2 pi times the peak stored energy in the element, divided by the energy dissipated per cycle. so we can talk about the Q of an inductor. Whereas with the damping factor, this is really defined in the context of a second order system. Now, this definition of Q is actually consistent with this one in the second order system, but the second order system definition is perhaps more restrictive. This is a nice form to use, though, with Q because we, we're going to see in a minute that, it has a very nice interpretation on the Bode plot. Okay, so, here was our original transfer function for the two pole case. we can equate the denominator polynomial to the denominator polynomial in our standard form and did analytical expressions then for the omega not and the q in terms of the original component elements. So, by equating the coefficient of s squared, we have s squared to the s squared divided by LC and this form is equal to one over ω nought squared in this form. So we can solve for ω nought and one ω nought will be ove over root LC or f nought, which is ω nought over 2π, then is one over 2π root LC. Knowing that we can also solve for the Q. So in out standard form, one over Q omega not, is equal to this term, L over R. So we can substitute in our omega not into there and then solve for Q, and if we do that, here's what we get is the expression for the Q for this circuit. Okay, given the standard form, let's work out what the magnitude in phase asymptotes do. So we can take the same approach as in the real pole case. That s equal j omega and work out the magnitude, and if you do that, here's what we find is the magnitude. This term on the left here comes from the real part squared and this term comes from the imaginary part squared. So, we can find the asymptotes. What happens for omega much less than omega naught at low frewuncy? Well, when omega is much less than omega naught then this term here is small compared to one and, this term is also small compared to one. If we let omega be low enough then the whole thing just goes to one. So our low frequency asymptote as in the case of the real pole is 0 db or 1 per frequency sufficiently less than the corner frequency of nought. In the high frequency case where omega is much greater than omega nought. Then which term dominates? Well, this term here goes like omega over omega nought squared, and then squared again, so to the 4th power. and if omega is much greater than omega nought, then this term will be large compared to the 1, so we can throw that out. This term here goes like omega over omega nought squared, it doesn't get large as fast as the omega over omega nought to the fourth power term though, so this one doesn't dominate, it's this one that dominates at high frequency. So the magnitude of G then, we'll go to one over square of just this term, omega over omega naught to the fourth power, which is omega over omega naught to the minus two power. Or in terms of f, we can write it as f over f naught to the minus two power. So this is an asymptote, with a slope of the minus 2 power times 20 dd per decade or minus 40 dd per decade. It intersects the low frequency asymptote at the corner, so I think this is a nice set of asymptotes. It isn't that consistent with what you would expect from two poles if they were real. And we're both at f nought. What remains is to investigate what happens in the vicinity of the corner frequency, and see the deviation from the curve. So as in the real pole case, what we can do is evaluate the exact transfer function, or the exact formula, at the corner frequency omega equals omega not. So at omega equals omega not, what we get. We just put omega naughts, in for the omegas. Omega naught over omega naught is one, so we will get one minus one squared which actually cancels out, and the real term that came from both of these adds up to zero, and we are left with the term that came from the complex part. And omega nought over omega nought quantity squared is 1. So the whole thing goes, then, to 1 over root 1 over Q squared, which is just Q. Okay, so that's an interesting answer. It says that the value of the exact function at the corner frequency is equal to Q. Well, how large is Q? The fact is that Q can be absolutely anything. And it's interesting that the quantities that determine the asymptotes, see, the low-frequency asymptote is determined by this term, high-frequency asymptote is determined by this term. And the exact value at, at the corner frequency is determined by this term. So Q doesn't matter anywhere except at near the corner frequency. And Q can in fact be absolutely anything. In a passive circuit as a positive number that's all we can say about it. So, in fact, if we go back here q appears only in this term, and it's equal to r root c over l. And we can make r be anything we want, from zero to infinity. Which means q can be anything. So, what determines the actual function at the corner frequency has nothing to do with the terms that determine the asymptotes. Away from the corner. If q is large, then the fun, say, q is a million. The function will go up here to a million, and it will look like this. If q is a small number, 0.01 down here, then the function will go down there. So we can get an exact function that deviates wildly from the asymptotes at the corner frequency. And this behavior is in fact characteristic of a resonant circuit, where at this resonant frequency, f naught, we get we get solutions that deviate substantially from what happens at other frequencies. Okay. How about the phase asymptotes? Well, we can find the phase of G by finding the arctangent of the imaginary over real part. As we did in the real case real pole case. So if we do that, here is what we get. we can plot the, this function as usual, and on the computer-drawn plot, here's what it looks like. The curves tend to a low-frequency asymptote of zero degrees. And the high-frequency asymptote of -180°, as you would expect two poles to do, and then in the vicinity of the corner frequency the the change from 0° to -180° becomes more sharp as you increase the Q factor. So, we can think about some mid-frequency asymptote, say that does this, to change from 0 to minus 180, and how steep this, asymptote is, or how steep the change is, depends on Q. All of the functions go through minus 90 degrees at the corner frequency. But then the slope here depends on the q-factor. As in the real poll case, we can try to dirive an asymptote for the mid-frequency phase. We might take the same approach as in the real pole case, where we can at first try to find what is the slope of this. The actual phase function at the corner frequency f nought. And if you do that and draw an asymptote like this that has the same slope, then what you find is that the brake frequencies here and here have this e to the pi over two factor like in the, the real pole case. Recall e to the pi over two is approximately 4.81. but then it's raised to the one over two cubed power. So this right frequency right here is f nought times 4.81 to the 1 over 2Q power. Likeways, this break frequency is f nought divided by that quantity. So we have the same choices as in the real pole case. We can use 4.81, or we can round it up to five, or we can round it up to ten. And what I would say here is that, to keep, keep our asymptotes consistent in the real and complex poll case, and in fact to get the same answer at Q of a half, where, which is the boundary between real and, and complex polls, we should use the same factor here. And so what we're going to do is shown here. We're not going to quite do this, we're going to use the approach shown here, where this break frequency right there is ten to the one over two cubed power times f nought. And this one is at f nought divided by 10 to the 1 over 2 cubed power. Here's a sketch of what that asymptote looks like compared to the real curve, and you can see that the actual curve crosses this asymptote in three places. And so the you know, the, this approach, or this way to draw the asymptote, is not a bad approximation. incidentally, what this choice implies, then, is that the slope of the asymptote turns out to be minus 180 times Q. Degrees per decade. So if we have a q of two, then we'll have a slope of, what, minus three hundred sixty degrees per decade. The larger q is, the more negative this slope is. And the closer we get to just a step change from zero to minus one eighty. In the vicinity of, of f nought. To summarize, here is the plot then of what the actual curves look like. These are nice computer-generated plots of the magnitude and the phase response for different values of Q and they're drawn into vicinity of of the corner frequency. So again as Q gets larger and larger, we get a response that has more peaking in it, which is called the resonant response, [NOISE] the phase changes more abruptly from zero to -100° in the vicinity of resonance. Again, as view is increased.