System Identification

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System Identification

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What is an Identification Problem? Why Use Frequency Methods? A More Complex Example Problems Return to Table of Contents

What Is An Identification Problem? You have a background in linear systems. Often when you are learning linear systems you encounter problems like the following. Given a system described by one of the following A transfer function, A linear differential equation with constant coefficients that relates the input and the output of the system, An impulse response, A set of state equations, And, given an input to the system, Determine the output (response) of the system. In other words, if you are given a system and an input, you should be able to find the output of the system. That's a problem that could be difficult, but which has a straight forward approach that will get you an answer. The approach might be different depending upon how the system is described, but the approach is always straight forward. It's not very likely that you will have that problem in your job. If you do, there will probably be some sort of complication - like a nonlinearity in the system - that makes your linear systems approach inapplicable. However, there are many times when you have an inverse problem. You may not know the system description, and you may need to figure out a description for the system transfer function, impulse response, differential equation, state equations, whatever knowing a sample of the input and the output. That's a whole different kettle of fish. If you think about linear systems, you probably got the idea that you could always figure out a transfer function for a circuit. But, there are many systems for which you can't get a good transfer function. Aircraft may have transfer functions that vary widely with different conditions (airspeed, altitude, fuel load, atmospheric conditions, etc.) and it may not be easy to compute those transfer functions from physical data. (The author of this page spent a few hours in an airplane with function generators on the floor of the plane feeding low frequency sine waves to the ailerons, etc. It's a good recipe for producing a queasy stomach.) Chemical plants are another example of something that you

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System Identification

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have to control but where you can't get a good handle on the transfer function of the system. In situations like that you may need to have some tools that will let you get a system description from a record of input and output signals. Let's think about how you would go about that. We'll start with a simple situation after we examine the goals for this lesson. The method we will described is shown pictorially on an another page. Click here for that.

Goals For This Lesson It's often possible to measure the frequency response of a system using a sinusoidal input. Those measurements can be used to produce a Bode' plot of the frequency response. In a situation like that you can use that measured data to calculate a transfer function for a system. In this lesson we look at some simple systems with these goals in mind. Given a Bode' plot for a system, To be able to determine the resonant frequency and damping ratio for complex poles in the system. To be able to determine the DC gain for the system. To be able to determine the resonant frequency and damping ratio for complex poles in the system. We assume that you are familiar with frequency response ideas and plots. (Click here to review that material.) At the end of this lesson you will be even more familiar with Bode' plots for first order systems and second order systems. At that point we will examine some higher order systems to see how what you have learned about first and second order systems can be applied there.

Why Use Frequency Techniques? Frequency response methods are widely used. Whenever you buy stereo equipment specifications for the equipment is probably phrased in terms of frequency response. Historically, sinusoidal signals were easiest to generate, so engineers started to use these methods, and a lot of good methods have been devised over the years. Frequency response plots - particularly Bode' plots - have parts that are fairly good straight lines and that makes it easy to measure system parameters like time constants, damping ratios, etc. Frequency response methods are widely used. Often you can get a Bode' plot for a system, and you need to determine what the system is. If you can achieve these goals you're on your way to doing that. One widely used way of getting a transfer function is the following.

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System Identification

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Measure the frequency response. Choose a set of frequencies - using experience, knowledge of the system, whatever - and input sine waves at the chosen frequencies. Measure amplitude gain and phase shift. Plot the data and from the shape of the curve(s) determine the transfer function. This is all more easily said than done. There are several choices for plots, but you will probably find that Bode' plots are the best choice because of their propensity to assume straight lines over different regions of the plot. In any event, that's the essence of the method. A First Order Example System Here's an example plot. Look at the features that would help you identify the system. Note that the complete plot includes both magnitude and frequency plots.

The magnitude plot - the red line - consists of two straight line segments joined by a curved section. The two straight line segments can be extended to meet at the corner frequency. Click the button in, then move the mouse outside the button and release to keep the line from disappearing in the plot. The low frequency asymptote lets you compute the DC gain. The high frequency asymptote drops off at -20 db/decade. The high frequency phase asymptote is -90o. Both of these observations support a conclusion that there is one more pole than zero in the system. A simpler conclusion is that the system is first order, but that conclusion is not one we can be absolutely certain of. Why can't we assume that this system is first order? (Actually it is first order, and the transfer function is 1/(s + 1) - a particularly simple one that lets us make a few points we need to make.) You don't know for sure if you have a pole and zero close together. You could have

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something like this. Here's the Bode' plot for a system with a transfer function of: G(s) = (.3 s + 1)/[s + 1)(.33 s + 10] Note that there is a pole at s = -1/.33 and a zero at s = -1/.3, and that the pole and zero are close together. The pole is near f = 0.5.

The pole and zero are close together so that there's only a slight bend in the magnitude plot and a small blip in the phase plot. When the pole and zero are really close you may see almost no effect. But they could be there. Here's another example. Let's examine this system the same way. This system is a little more complex and presents a few other interesting features.

The magnitude plot - the red line - consists of two straight line segments joined by a curved section - and it is a very funny curve. The two straight line segments can be extended to meet at a frequency that turns out to be the frequency of the resonant peak. That frequency seems to be about fn = 0.6. Where the lines meet is considerably below the peak, and the lines do not generate a good approximation to the actual curve because of the peak.

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System Identification

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There are other things we can learn about this example system. The low frequency asymptote lets you compute the DC gain. Here the asymptote is at -6 db, so the DC gain is 0.5. The high frequency asymptote drops off at -40 db/decade. The high frequency phase asymptote is -180o. Both of these observations support a conclusion that there are two more poles than zeros in the system. A simpler conclusion is that the system is second order, but that conclusion is not one we can be absolutely certain of - for the same reasons we saw earlier for the first order system. Since this system is a second order system, it not only has a natural frequency, but there is a damping ratio to be determined. We can use the resonant peak to get the damping ratio. The resonant peak is 14 db high. That corresponds to a gain of 4.47 In an earlier lesson we got a formula for the gain at the resonant peak. 4.47 = 1/(2z) That implies z = 1/8.94 = .112 Is that right or wrong? It's not a correct computation. The correct formula is: Height above DC gain = 1/(2z). In the original lesson, the expression is: 4.47 = Gdc/(2z). The DC gain is more like -6db. There is a 20 db difference from the DC gain to the peak - a gain of 10 - so we have: 10 = 1/(2z) So z = 1/20 = .05 And that's right! Now, we have enough information to compute the transfer function. The DC gain was found to be 0.5, so Gdc = 0.5 We also found the natural frequency, fn = 0.6. That corresponds to wn = 2p0.6 ~= 3.77 r/sec. Finally, z = .05. That gives us a transferfunction of: G(s) = 7.11/(s2 + 3.77 s + 14.29) A More Complex Example

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Here's another example. This one looks fairly horrible. What can we conclude? What evidence do we see?

The high frequency phase asymptote is -270o. This is at least a third order system. If it's a third order system, we still need a strategy to figure out what it is. We can note the following. There must be a second order set of poles that are undamped. They produce the resonant peak. That peak is at f = 0.159 Hz, so that's the natural frequency. Convert it to wn. and you get: wn = 1.0 There may be two line segments above and below the peak frequency. Check that by clicking the button in the picture below. Looks good doesn't it? Is there another corner where the lower frequency line touches zero db?

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Now, ponder the question of the slope of the lower frequency line there. Is it -20 db/dec? We need to answer the question about the magnitude slope. It doesn't really look like -20 db/dec. Check the lines again. However, it may be that there's not a large enough range of frequency for the slope to get to its' asymptotic value - presumably -20 db/dec. At high frequencies, the slope looks more like -60 db/dec, and the phase looks like it is approaching -270o. It's a pretty good assumption that this is a third order system. It's a pretty good assumption, but don't bet your life on it. The exact conclusion is that it is at least a third order system. Check the example above where we had a pole and zero close together where they tended to mask each other even though they didn't exactly cancel out. We could have that situation here too. The best bet is that there is a real pole that causes the phase to begin getting negative around f = .01Hz, and the magnitude to dip there as well. To find the system here, you may need to try a few transfer functions and try to match this data. An educated guess would be to look for the point where the magnitude is 3 db down from the DC gain. That's -9 db, and that's clearly near the midpoint between f = .01 and f = .02 Hz. That's probably the corner frequency for the real pole. Finish this. Find G(s). For a final example, here's a deceptive system.

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System Identification

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This system looks like it is second order. But there's supporting and negating evidence for that conclusion. There's a peak that looks like a resonant peak. Phase goes to -180o at high frequencies. High frequency rolloff is -40 db/dec. However, phase is positive at f = .02Hz. That can't happen in a second order system. And, if you try to fit this data with a second order system, the peak here is too broad. This system is really a third order system. Here's the transfer function. G(s) = (5 s + 1)/[(s + 1)3] The zero is what causes the rise in the frequency response. The three poles at 1, together with the zero produce the effects noted above. Because the zero is at a lower frequency than the poles, the phase gets slightly positive for a while, but later tends to -180o at high frequencies. Can we conclude anything from this? Here are some items to note. If you have frequency response data, you may be able to figure out what the transfer function is. Closely spaced poles and zeroes may be missed. There is the possibility of confusing real poles and zeroes with a pair of complex poles. There are surely other things we have overlooked. What If? There are some questions to ponder?

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What if you only have a magnitude plot - as sometimes happens? You may still be able to determine the transfer function. However, you will not have the phase as a check, and there's always the possibility of a non- minimum phase factor in the transfer function. What if the data is not taken at points that are dense enough? It's possible to miss highly resonant peaks. This is another reason you should have some a priori information about the system, or you should know the form of the transfer function from fundamental physical principles. Problems Identification Problem - 2.1 Identification Problem - 2.2 Identification Problem - 2.3 Identification Problem - 2.4 Return to Table of Contents Send us your comments on these lessons.

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