Topic # Feedback Control Systems

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Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response

Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the system has some desired characteristics. Typical objectives: Stabilize the system (Stabilization) Regulate the system about some design point (Regulation) Follow a given class of command signals (Tracking) Reduce response to disturbances. (Disturbance Rejection) Typically think of closed-loop control so we would analyze the closed-loop dynamics. Open-loop control also possible (called feedforward ) more prone to modeling errors since inputs not changed as a result of measured error. Note that a typical control system includes the sensors, actuators, and the control law. The sensors and actuators need not always be physical devices (e.g., economic systems). A good selection of the sensor and actuator can greatly simplify the control design process. Course concentrates on the design of the control law given the rest of the system (although we will need to model the system).

Fall 2007 16.31 1 2 Why Use Control Typically easy question to answer for aerospace because many vehicles (spacecraft, aircraft, rockets) and aerospace processes (propulsion) need to be controlled just to function Example: the F-117 does not even fly without computer control, and the X-29 is unstable But there are also many stable systems that simply require better performance in some sense (e.g., faster, less oscillatory), and we can use control to modify this behavior.

Fall 2007 16.31 1 3 Feedback Control Approach Establish control objectives Qualitative don t use too much fuel Quantitative settling time of step response <3 sec Typically requires that you understand the process (expected commands and disturbances) and the overall goals (bandwidths). Often requires that you have a strong understanding of the physical dynamics of the system so that you do not fight them in inappropriate (i.e., inefficient) ways. Select sensors & actuators What aspects of the system are to be sensed and controlled? Consider sensor noise and linearity as key discriminators. Cost, reliability, size,... Obtain model Analytic (FEM) or from measured data (system ID) Evaluation model reduce size/complexity Design model Accuracy? Error model? Design controller Select technique (SISO, MIMO), (classical, state-space) Choose parameters (ROT, optimization) Analyze closed-loop performance. Meet objectives? Analysis, simulation, experimentation,... Yes done, No iterate...

Fall 2007 16.31 1 4 Example: Blimp Control Control objective Stabilization Red blimp tracks the motion of the green blimp Sensors GPS for positioning Compass for heading Gyros/GPS for roll attitude Actuators electric motors (propellers) are very nonlinear. Dynamics rigid body with strong apparent mass effect. Roll modes. Modeling Analytic models with parameter identification to determine mass. Disturbances wind

Fall 2007 16.31 1 5 State-Space Approach Basic questions that we will address about the state-space approach: What are state-space models? Why should we use them? How are they related to the transfer functions used in classical control design? How do we develop a state-space model? How do we design a controller using a state-space model? Bottom line: 1. What: representation of the dynamics of an n th -order system using n first-order differential equations: m q + c q + kq = F [ ] [ q 0 1 = q k/m c/m ẋ = Ax + Bu ] [ q q ] + [ 0 1/m ] u 2. Why: State variable form convenient way to work with complex dynamics. Matrix format easy to use on computers. Transfer functions only deal with input/output behavior, but state-space form provides easy access to the internal features/response of the system. Allows us to explore new analysis and synthesis tools. Great for multiple-input multiple-output systems (MIMO), which are very hard to work with using transfer functions.

Fall 2007 16.31 1 6 3. How: There are a variety of ways to develop these state-space models. We will explore this process in detail. Linear systems theory 4. Control design: Split into 3 main parts Full-state feedback fictitious since requires more information than typically (ever?) available Observer/estimator design process of estimating the system state from the measurements that are available. Dynamic output feedback combines these two parts with provable guarantees on stability (and performance). Fortunately there are very simple numerical tools available to perform each of these steps Removes much of the art and/or magic required in classical control design design process more systematic. Word of caution: Linear systems theory involves extensive use of linear algebra. Will not focus on the theorems/proofs in class details will be handed out as necessary, but these are in the textbooks. Will focus on using the linear algebra to understand the behavior of the system dynamics so that we can modify them using control. Linear algebra in action Even so, this will require more algebra that most math courses that you have taken....

Fall 2007 16.31 1 7 My reasons for the review of classical design: State-space techniques are just another way to design a controller, and it is essential that you understand the basics of the control design process. Otherwise these are just a bunch of numerical tools. To truly understand the output of the state-space control design process, I think it is important that you be able to analyze it from a classical perspective. Try to answer why did it do that? Not always possible, but always a good goal. Feedback: muddy cards and office hours. Help me to know whether my assumptions about your backgrounds is correct and whether there are any questions about the material. Matlab will be required extensively. If you have not used it before, then start practicing.

Fall 2007 16.31 1 8 System Modeling Investigate the model of a simple system to explore the basics of system dynamics. Provide insight on the connection between the system response and the pole locations. X 1 X 2 Μ 1 Μ 2 F Figure by MIT OpenCourseWare. Consider the simple mechanical system (2MSS) derive the system model 1. Start with a free body diagram 2. Develop the 2 equations of motion m 1 ẍ 1 = k(x 2 x 1 ) m 2 ẍ 2 = k(x 1 x 2 ) + F 3. How determine the relationships between x 1, x 2 and F? Numerical integration - good for simulation, but not analysis Use Laplace transform to get transfer functions Fast/easy/lots of tables Provides lots of information (poles and zeros)

Fall 2007 16.31 1 9 Laplace transform L{f(t)} 0 f(t)e st dt Key point: If L{x(t)} = X(s), then L{ẋ(t)} = sx(s) assuming that the initial conditions are zero. Apply to the model L{m 1 ẍ 1 k(x 2 x 1 )} = (m 1 s 2 + k)x 1 (s) kx 2 (s) = 0 L{m 2 ẍ 2 k(x 1 x 2 ) F } = (m 2 s 2 + k)x 2 (s) kx 1 (s) F (s) = 0 [ m1 s 2 + k k k m 2 s 2 + k ] [ X1 (s) X 2 (s) ] = [ 0 F (s) ] Perform some algebra to get X 2 (s) F (s) = m 1 s 2 + k m 1 m 2 s 2 (s 2 + k(1/m 1 + 1/m 2 )) G 2(s) G 2 (s) is the transfer function between the input F and the system response x 2

Fall 2007 16.31 1 10 Given that F G 2 (s) x 2. If F (t) known, how find x 2 (t)? 1. Find G 2 (s) 2. Let F (s) = L{F (t)} 3. Set X s (s) = G 2 (s) F (s) 4. Compute x 2 (t) = L 1 {X 2 (s)} Step 4 involves an inverse Laplace transform, which requires an ugly contour integral that is hardly ever used. x 2 (t) = 1 2πi σc +i σ c i X 2 (s)e st ds where σ c is a value selected to be to the right of all singularities of F (s) in the s-plane. Partial fraction expansion and inversion using tables is much easier for problems that we will be dealing with. Example with F (t) = 1(t) F (s) = 1/s X 2 (s) = m 1 s 2 + k m 1 m 2 s 3 (s 2 + k(1/m 1 + 1/m 2 )) = c 1 s + c 2 s 2 + c 3 s 3 + c 4 s + c 5 s 2 + k(1/m 1 + 1/m 2 ) Solve for the coefficients c i Then inverse transform each term to get x 2 (t).

Fall 2007 16.31 1 11 Note that there are 2 special entries in the tables 1. 1 (s+a) e at which corresponds to a pole at s + a = 0, or s = a 2. ω 2 n (s 2 +2ζω n s+ω 2 n) e ζω nt sin(ω n 1 ζ2 t) Im ζ > 0 y(t) Envelope t W n - ζw n Wd Re Corresponds to a damped sinusoidal response ζ is the damping ratio ω n is the natural frequency ω d = ω n 1 ζ2 is the damped frequency. Figure by MIT OpenCourseWare. Results show the strong connection between the pole locations and the time response of the system But there are other factors that come into play, as we shall see.

Fall 2007 16.31 1 12 For a second order system, we can be more explicit and relate the main features of the step response (time) and the pole locations (frequency domain). G(s) = with u(t) a step, so that u(s) = 1/s ω 2 n (s 2 + 2ζω n s + ω 2 n) Then y(s) = G(s)u(s) = s(s 2 +2ζω n s+ωn) which gives (σ = ζω n) 2 y(t) = 1 e (cos(ω σt d t) + σ ) sin(ω d t) ω d ω 2 n Several key time domain features: Rise time t r (how long to get close to the final value?) Settling time t s (how long for the transients to decay?) Peak overshoot M p, t p (how far beyond the final value does the system respond, and when?) Can analyze the system response to determine that: 1. t r 2.2/ω h ω h = ω n ( 1 2ζ 2 + 2 4ζ 2 + 4ζ 4 ) 1/2 or can use t r 1.8/ω n 2. t s (1%) = 4.6/(ζω n ) πζ 1 ζ 3. Peak of time response M p = e 2 and t p = π/ω d

Fall 2007 16.31 7 13 Formulas relate time response to pole locations. Can easily evaluate if the closed-loop system will respond as desired. Use to determine acceptable locations for closed-loop poles. Examples: Max rise time - min ω n Max settling time min σ = ζω n Max overshoot min ζ Usually assume that the response of more complex systems (i.e. ones that have more than 2 poles) is dominated by the lowest frequency pole pair. Then the response is approximately second order, but we must check this These give us a good idea of where we would like the closed-loop poles to be so that we can meet the design goals. Feedback control is all about changing the location of the system poles from the open-loop locations to the closed-loop ones. This course is about a new way to do these control designs

Index Example: Blimp control, 4 Feedback Control Approach, 3 Introduction, 1 State-Space Approach, 5 Time response, 12 14