ME 132, Dynamic Systems and Feedback. Class Notes. Spring Instructor: Prof. A Packard
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1 ME 132, Dynamic Systems and Feedback Class Notes by Andrew Packard, Kameshwar Poolla & Roberto Horowitz Spring 2005 Instructor: Prof. A Packard Department of Mechanical Engineering University of California Berkeley CA,
2 ME 132, Spring 2005, UC Berkeley, A. Packard 1 1 Introduction In this course we will learn how to model and control engineering systems. Some examples of systems which benefit from well-designed control systems are Airplanes, helicopters, rockets, missiles: flight control systems including autopilot, pilot augmentation Cruise control for automobiles. Lateral/steering control systems for future automated highway systems Position and speed control of mechanical systems: 1. AC and/or DC motors, for machines, including Disk Drives/CD, robotic manipulators, assembly lines. 2. Elevators 3. Magnetic bearings, MAGLEV vehicles, etc. Pointing control (telescopes) Chemical and Manufacturing Process Control: temperature; pressure; flow rate; concentration of a chemical; moisture content; thickness. The two most basic objectives of a control system are: The automatic regulation (or tracking) of certain variables in the controlled plant to desired values (or trajectories), in the presence of unforseen disturbances The stabilization or improved stability of the controlled plant. A key realization is the fact that most of the systems that we will attempt to model and control are dynamic. We will later develop a formal definition of a dynamic system. However, for the moment it suffices to say that dynamic systems have memory, i.e. the present output of the system is generally a function of previous inputs, as well as the current input to the system. This means that the control actions have impact both when they are applied, and later. Another key concept idea in this course is the use of feedback to control engineering systems. Let us briefly compared two types of control systems: open loop and closed loop. 1. Open-loop control systems: In an open-loop system, the input to the plant does not in any way depend on the current and past values of the output of the plant. The design of the open-loop controller is based on inversion. In some sense, the controller should be an (usually electrical) inverse of the plant. The biggest problem
3 ME 132, Spring 2005, UC Berkeley, A. Packard 2 with open-loop control systems is that they rely totally on calibration, and cannot effectively deal with exogenous disturbances. Moreover, they cannot effectively deal with changes in the process, due to various effects, such as aging components. They require recalibration. Essentially, they cannot deal with uncertainty. Another disadvantage of open-loop control systems is that they cannot stabilize an unstable system, such as balancing an inverted pendulum. 2. Closed-loop control systems: In order to make the control system robust to uncertainty and disturbances, we design control systems which sense the output of the system, and adjust the control input, using feedback rules, which are based on how the system output deviates from its desired value. These feedback rules are based on a model of how the system behaves. If the system behaves slightly differently than the model, usually the feedback helps compensate for these differences. However, if the system actually behaves significantly different than the model, then feedback might even cause instability. This is a drawback of feedback systems. 1.1 Structure of a closed-loop control system The physical system to be controlled is called the plant. This term has its origins in chemical engineering where the control of chemical plants or factories is of concern. A sensor is a device that measures a physical quantity like pressure, acceleration, humidity, or chemical concentration. Very often, sensors produce an electrical signal whose voltage is proportional to the physical quantity being measured. This is very convenient, because these signals can be readily processed with electronics, or can be stored on a computer for analysis or for real-time processing. An actuator is a device that has the capacity to affect the behavior of the plant. An electrical signal is applied to the actuator, which results in some mechanical motion such as the opening of a valve, or the motion of a motor, which in turn induces changes in the plant dynamics. Sometimes, as for example electrical heating coils in a furnace, the applied voltage directly affects the plant behavior without mechanical motion being involved. The controlled variables are the physical quantities we are interested in controlling and/or regulating. The reference or command is an electrical signal that represents what we would like the regulated variable to behave like. Disturbances are phenomena that affect the behavior of the plant being controlled. Disturbances are often induced by the environment, and often cannot be predicted in advance or measured directly. The controller is a device that processes the measured signals from the sensors and the reference signals and generates the actuated signals which in turn, affects the behavior of
4 ME 132, Spring 2005, UC Berkeley, A. Packard 3 the plant. Controllers are essentially strategies that prescribe how to process sensed signals and reference signals in order to generate the actuator inputs. Finally, noises are present at various points in the overall system. We will have some amount of measurement noise (which captures the inaccuracies of sensor readings), actuator noise (due for example to the power electronics that drives the actuators), and even noise affecting the controller itself (due to quantization errors in a digital implementation of the control algorithm). Throughout these notes, we will attempt to consistently use the following symbols: P plant K controller u input y output d disturbance n noise r reference Based on our discussion above, we can draw the block diagram of Figure 1 that reveals the structure of many control systems. Again, the essential idea is that the controller processes measurements together with the reference signal to produce the actuator input u(t). In this way, the plant dynamics are continually adjusted so as to meet the objective of having the plant outputs y(t) track the reference signal r(t). Actuator noise Disturbances Measurement noise Actuators Plant Sensors Controller Controller noise Commands Figure 1: Basic structure of a control system. 1.2 Example: Temperature Control in Shower A simple, slightly unrealistic example of some important issues in control systems is the problem of temperature control in a shower. The components which make up the plant in the shower are
5 ME 132, Spring 2005, UC Berkeley, A. Packard 4 Hot water supply (constant temperature, T H ) Cold water supply (constant temperature, T C ) Adjustable valve that mixes the two; use θ to denote the angle of the valve, with θ = 0 meaning equal amounts of hot and cold water mixing. In the units chosen, assume that 1 θ 1 always holds. 1 meter (or so) of piping from valve to shower head If we assume perfect mixing, then the temperature of the water just past the valve is T v (t) := T H+T C = c 1 + c 2 θ(t) 2 + T H T C 2 θ(t) The temperature of the water hitting your skin is the same (roughly) as at the valve, but there is a time-delay based on the fact that the fluid has to traverse the piping, hence where is the time delay, about 1 second. T (t) = T v (t ) = c 1 + c 2 θ(t ) Let s assume that the valve position only gets adjusted at regular increments, every seconds. Similarly, lets assume that we are only interested in the temperature at those instants as well. Hence, we can use a discrete notion of time, indexed by a subscript k, so that for any signal, v(t), write v k := v(t) t=k In this notation, the model for the Temperature/Valve relationship is T k = c 1 + c 2 θ k 1 (1) Now, taking a shower, you have a desired temperature, T des, which may even be a function of time T des,k. How can the valve be adjusted so that the shower temperature approaches this? Open-loop control: pre-solve for what the valve position should be, giving θ k = T des,k c 1 c 2 (2) and use this basically calibrate the valve position for desired temperature. This gives T k = T des,(k 1) which seems good, as you achieve the desired temperature one time-step after specifying it. However, if c 1 and/or c 2 change (hot or cold water supply temperature changes, or valve
6 ME 132, Spring 2005, UC Berkeley, A. Packard 5 gets a bit clogged) there is no way for the calibration to change. If the plant behavior changes to T k = c 1 + c 2 θ k 1 (3) but the control behavior remains as (2), the overall behavior is T k+1 = c 1 + c 2 c 2 (T des,k c 1 ) which isn t so good. Any percentage variation in c 2 is translated into a similar percentage error in the achieved temperature. How do you actually control the temperature when you take a shower: Again, the behavior of the shower system is: T k+1 = c 1 + c 2 θ k Closed-loop Strategy: If at time k, there is a deviation in desired/actual temperature of T des,k T k, then since the temperature changes c 2 units for every unit change in θ, the valve angle should be increased by an amount 1 c 2 (T des,k T k ). That might be too aggressive, trying to completely correct the discrepancy in one step, so choose a number λ, 0 < λ < 1, and try θ k = θ k 1 + λ c 2 (T des,k T k ) (4) (of course, θ is limited to lie between 1 and 1, so the strategy should be written in a more complicated manner to account for that - for simplicity we ignore this issue here, and return to it later in the course). Substituting for θ k gives which simplifies down to 1 c 2 (T k+1 c 1 ) = 1 c 2 (T k c 1 ) + λ c 2 (T des,k T k ) T k+1 = (1 λ) T k + λt des,k Starting from some initial temperature T 0, we have T 1 = (1 λ)t 0 + λt des,0 T 2 = (1 λ)t 1 + λt des,1 = (1 λ) 2 T 0 + (1 λ)λt des,0 + λt des,1. =. T k = (1 λ) k T 0 + k 1 n=0(1 λ) n λt des,k 1 n If T des,n is a constant, T, then the summation simplifies to T k = (1 λ) k T 0 + [ 1 (1 λ) k] T = T + (1 λ) k [ T 0 T ] which shows that, in fact, as long as 0 < λ < 2, then the temperature converges (convergence rate determined by λ) to the desired temperature.
7 ME 132, Spring 2005, UC Berkeley, A. Packard 6 Assuming your strategy remains fixed, how do unknown variations in T H and T C affect the performance of the system? Shower model changes to (3), giving T k+1 = ( 1 λ ) T k + λt des,k where λ := c 2 c 2 λ. Hence, the deviation in c 1 has no effect on the closed-loop system, and the deviation in c 2 only causes a similar percentage variation in the effective value of λ. As long as 0 < λ < 2, the overall behavior of the system is acceptable. This is good, and shows that small unknown variations in the plant are essentially completely compensated for by the feedback system. On the other hand, large, unexpected deviations in the behavior of the plant can cause problems for a feedback system. Suppose that you maintain the strategy in equation (4), but there is a longer time-delay than you realize? Specifically, suppose that there is extra piping, so that the time delay is not just, but m. Then, the shower model is T k+m 1 = c 1 + c 2 θ k 1 (5) and the strategy (from equation 4) is θ k = θ k 1 + λ c 2 (T des,k T k ). Combining, gives T k+m = T k+m 1 + λ (T des,k T k ) This has some very undesirable behavior, which is explored in problem 4 at the end of the section. 1.3 Problems 1. In this class, we will deal with differential equations having real coefficients, and real initial conditions, and hence, real solutions. Nevertheless, it will be useful to use complex numbers in certain calculations, simplifying notation, and allowing us to write only 1 equation when there are actually two. Recall that if γ is a complex number, then γ = γr 2 + γi 2, where γ R := Real(γ) and γ I := Imag(γ). If γ 0, then the angle of γ, denoted γ, satisfies cos γ = γ R γ, sin γ = γ I γ and is uniquely determinable from γ (only to within an additive factors of 2π). (a) Draw a 2-d picture (horizontal axis for Real part, vertical axis for Imaginary part) (b) Suppose A and B are complex numbers. Using the numerical definitions above, carefully derive that AB = A B, (AB) = A + B
8 ME 132, Spring 2005, UC Berkeley, A. Packard 7 2. For any β C and integer N, consider the summation N α := β k k=0 If β = 1, show that α = N + 1. If β 1, show that α = 1 βn+1 1 β If β < 1, show that k=0 β k = 1 1 β 3. Consider the difference equation p k+1 = αp k + βu k (6) with the following parameter values, initial condition and terminal condition: α = ( 1 + R ), β = 1, u k = M for all k, p 0 = L, p 360 = 0 (7) 12 where R, M and L are constants. (a) In order for the terminal condition to be satisfied (p 360 = 0), the quantities R, M and L must be related. Find that relation. Express M as a function of R and L, M = f(r, L). (b) Is M a linear function of L (with R fixed)? If so, express the relation as M = g(r)l, where g is a function you can calculate. (c) Note that the function g is not a linear function of R. Calculate dg dr R=0.065 (d) Plot g(r) and a linear approximation, defined below g l (R) := g(0.065) + [R 0.065] dg dr R=0.065 for R is the range 0.01 to 0.2. Is the linear approximation relatively accurate in the range to 0.075? (e) On a 30 year home loan of $250,000, what is the monthly payment, assuming an interest rate of 6.75%. Hint: The amount owed on a fixed-interest-rate mortgage from month-to-month is represented by the difference equation in equation (6). The parameters in (7) all have appropriate interpretations.
9 ME 132, Spring 2005, UC Berkeley, A. Packard 8 4. Consider the shower example. Suppose that there is extra delay in the shower s response, but that your strategy is not modified to take this into account. We derived that the equation governing the closed-loop system is T k+m = T k+m 1 + λ (T des,k T k ) where the time-delay from the water passing through the mixing value to the water touching your skin is m. Using calculators, spreadsheets, computers (and/or graphs) or analytic formula you can derive, determine the values of λ for which the system is stable for the following cases: (a) m = 2, (b) m = 3, (c) m = 5. Remark 1: Remember, for m = 1, the allowable range for λ is 0 < λ < 2. Hint: For a first attempt, assume that the water in the piping at k = 0 is all cold, so that T 0, T 1,..., T m 1 = T C, and that T des,k = 1 2 (T H + T C ). Compute, via the formula, T k for k = 0, 1,..., 100 (say), and plot the result. 5. Given a complex number G, and a real number θ, show that (here, j := 1) Re ( Ge jθ) = G cos (θ + G) 6. Given a real number ω, and real numbers A and B, show that A sin ωt + B cos ωt = ( A 2 + B 2) 1/2 sin (ωt + φ) for all t, where φ is an angle that satisfies cos φ = A (A 2 + B 2 ), sin φ = B 1/2 (A 2 + B 2 ) 1/2 Note: you can only determine φ to within an additive factor of 2π. How are these conditions different from saying just tan φ = B A 7. For a function F of a many variables (say two, for this problem, labled x and y), the sensitivity of F to x is defined as the percentage change in F due to a percentage change in x. Denote this by Sx F. Show that for infinitesimal changes in x, the sensitivity is Sx F x F = F (x, y) x Let F (x, y) = xy 1+xy. What is SF x. 8. Draw the block diagram for temperature control in a refrigerator. What disturbances are present in this problem?
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