The basic principle to be used in mechanical systems to derive a mathematical model is Newton s law,

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1 Chapter. DYNAMIC MODELING Understanding the nature of the process to be controlled is a central issue for a control engineer. Thus the engineer must construct a model of the process with whatever information is available. Once the model is obtained, the task is to design a controller and then to analyze the process behavior under such a design. In many cases the modeling of complex systems is difficult, expensive, and timeconsuming. But, as this is an essential task, modeling will be done using various tools in order to find an approximated model which can reasonably well represent the process for our purposes. The most common and reliable way of finding a mathematical model of a process is to use the basic laws of Physics. Though this is not a trivial task at some circumstances, it is readily possible and useful at some occasions. This type of modeling results in some dynamic equations, usually ordinary differential equations, which later on could be solved to learn the nature of the response of the system... Dynamics of Mechanical Systems The basic principle to be used in mechanical systems to derive a mathematical model is Newton s law, F = ma where F : Vector sum of all forces applied to each body in a system, N, a : vector acceleration of each body, m / s, m : mass of the body, kg. Suppose we have a cart as shown in Figure.. For simplicity, assume that rotational inertia of the wheel is negligible and friction is proportional to the cart speed. Figure. also shows the free-body diagram and coordinate of the system. Figure.. A simple mechanical system. The equation of motion is then, b u b x = M x x+ x = u, with a = x M M Another mechanical system is depicted in Figure.. The aim is to find the equations of the motion for m and m due to a displacement u at the bottom point of the system. This system is close to the operation of bus suspension system. 8

2 Chapter Figure.. A mechanical system. Considering the above figure using Newton's law, we have for m: And for m: k y by& + k x + bx& = m && y > m && y = k ( x y) + b( x& y& ) OR m && y + by& + k y = bx& + k x k y + by& k x bx& k u = m && y > OR m && y + bx& + ( k + k ) x = by& + k y + k u m && y = k ( u x) + k ( y x) + b( y& x& ) For a one dimensional rotational systems, Newton s law is modified to M = Iα, where M : sum of all torques on a body, N, I : body s moment of inertia about its center of mass, kg m α : angular acceleration of the body, rad / s. An example of application of such equations is in model of attitude motion of a satellite as shown in Figure.3. Satellites usually require attitude control so that antennas, sensors, and solar panels are properly oriented. To understand the full three-axis attitude control system, one axis at a time is often considered (see Figure.3). 9

3 Chapter Fig..3: Attitude control in a Satellite is an example of a rotational system. The angle θ that describes the satellite orientation must be measured with respect to an inertial reference (i.e. a reference that has no angular acceleration). The control force comes from reaction jets that produce a torque ( Fd c ) about the mass center. There may also be small disturbance torques ( M D ) on the satellites. Then the equation of motion will be Fd M I I M c + D = θ θ = c + MD where Mc = Fd c I. Dynamics of Electromechanical Systems In electromechanical systems, the laws of Kirchhoff, laws of motors (or electromagnetic fields), and laws of mechanics are the basic tools to find a mathematical model of the systems. For control of rotary motion, a common device is a DC (direct current) motor. Figure.4 depicts a schematic diagram of an armature-controlled dc motor. Fig..4; Schematic diagram of an armature-controlled dc motor. 0

4 Chapter where R a = armature-winding resistance, ohms L a = armature-winding inductance, henrys i a = armature-winding current, amperes i f = field current, amperes e a = applied armature voltage, volts e b = induced voltage (ie. emf), volts θ = angular displacement of the motor shaft, radians T = torque delivered by the motor, N.m J = equivalent moment of inertia of the motor and load referred to the motor shaft, kg m f = equivalent viscous-friction coefficients of the motor and load referred to the motor shaft, N m/ rad /sec The torque T delivered by the motor is proportional to the product of the armature current i a and the air gap flux ψ (i.e. ψ = Kf ia), T = K if Kf ia and since the field current is usually constant, the torque becomes directly proportional to the armature current, T = Ki a where K is motor torque constant. On the other hand, when the armature is rotating, a voltage proportional to the product of the flux and angular velocity is induced in the armature. For the constant flux the induced voltage e b is directly proportional to the angular velocity d θ, e K d θ b = b with K b is the emf constant. The speed of an armature -controlled dc motor is controlled by the armature voltage e a. The armature voltage is supplied by an amplifier. The differential equation for the armature circuit is L di a a + Ri a a+ eb= ea The produced torque is J d θ f d θ + = T = Ki a These last two equations describe our electromechanical system..3 Dynamics of Other Systems Electrical, heat flow, and fluid flow systems are among other systems can be discussed here. However, only the fluid flow systems are considered in the following. Fluid flows are common in many control systems components or control applications. The physical laws of force equilibrium, resistance, and energy and mass conservations are examples of the governing relations.

5 Chapter Consider the water tank example in Figure.5. Figure.5. A typical liquid level system. The total mass balance for the system is d( ρah) = ρfi ρfout where F i = Inlet volumetric flow rate F out = Outlet volumetric flow rate A = Cross sectional area of the tank ρ = Density of water h = Height of water m = Mass of water in the tank Assuming ρ is constant and expressing the output flow rate as Fout =α h A dh = Fi α h, we will have.4 Further Studies* There are many other examples of dynamic modeling of systems. Following are some examples. To get more knowledge please visit the websites of other well known universities, for instance, see: o MIT Free Courses: Astronautics/6-3Feedback-Control SystemsFall00/CourseHome/index.htm o University of Michigan: The following examples are from the University of Michigan web site mentioned above. Example: Modeling a Bus Suspension System using Transfer Function Physical setup Photo courtesy: Eisenhower Center Designing an automatic suspension system for a bus turns out to be an interesting control problem. When the suspension system is designed, a /4 bus model (one of the four wheels) is used to simplify the problem to a one dimensional spring-damper system. A diagram of this system is shown below:

6 Chapter Where: * body mass (m) = 500 kg, * suspension mass (m) = 30 kg, * spring constant of suspension system(k) = 80,000 N/m, * spring constant of wheel and tire(k) = 500,000 N/m, * damping constant of suspension system(b) = 350 Ns/m. * damping constant of wheel and tire(b) = 5,00 Ns/m. * control force (u) = force from the controller we are going to design. Design requirements: A good bus suspension system should have satisfactory road holding ability, while still providing comfort when riding over bumps and holes in the road. When the bus is experiencing any road disturbance (i.e. pot holes, cracks, and uneven pavement),the bus body should not have large oscillations, and the oscillations should dissipate quickly. Since the distance X-W is very difficult to measure, and the deformation of the tire (X-W) is negligible, we will use the distance X-X instead of X-W as the output in our problem. Keep in mind that this is an estimation. The road disturbance (W) in this problem will be simulated by a step input. This step could represent the bus coming out of a pothole. We want to design a feedback controller so that the output (X-X) has an overshoot less than 5% and a settling time shorter than 5 seconds. For example, when the bus runs onto a 0 cm high step, the bus body will oscillate within a range of +/- 5 mm and return to a smooth ride within 5 seconds. Equations of motion: From the picture above and Newton's law, we can obtain the dynamic equations as the following: Example: Modeling the Ball and Beam Experiment 3

7 Chapter Problem Setup A ball is placed on a beam, see figure below, where it is allowed to roll with degree of freedom along the length of the beam. A lever arm is attached to the beam at one end and a servo gear at the other. As the servo gear turns by an angle theta, the lever changes the angle of the beam by alpha. When the angle is changed from the vertical position, gravity causes the ball to roll along the beam. A controller will be designed for this system so that the ball's position can be manipulated. For this problem, we will assume that the ball rolls without slipping and friction between the beam and ball is negligible. The constants and variables for this example are defined as follows: M mass of the ball 0. kg R radius of the ball 0.05 m d lever arm offset 0.03 m g gravitational acceleration 9.8 m/s^ L length of the beam.0 m J ball's moment of inertia 9.99e-6 kgm^ r ball position coordinate alpha beam angle coordinate theta servo gear angle The design criteria for this problem are: Settling time less than 3 seconds Overshoot less than 5% System Equations The Lagrangian equation of motion for the ball is given by the following: 4

8 Chapter Linearization of this equation about the beam angle, alpha = 0, gives us the following linear approximation of the system: The equation which relates the beam angle to the angle of the gear can be approximated as linear by the equation below: Substituting this into the previous equation, we get: Example 3: Modeling a Pitch Controller Photo courtesy: Boeing Physical setup and system equations The equations governing the motion of an aircraft are a very complicated set of six non-linear coupled differential equations. However, under certain assumptions, they can be decoupled and linearized into the longitudinal and lateral equations. Pitch control is a longitudinal problem, and in this example, we will design an autopilot that controls the pitch of an aircraft. The basic coordinate axes and forces acting on an aircraft are shown in the figure below: 5

9 Chapter Assume that the aircraft is in steady-cruise at constant altitude and velocity; thus, the thrust and drag cancel out and the lift and weight balance out each other. Also, assume that change in pitch angle does not change the speed of an aircraft under any circumstance (unrealistic but simplifies the problem a bit). Under these assumptions, the longitudinal equations of motion of an aircraft can be written as: Please refer to any aircraft-related textbooks for the explanation of how to derive these equations. Also, click Variables to see what each variable represents. For this system, the input will be the elevator deflection angle, and the output will be the pitch angle. Design requirements The next step is to set some design criteria. We want to design a feedback controller so that the output has an overshoot of less than 0%, rise time of less than seconds, settling time of less than 0 seconds, and the steady-state error of less than %. For example, if the input is 0. rad ( degress), then the pitch angle will not exceed 0. rad, reaches 0. rad within seconds, settles % of the steady-state within 0 seconds, and stays within 0.96 to 0.04 rad at the steady-state. Overshoot: Less than 0% Rise time: Less than seconds Settling time: Less than 0 seconds Steady-state error: Less than % () 6