The Behavior of Dynamic Systems Professor Sanjay Sarma December 4, 2007 1.0 Where are we in the course? In the first two-thirds of the course, we concentrated on the generation of differential equations for systems consisting of rigid bodies, springs and dashpots. In the last third of the course, we look at the solutions of these equations. System Particle System of particles Rigid Bodies Lagrangian formulation Oscillations Kinematics September September September Kinetics & Constitutive September October October November December 2.0 Solving the Differential Equations: Overview Generally, dynamic systems yield non-linear 2nd-order ODE’s. That’s the good news: no partial differential equations and no third order or higher equations. Non-linear systems hide insights. The bad news is that these equations are often non-linear. That would have been especially bad news fifty years ago when computing was still in its infancy. Today, we can rewrite any ODE in such a way that it can be solved by brute-force methods using advanced computers. The only problem is it is hard to draw generalizable insights from these brute-force numerical techniques. (Phase plots are a feeble attempt to draw some insights from non-linear equations.) Local linear analysis yields rich insights. A theorem called the Hartman-Grobman The-orem (which you can forget about now that you have read this) tells us that if we linearize a non-linear equation about an equilibrium point, then insights about stability etc. that we can draw from local behavior (read small perturbations) are valid for the underlying non-linear system. This fact gives us a path forward and makes available the vast landscape of tools available for the analysis of linear systems. Linearization involves two steps. First, find all the equilibrium points. Second, intro-duce small perturbations about each equilibrium point that you are interested in and rewrite the equations of motion in terms of these perturbations, remembering along the way to strike out any quadratic terms. Solve the linear system. There is a large body of work for solving systems of linear equa-tions with constant coefficients. The way forward is easy. The Behavior of Dynamic Systems December 4, 2007 1 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].1. Regardless of whether there is a forcing function, start with the characteristic equation. • Inserting an exponential solution as a “guess” always yields a characteristic equa-tion. The roots of the characteristic equations are all you need to proceed forward. They tell you if the system is stable, marginally stable or unstable. They tell you if the system is oscillatory and if so, what the frquency is. Oh, and they tell you how quickly the system dies down (i.e., the damping factor). • If the system is free, reconstitute the exponential solution with the roots in place in the exponent and solve for the initial conditions.Write out the solution of the homog-enous equation. • If the system is oscillatory, it will come out automatically in the equation because there will be imaginary parts in the exponent. 2. If the system is forced, determine the roots from the characteristic equation as before. • But then, guess a particular solution. We will limit our analysis to harmonic excita-iΩt tion of the form poe . The guess will always be of the same form with a “phase iΩt – iα lag” thrown in: x( ) t = Ue . • Add the homogeneous and solution and the particular solution, plug in the initial conditions, and you are ready to go! 3.0 Linearization Consider the inverted pendulum system we discussed in class, shown in Figure 1. The equation of motion is: 2·· 2 ml θ + ka sin θ cosθ – mgl sinθ = 0. (EQ 1) 3.1 Equilibrium points · ·· We find the equilibrium points by settingθ andθ to eq zero and solving for θ. (That is the definition of an equilibrium point.) Solving gives us the following equi-librium points: FIGURE 1. Inverted pendulum eq • θ1= nπ . In other words, when the pendulum is perfectly vertical pointing either upwards or downwards. θk a l m The Behavior of Dynamic Systems December 4, 2007 2 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].eq mgl • θ2 3= ±acos ---------. in other words, when the pendulum leans to the left or right enough , 2 ka that torque from gravity exactly counterbalances the torque from the spring. Keep in mind that we are assuming that: mgl ---------≤ 1 ; (EQ 2)2 ka otherwise the spring is too weak to ever counter gravity. In class, we guessed at the phase plot (correctly) as shown in Figure 2 mglmgl acos -------- -–acos -------- -22 ka π/2π/2 ππ ka 0 FIGURE 2. Phase plot for the spring-loaded inverted pendulum 3.2 Small Perturbations As described before, linearization is useful around equilibrium points. Simply put, for each equilibrium point, create a new perturbation variable which we will asume to be small. Rewrite the equations of motion in terms of these perturbation variables, perform-ing the following steps along the way: a)Use Taylor Series to expand any function to the first order (linear terms). b)Delete any quadratic terms that arise in terms of these perturbation variables or their derivatives. In the case of the inverted pendulum, this is straightforward. Consider any equilibrium point θeq i and introduce a perturbation variable δi such that θ = θeq i + δi . Here are some simplifications for any equilibrium point: · · θ ≅ δ (EQ 3) ·· ·· θ ≅ δ (EQ 4) sin(θeq + δ) ≅ sin θeq + δ cosθeq (EQ 5) The Behavior of Dynamic Systems December 4, 2007 3 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].cos(θeq + δ) = cosθeq –δ sinθeq (EQ 6) When θeq = 0 , we can see that sin(0 + δ) ≅ sin 0 + δ cos0 ≅ δ (EQ 7) and cos(θeq + δ) = cos 0–δ sin0 ≅ 1 . (EQ 8)