MECH : a Primer for Matlab s ode suite of functions

Transcription

1 Objectives MECH 4-563: a Primer for Matlab s ode suite of functions. Review the fundamentals of initial value problems and why numerical integration methods are needed.. Introduce the ode suite of numerical integrators that are built into MATLAB, in particular ode Learn how to transform a dynamic MDOF (multi-degree of freedom) system, involving acceleration terms, into a first order state space form suitable for any of the MATLAB ode commands. 4. Execute an ode based numerical integration in Matlab for a simple multi-body system. Part I A (very) short introduction to initial value problems. Initial value problems were introduced in MATH and CSC 349A. Consider a single ordinary differential equation in terms of time: y y t t y y y y () ; for subject to () and () The domain of the problem is not necessarily defined (t could grow to any value), but the initial value of the function y (otherwise known as the initial condition) is, and must be, provided. Referring back to MATH, you could apply the annihilator method to get an exact solution a function yt () that satisfies eq. () for any value of t. However, what if the problem was: y y ln(4t )+ ; for t subject to y() y.563 () ( 4 y) Anyone interested in finding an exact solution y(t) to eq. ()? To solve eq. (), one usually turns to a numerical integrator. Numerical integrators are not designed to calculate the value of a definite integral (they don t integrate functions between fixed bounds to find the area under the function over the interval). Rather, a numerical integrator:. recognizes the current value of time and the function y (beginning with the first given value y at time t ). applies some knowledge about how y is changing (by calculating the current value of y and y ) 3. uses the current y, y and y information to make an intelligent guess at what the value of y will be one time step ( t ) later at t t t. To execute step, numerical integrators have to be provided with a function that defines the rates of change of the function y. This function must be in terms of y and t. Obviously, this function can be constructed from eq. (), but this construction is complicated by the fact that most numerical integrators

2 (including the entire suite of MATLAB ode integrators) are designed for first order differential equations. As such, the y that appears at the LHS of eq. () is a little problematic. Note that as a mechanical engineer, most of the ODEs you will encounter will be second order differential equations (blame it on Newton). So knowing how to convert nd order ODEs into an equivalent first order form is pretty valuable. Part II Converting nd order (dynamic) systems to first order form. We rewrite the ODE in eq. () to get an explicit expression for y : y y t ln(4 )+ ( 4 ).563 y Now we get creative and introduce two new variable names, q and q : q y q q y Now, we are trying to find how q and q vary with time. Obviously these two functions of time are coupled since one is just the differential of the other. We will enforce this relationship in the next step The first order rate of change of the state vector q is given by: q y q q ln(4t ) ( 4 q) h(, t) y q q (3) q The bottom equation of eq. (3) is a nearly trivial equation, that simply enforces the fact that the second variable q is an indefinite integral of the first variable q. The final form of the problem is now a pair of initial value problems that are first order. Simply: q h (, t) q (4) A consequence of adding a second variable is that there now must be two initial conditions: q () y q() q () y (5) where y and y are the initial velocity and position values respectively. Together, equations (4) and (5) form a complete first order numerical integration problem.

3 Part III - Solving the First Order Initial Value Problem Create a Matlab function called derivs (a common name that is short for derivatives) in a file derivs.m The function derivs has to return the x matrix h on the RHS of eq. (4) given the current values of time t and the state variables q and q. The derivs function should appear as: function [q_dot] = derivs(t,q) % return the time rate of change of q and q according to eq. (3) of % the Asg #5 Part I Computational Help File. % q is the current value of the variable q (velocity) % q is the current value of the variable q (position) % the top entry in q_dot must be the time rate of change of q. % the bottom entry in q_dot must be the time rate of change of q. q = q(,); q = q(,); q_dot = zeros(,); % clear any possible values in q_dot. q_dot(,) = -*q+(-sqrt(*log(4*t^+))+/(+4*q)^.563); q_dot(,) = q; % done. Note that the ode45 integrator has very specific requirements for the order of the input arguments: the first argument must be the independent variable (in this case time t) and the second variable must be the state vector (in this case the vector q ). Also, note that the convention in Matlab is to make state vectors as columns, NOT rows. The integration can be carried out until any value of t you desire at least 7 s is recommended. Longer durations mean longer run times. If you go through the help menus, you will find that there are several integrators for ordinary differential equations. All are named with an ode prefix. We will use ode45 here. If you can locate the description of ode45 in the Matlab help you will find that ode45 is an explicit single step technique that has an adaptive time step it is a Runge Kutta 4 th /5 th order interpolation method that belongs to the family discussed previously in your BEng program in CSC 349a. The time step is adjusted within ode45 to minimize the estimated truncation error. From the command prompt try: 3

4 options = odeset( RelTol,e 4, MaxStep,.) [t_out, q_out] = ode45(@derivs,[:.:7.],[;],options) You can adjust the second argument to ode45 to control the resolution of the intermediate output to the t_out and q_out variables. The third input is the vector of the initial values of q. Note that the dimension of the initial conditions (here a x array) must match the dimension of q and q used in derivs. Again, common practice is to make states vectors into column arrays. Note that if you use a smaller output interval you get a larger record of q and q values which helps with making clearer plots. However, at some point you will slow the system down slightly for more difficult problems. After you have computed the entries in t_out and q_out, above plot the results. At the command prompt you can use the following commands (or contain them in a Matlab script that you can save). plot(t_out,q_out(:,),'k-','linewidth',.); grid on; xlabel('time (s)'); ylabel('position y'); You should see something like:..5 Position y Time (s) 4

5 Part IV Setting up Problem 6.6 for Solution with ode45. The only significant difference between the sample problem above and Problem 6.5 is that the external force F(t) is a piecewise continuous function of time. Your derivs function for problem 6.6 will need to use a if-then-else or case-switch statement to properly complete the evaluation of the acceleration term q, x where x is the horizontal position of the cart mass. Be sure for your completion of assignment problem 6.6 that you:. use odeset to enforce a RelTol value of e-4, and a MaxStep value of... plot the position of the cart for s t.5 s. 3. Overlay your three calculated cart positions (at t.5 s, t. s, and t.5 s ) on top of your plot of ode45 results. 5