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MIT2_017JF09_p38

# MIT2_017JF09_p38 - 38 MONTE CARLO AND GRID-BASED TECHNIQUES...

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38 MONTE CARLO AND GRID-BASED TECHNIQUES FOR STOCHASTIC SIMULATION 145 38 Monte Carlo and Grid-Based Techniques for Stochas­ tic Simulation In this problem you will compare the performance of random vs. regular sampling on a specific stochastic dynamics problem. The system we are considering is a simple rotary mass, controlled by a motor: ¨ = τ = k t i, where J is the mass moment of inertia, φ is its angular position, τ is the control torque, k t is the torque constant of the motor, and i is the electrical current applied. While this is a simple control design problem for given values of J and k t , the situation we study here is when these are each only known within a range of values. In particular, J is described as a uniform random variable in the range [5 , 15] kg · m 2 , and k t is a uniform random variable in the range [4 , 6] Nm/A . The basic question we ask is: if the control system is designed for a nominal condition, say J = 10 kg · m 2 and k t = 5 Nm/A , how will the closed-loop system vary in its response, for all the possible J and k t ? This is a question of stochastic simulation, that is, finding the statistics of a function output, given the statistics of its input. The code fragment provided below applies Monte Carlo and grid-based approaches to find the mean and variance of the function cos( y ), when y is uniformly distributed in the range [2 , 5]. Try running this a few times and notice the effects of changing N . The grid-based approach is clearly giving a good result with far less work than MC - for this example with only one random dimension. In general, the grid-based methods suffer greatly as the d dimension increases; for trapezoidal integration, the error goes as 1 /N 2 /d , whereas for Monte Carlo it is simply 1 /N 1 / 2 for any d ! 1. For the nominal system model (as above) design a proportional-derivative controller so that the closed-loop step response reaches the commanded angle for the first time in about one second and the maximum overshoot is twenty percent. The closed-loop system equation is ¨ = k t ( k p ( φ φ desired ) k d φ ˙ ) −→ ¨ + k t k d φ ˙ + k t k p φ = k t k p φ desired . Remember that if you write the left-hand side of the equation as φ ¨ +2 ζω n + ω n 2 , you can tune this up quite easily because the overshoot scales directly with damping ratio ζ , and you can then adjust ω n to get the right rise time. Show a plot of the step response and list your two gains k p and k d .

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