3 Modeling and Simulation

Transcription

1 ME 132, Spring 2005, UC Berkeley, A Packard 13 3 Modeling and Simulation 31 Systems of 1st order, coupled differential equations Consider an input/output system, with m inputs (denoted by d), q outputs (denoted by e), and governed by a set of n, 1st order, coupled differential equations, of the form ẋ 1 (t) ẋ 2 (t) ẋ n (t) e 1 (t) e 2 (t) e q (t) = f 1 (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) f 2 (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) f n (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) h 1 (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) h 2 (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) h q (x 1 (t), x 2 (t),, x n (t), d 1 (t), d 2 (t),, d m (t), t) where the functions f i, h i are given functions of the n variables x 1, x 2,, x n, the m variables d 1, d 2,, d m and also explicit functions of t For shorthand, we write (79) as (8) ẋ(t) = f (x(t), d(t), t) e(t) = h (x(t), d(t), t) (9) Given an initial condition vector x 0, and a forcing function d(t) for t t 0, we wish to solve for the solutions x 1 (t) e 1 (t) x 2 (t) e 2 (t) x(t) = x n (t), e(t) = on the interval [t 0, t F ], given the initial condition and the input forcing function d( ) x (t 0 ) = x 0 e q (t) simulink solves for this using numerical integration techniques, such as 4th and 5th order Runge-Kutta formulae You can learn more about numerical integration by taking Math 128 We will not discuss this important topic in detail in this class However, to understand in a very crude manner how ODE solvers work, consider the Euler method of solution If the functions f and u are reasonably well behaved in x and t, then the solution exists, is continuous, and differentiable at all points Hence, it is reasonable that a Taylor series for x at a given time t will be predictive of the values of x around t

2 ME 132, Spring 2005, UC Berkeley, A Packard 14 If we do a Taylor s expansion on a function x, and ignore the higher order terms, we get an approximation formula x(t + δ) x(t) + δẋ(t) = x(t) + δf(x(t), d(t), t) Roughly, the smaller δ is, the closer the left-hand-side is to the actual value of x(t + δ) Euler s method propogates a solution to (79) by using this approximation repeatedly for a fixed δ, called the stepsize Hence, Euler s method gives that for any integer k >= 0, the solution to (79) approximately satisfies x((k + 1)δ) = x(kδ) + δ f(x(kδ), d(kδ), kδ) Writing out the first 4 time steps (ie, t = 0, δ, 2δ, 3δ, 4δ) gives x(δ) = x(0) + δf (x(0), d(0), 0) x(2δ) = x(δ) + δf (x(δ), d(δ), δ) x(3δ) = x(2δ) + δf (x(2δ), d(2δ), 2δ) x(4δ) = x(3δ) + δf (x(3δ), d(3δ), 3δ) (10) and so on So, as long as you have a subroutine that can evaluate f(x, d, t), given x and t, you can quickly propogate an approximate solution simply by calling the subroutine once for every timestep Computing the output, e(t) simply involves evaluating the function h(x(t), d(t), t) at the solution points In the Runge-Kutta method, a more sophisticated approximation is made, which results in more computations (4 function evaluations of f for every time step), but much greater accuracy In effect, more terms of the Taylor series are used, involving matrices of partial derivatives, and even their derivatives, df dx, d2 f dx 2, d3 f dx 3 but without actually requiring explicit knowledge of these derivatives of the function f 32 Remarks about Integration Options in simulink The Simulation SimulationParameters Solver page is used to set additional optional properties, including integration step-size options, and may need to be used to obtain smooth plots Additional options are

3 ME 132, Spring 2005, UC Berkeley, A Packard 15 Term RelTol AbsTol MaxStep Meaning Relative error tolerance, default 1e-3, probably leave it alone, though if instructions below don t work, try making it a bit smaller Absolute error tolerance, default 1e-6, probably leave it alone, though if instructions below don t work, try making it a bit smaller maximum step size I believe the default is (StopTime- StartTime)/50 In general, make it smaller than (StopTime-StartTime)/50 if your plots are jagged 33 Problems 1 A sealed-box loudspeaker is shown below From Amplifier Pole Piece Voice Coil Vent Magnet Back Plate Top Plate Spider Voice Coil Former Basket Subwoofer Electrical Leads Dust Cap Woofer Cone Surround Signal From Accelerometer Sealed Enclosure Ignore the wire marked accelerometer mounted on cone We will develop a model for accoustic radiation from a sealed-box loudspeaker The equations are Speaker Cone: force balance, Voice-coil motor (a DC motor): Magnetic flux/length Factor: m z(t) = F vc (t) + F k (t) + F d (t) + F b + F env (t) V in (t) LI(t) RI(t) B l (t)ż(t) = 0 F vc (t) = B l (t)i(t) B l (t) = BL BL 1 z(t) 4

4 ME 132, Spring 2005, UC Berkeley, A Packard 16 Suspension/Surround: Sealed Enclosure: F k (t) = K 0 z(t) K 1 z 2 (t) K 2 z 3 (t) F d (t) = R S ż(t) F b (t) = P 0 A Environment (Baffled Half-Space) where x(t) is 6 1, and A e := ( 1 + Az(t) V 0 ) γ ẋ(t) = A e x(t) + B e ż(t) F env (t) = AP 0 C e x(t) D e ż(t) B e := C e := [ ] D e = 4662 This is an approximate model of the impedance seen by the face of the loudspeaker as it radiates into an infinite half-space The values for the various constants are

5 ME 132, Spring 2005, UC Berkeley, A Packard 17 Symbol Value A meters 2 V meters 3 P Pa m 0117 kg L H R 3 Ω BL Tesla meters BL meters 4 K N/meter K 1 0 K N/meter 3 R S 128 N sec/meter (a) Build a Simulink model for the system Use the Subsystem grouping capability to manage the complexity of the diagram Each of the equations above should represent a different subsystem Make sure you think through the question what is the input(s) and what is the output(s)? for each subsystem (b) Write a function that has two input arguments (denoted V and Ω) and two output arguments, z max,pos and z max,neg The functional relationship is defined as follows: Suppose V in (t) = V sin Ω 2πt, where t is in seconds, and V in volts In other words, the input is a sin-wave at a frequency of Ω Hertz Simulate the loudspeaker behavior for about 25/Ω seconds The displacement z of the cone will become nearly sinusoidal by the end of the simulation Let z max,pos and z max,neg be the maximum positive and negative values of the displacement in the last full cycle (ie, the last 1/Ω seconds of the simulation, which we will approximately decree as the steady-state response to the input ) (c) Using bisection, determine the value (approximately) of V so that the steadystate maximum (positive) excursion of z is 0007meters if the frequency of the excitation is Ω = 25 What is the steady-state minimum (negative) excursion of z?