Solution for Assignment 4

Transcription

1 SE3XO3/CS4X3 11F4.1 Solution for Assignment 4 Due. Dec. 1, Thursday, 13:3. 1. (4 marks) The integral defining the error function erf(x) = 2 x e t2 dt π is fairly easy to evaluate numerically. Write a MATLAB/Octave program which uses QUADS to print a table of erf(x) for x =.,.1,...,1.9,2.. Compare your table with MATLAB/Octave function erf. Since the calculation of each entry in your table is done independently of the other entries, this is not a particularly efficient way to generate such a table. Solution: The error is about 1 7 and running time about.12 seconds. x = linspace(,2,21); y = zeros(21,1); t = cputime; s = 2/sqrt(pi); % scaling factor F = inline( exp(-t.*t) ); for k=2:length(x) [y(k),ek] = QUADS(F,., x(k),.1); end y = s*y; tq = cputime - t; % print and compare y - erf(x ), tq, 2. (4 marks) The error function is usually defined by an integral erf(x) = 2 x e t2 dt, π but it can also be defined as the solution to the differential equation y (x) = 2 π e x2, y() =. Write a MATLAB/Octave program which uses ode45/lsode to print a table of erf(x) for x =.,.1,.2,...,1.9, 2.. Compare your table with the one produced by the quadrature method. Compare their execution times. Solution: The error is about 1 7 and running time about.3 seconds.

2 SE3XO3/CS4X3 11F4.2 xspan = linspace(,2,21); %.,.1,.2,..., 1.9, 2. y =.; % initial value t = cputime; s = 2/sqrt(pi); F = inline( exp(-t*t), t, y ); [xout, yout] = ode45(f, xspan, y); yout = (2/sqrt(pi))*yout; td = cputime - t; % print yout - erf(xspan ), td, 3. (8 marks) Write a MATLAB/Octave function for an adaptive quadrature using the Rectangle rule: [numi, esterr] = QUADR(fname, a, b, tol, maxlev) by modifying QUADS, which uses the Simpson s rule. Note that In the Rectangle rule, when the number of panels is doubled, the accuracy is quadrupled. Suppose that R 1 is the one panel result using the Rectangle rule, R 2 is the two panel result, and I is the exact result, then I R 1 4(I R 2 ). It then follows that I R 2 (R 2 R 1 )/3, which gives an error estimation for R 2. Similar to QUADS, the function values computed at one recursion level can be passed to the next level for reuse. Thus QUADR is a wrapper function to hide the implementation. Apply your QUADR to compute Si(2), where Si(z) is the sine integral, Si(z) = What would happen if QUADS were applied? Solution: Si(2) z sin(x) x. If QUADS were applied, the integrand sin(x)/x would be evaluated at, resulting NaN. 4. A vehicle of mass M, suspended by a lumped spring-dashpot system as shown, travels with constant horizontal velocity. At time t = the vehicle is traveling with its center of gravity on the ground and with no vertical velocity. For subsequent times, the vertical displacement of the road above the ground is given by a function x (t). Suppose the spring is linear with a constant of proportionality k and that the damping coefficient of the dashpot r is a nonlinear function of the relative velocities of the two ends of the dashpot: r = r (1 + c dt ) dt. It is easily shown that the displacement of the center of gravity of the vehicle x(t) is the solution of the second-order ordinary differential equation ( M d2 x dt 2 = k(x x ) r dt ) dt

3 SE3XO3/CS4X3 11F4.3 with the initial conditions x() =, dt =. t= Write a MATLAB/Octave program calling ode45/lsode to calculate x(t), t t max for a road contour given by x (t) = A(1 cos ωt), where 2A is the maximum road displacement above the ground. Note that for the linear case (c = ) the underdamped, critically damped, and overdampted cases correspond to r ξ = 2 km, being less than, equal to, and greater than unity, respectively. The program should read values for M, r, c, k, A, w, and t max, where w corresponds to ω, and plot x, /dt, d 2 x/dt 2, and ξ. Suggested test data: Investigate values of M = 1, A = 2, w = 7, t max = 5, k = 64. r c Solution: Plot for r = 24 and c = 1:

4 SE3XO3/CS4X3 11F4.4 % global parameters global R % linear damping coefficient global C % nonlinear damping coefficient R = input( R = ); C = input( C = ); odetol =.1; % tolerance in ode45 % global constants global M % mass global A % 2A is max road displacement global W % omega global TMAX % final time global K % spring proportionality M = 1; A = 2; W = 7; TMAX = 5; K = 64; % solve the differential equation tspan = [; TMAX]; % time span yinit = [;]; % initial values [t,x] = ode45(@msqn, tspan, yinit, odetol); % plot displacement subplot(221), plot(t,x(:,1)) % plot road contour x = A*(ones(size(t)) - cos(w*t)); subplot(222), plot(t,x) % plot acceleration x = A*(1. - cos(w*t)); = A*W*sin(W*t); r = R*(ones(size(t)) + C*abs(x(:,2) - )); % damping coefficient a = -(K*(x(:,1) - x) + r.*(x(:,2) - ))/M; subplot(223), plot(t,a) % plot xi xi = r/(2*sqrt(k*m)); subplot(224), plot(t,xi) %%%%%%%%%%%%%% % diff. eqn. % %%%%%%%%%%%%%% function yp = msqn(t,y) % yp = msqn(t,y) % mass-spring differential equation function % % y(1) = x % y(2) = % % [y(1)] = [yp(1)] % [y(2)] [yp(2)] global R global C global M

5 SE3XO3/CS4X3 11F4.5 global A global W global K yp = zeros(2,1); yp(1) = y(2); x = A*(1. - cos(w*t)); = A*W*sin(W*t); R = R*(1. + C*abs(y(2) - )); yp(2) = -(K*(y(1) - x) + R*(y(2) - ))/M;