Consider the following differential equation x 025X yy 01

Consider the following differential equation: x + 0.25X + yy + 0.1 y = u, X(0) = 0, x(0) = 1 y = 0.5 + yx. y(0) = 1, y(0) = 1 where u is a unit step input. Develop the update equation in terms of dt based on the forward difference method Develop a complete Matlab script that plots the numerical solution x(f) and y(r) for the equation as a function of time. Consider dt = 0.01 second.

Solution

A momentum balance around the x-axis gives J d2 dt2 = mnl sin + Fl cos (4.8) The force F has two components, a centripetal force and an inertia force due to the acceleration of the coordinate system. The force can be determined from kinematic relations, see Figure 4.5. To describe these we introduce the steering angle , and the forward velocity V0. Furthermore the distance between the contact point of the front and rear wheel is b and the distance between the contact point of the rear wheel and the projection of the center of mass of bicycle and rider is a. To simplify the equations it is assumed that the angles and are so small that sines and tangent are equal to the angle and cosine is equal to the one. Viewed from the top as shown in Figure 4.5 the bicycle has its center of rotation at a distance b/ from the rear wheel. The centripetal force is Fc = mV2 0 b The y-component of the velocity of the center of mass is Vy = V0 = aV0 b where a is the distance from the contact point of the back wheel to the projection of the center of mass. The inertial force due to the acceleration of the coordinate system is thus Fi = amV0 b d dt Inserting the total force F = Fc + Fi into (4.8) we find that the bicycle can be described by J d2 dt2 = mnl + amV0l b d dt + mV2 0 l b (4.9) This equation has the characteristic equation Js2 mnl = 0 which has the roots s = ±rmnl J The system is unstable, because the characteristic equation has one root in the right half plane. We may therefore believe that the rider must actively stabilize the bicycle all the time.