System equations

The second derivative of the input angle actually affects the second derivative of . However, we will ignore this contribution. The Lagrangian equation of motion for the ball is then given by the following:

(1)

Linearization of this equation about the beam angle, , gives us the following linear approximation of the system:

(2)

The equation which relates the beam angle to the angle of the gear can be approximated as linear by the equation below:

(3)

Substituting this into the previous equation, we get:

(4)

Transfer Function

Taking the Laplace transform of the equation above, the following equation is found:

(5)

Rearranging we find the transfer function from the gear angle () to the ball position ().

(6)

It should be noted that the above plant transfer function is a double integrator. As such it is marginally stable and will provide a challenging control problem.

The transfer function can be implemented in MATLAB as follows:

m = 0.111;

R = 0.015;

g = -9.8;

L = 1.0;

d = 0.03;

J = 9.99e-6;

s = tf('s');

P_ball = -m*g*d/L/(J/R^2+m)/s^2

P_ball =

0.21

----

s^2

State-Space

The linearized system equations can also be represented in state-space form. This can be done by selecting the ball's position () and velocity () as the state variable and the gear angle () as the input. The state-space representation is shown below:

(7)

However, for our state-space example we will be using a slightly different model. The same equation for the ball still applies but instead of controlling the position through the gear angle, , we will control the torque of the beam. Below is the representation of this system:

(8)

(9)

Note: For this system the gear and lever arm would not be used, instead a motor at the center of the beam will apply torque to the beam, to control the ball's position.

The state-space equations can be represented in MATLAB with the following commands (these equations are for the torque control model).

Train system

In this example, we will consider a toy train consisting of an engine and a car. Assuming that the train only travels in one direction, we want to apply control to the train so that it has a smooth start-up and stop, along with a constant-speed ride.

The mass of the engine and the car will be represented by M1 and M2, respectively. The two are held together by a spring, which has the stiffness coefficient of k. F represents the force applied by the engine, and the Greek letter, mu (which will also be represented by the letter u), represents the coefficient of rolling friction.

Photo courtesy: Dr. Howard Blackburn

Free body diagram and Newton's law

The system can be represented by following Free Body Diagrams.

From Newton's law, you know that the sum of forces acting on a mass equals the mass times its acceleration. In this case, the forces acting on M1 are the spring, the friction and the force applied by the engine. The forces acting on M2 are the spring and the friction. In the vertical direction, the gravitational force is canceled by the normal force applied by the ground, so that there will be no acceleration in the vertical direction. The equations of motion in the horizontal direction are the followings:

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