The “Quarter-Car” Model.doc

The “Quarter-Car” Model Equations of Motion Input-Output Equations State Equations Output Equations The figure to the right shows the diagram of the quarter-car model. Note that I’ve changed the nomenclature slightly, using “u(t)” as the input since the previous symbol (y(t)) is often used to represent output variables. Equations of Motion: In class, we went through the process of drawing free body diagrams and applying Newton’s Second Law to arrive at the differential equations. I won’t reproduce those steps here. I will also assume that our reference positions for the displacements are the static equilibrium point, so we can safely drop the weights out of the equations, leaving the equations in this form: Input/Output Equations: It’s important to realize that before we can proceed to put these equations in the form of an input/output model, we need to identify the output. In this situation, there are two physical variables that are obvious choices, xs and xu. We will derive two different input/output models, one for each output. Keep in mind that other models are possible. For example, suppose you wanted to know how the force in the suspension spring behaved in response to the road input. You would derive a separate model for that as well. To combine these two equations of motion to a single input/output model, we have to manipulate the equations to eliminate the unwanted variable. We will use the “D-operator” method we discussed in class. Recall the “D-operator” is used to represent differentiation with respect to time. Putting both equations in “D-operator” form gives us: The first model we’ll derive is the model for the sprung mass. We’ll do this my eliminating the unneeded variable (xu). Therefore, well solve the bottom equation for xu, as shown: Substitute this into the top equation: Now its a matter of cranking out the algebra to get rid of the denominators. The quickest approach is to multiply both sides of the equation with the denominator