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The “Quarter-Car” Model Equations of Motion Input-Output Equations State Equations xs ms Output Equations ks bs The figure to the right shows the diagram of the quarter- xu car model. Note that I’ve changed the nomenclature mu slightly, using “u(t)” as the input since the previous symbol (y(t)) is often used to represent output variables. kt bt Equations of Motion: u(t) 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: ms s bs x s k s x s bs xu k s xu x mu u (bs bt ) xu (k s k t ) xu bs x s k s x s bt u k t u x 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: ms D 2 x s bs Dx s k s x s bs Dx u k s xu mu D 2 xu (bs bt ) Dx u (k s k t ) xu bs Dx s k s x s bt Du k t u 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 x u, as shown: bs D k s bt D k t xu xs u mu D (bs bt ) D (k s k t ) 2 mu D (bs bt ) D (k s k t ) 2 Substitute this into the top equation: bs D k s bt D k t (ms D 2 bs D k s ) x s (bs D k s ) xs u mu D (bs bt ) D (k s k t ) mu D (bs bt ) D (k s k t ) 2 2 Now it's 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 and multiply it out. The next step looks like this: m D u 2 (bs bt ) D (k s k t ) ms D 2 bs D k s x s (bs D k s )(bs D k s ) x s (bs D k s )(bt D k t )u Multiply the polynomials and simplify: {mu ms D 4 [mu bs ms (bs bt )]D 3 [mu k s bs (bs bt ) ms (k s k t )]D 2 [k s (bs bt ) bs (k s k t )]D k s (k s k t )}x s {bs2 D 2 2bs k s D k s }x s {bs bt D 2 (bs k t bt k s ) D k s k t }u Easy, right? OK, so it’s kind of messy but the end is in sight. Grouping all of the x s terms on the left hand side: {mu ms D 4 [mu bs ms (bs bt )]D 3 [mu k s bs bt ms k s ms k t )]D 2 [bt k s bs k t ]D k s k t }x s {bs bt D 2 (bs k t bt k s ) D k s k t }u Now we simply bring it back from the “D-operator” format to the traditional derivative format. Using the form of the input/output model given in the text: d 4 xs d 3 xs d 2 xs d xs d 2u d u a4 4 a3 3 a2 2 a1 a0 x s b2 2 b1 b0 u dt dt dt dt dt dt where: a 4 mu m s a3 mu bs m s bs m s bt a 2 mu k s m s k s m s k t bs bt a1 bt k s bs k t a0 k s k t b2 bs bt b1 bs k t bt k s b0 k s k t You can go through a very similar process, which I will not detail here, to arrive at the input/output model relating the input (u(t)) with the motion of the unsprung mass. The form of the model will be similar, with a significant change on the input side (a third derivative of the input): d 4 xu d 3 xu d 2 xu d xu d 3u d 2u d u a4 4 a3 3 a2 2 a1 a0 xu b3 3 b2 2 b1 b0 u dt dt dt dt dt dt dt But the coefficients will be slightly different: a 4 mu m s a3 mu bs m s bs m s bt a 2 mu k s m s k s m s k t bs bt a1 bt k s bs k t a0 k s k t b3 m s bt b2 m s k t bs bt b1 bs k t bt k s b0 k s k t IMPORTANT POINT HERE!!! Note that the coefficients of the output variable are the same in both cases (the a’s are identical). This is not a coincidence. These coefficients make up the characteristic polynomial of the system and will be the same for any output variable. This is a valuable check to make sure you’ve done your algebra correctly. State Equations: In converting the input/output models into state equations, it’s best to follow the somewhat formal method outlined in Section 3.4 of the text. We’ll begin by looking at the input/output model we derived between the sprung mass displacement (x s) and the input profile (u). Recall that the form of the model is: d 4 xs d 3 xs d 2 xs d xs d 2u d u a4 4 a3 3 a2 2 a1 a0 x s b2 2 b1 b0 u dt dt dt dt dt dt The various coefficients are defined above. Following the method outlined in the text, we write the auxiliary equation in a new time- varying variable that we’ll simply call x. This new equation is obtained by substituting our new variable for the output variable (xs in this case) and equating it to the input variable (with no derivatives). d 4x d 3x d 2x d x a4 a3 a2 a1 a0 x u dt 4 dt 3 dt 2 dt An appropriate set of state variables can be found in the following manner: q1 x q2 x q 3 x q 4 x The state equations are easily derived as follows: q1 q 2 q 2 q3 q3 q 4 a0 a a a 1 q4 q1 1 q 2 2 q3 3 q 4 u a4 a4 a4 a4 a4 Everything works out when we derive the output equations. The method can be demonstrated if we look at the original input/output equation and the auxiliary equation in “D” operator form: When we substitute the second equation into the first, eliminating u, we get: x s (b2 D 2 b1 D b0 ) x This becomes a valid output equation for the state space model when we realize that the auxiliary variable, x, and it’s first three derivatives are the state variables: x s b2 q3 b1 q 2 b0 q1 Finally, we complete this part by putting the state and output equations in matrix form and going back to substitute the physical parameters for the generic coefficients. q1 q 0 0 1 0 0 q 0 0 1 0 1 0 2 q 2 q3 0 0 0 1 q 0 u (t ) k s kt (bt k s bs k t ) (mu k s ms k s ms k t bs bt ) (mu bs ms bs ms bt ) 3 1 q 4 mu ms mu ms mu ms mu ms q 4 mu ms q1 q x s k s k t (bs k t bt k s ) bs bt 0 2 q3 q 4

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posted: | 2/26/2010 |

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