# The “Quarter-Car” Model

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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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