Our Ferris Wheel system is a second order system governed by the transfer function in the form, G(s) = a/(s^2+bs+c) = Kw_n²/(s²+2zw_ns+w_n²):  

Q(s)/T(s) = 1/[Is² + bs + k] 

where I = 1500000, b = 200000, and k = 4000000, so

a = 1/J = 1/1500000 = 6.667 E-07

b = b/J = 200000/1500000 = 0.1333

c = k/J = 4000000/1500000 = 2.667

For the sake of simplicity the following example will use a new system name, second_sys, which is interchangable with both the equivalent transfer function system, second_tf, and the state space system, second_ss, defined previously.

second_sys = second_tf = second_ss

 

NATURAL FREQUENCY

The natural frequency is w_n= sqrt(c).  

>> wn = sqrt(2.667)

wn =

    1.6331

 

DAMPING RATIO

There are two methods for determining the damping ratio using MATLAB.  The first is direct calculation from the relation z = b/(2sqrt(c)).

>> zeta = b/(2*sqrt(c))

zeta =

    0.0408

Another approach is to use MATLAB's damp command.

>> damp(second_sys)

The MATLAB output will be:

       Eigenvalue          Damping  Freq. (rad/s) 

-6.67e-002 + 1.63e+000i   4.08e-002   1.63e+000 
-6.67e-002 - 1.63e+000i   4.08e-002   1.63e+000 

The second column is the damping ratio of the two poles of the system.

 

POLES

There are two methods to calculated the poles.  The first is direct calculation from the formula.  A second order system has two poles at -zw_n ± w_n sqrt(z²-1).

Each pole must be calculated separately:

>> p1 = -zeta * wn + wn * sqrt(zeta^2 - 1)

p1 =

   -0.0667 + 1.6317i

The second pole is:

>> p2 = -zeta * wn - wn * sqrt(zeta^2 - 1)

p2 =

-0.0667 - 1.6317i

 

You can also determine the poles using MATLAB's pole function.  

>> pole(second_sys)

The MATLAB output will be:

ans =

-0.0667 + 1.6316i
-0.0667 - 1.6316i

This second order system has two complex poles with negative real parts, so this system is underdamped and stable.

 

ZEROS

The zero of a system is when the numerator of the transfer function equals zero, which makes the value of the transfer function zero.  By inspection the numerator of this example problem equals 1, so no value of s will yield a numerator equaling zero.

To find the zero(s) of the system using MATLAB's tzero function.

>> tzero(second_sys)

ans =

   Empty matrix: 0-by-1

This system has no zero, meaning that no value of s will make the numerator equal zero.

 

SETTLING TIME

The settling time is the time it takes to fall within a certain percentage of the steady state value for a step input or equivalently to decrease to a certain percentage of the initial value for an impulse input.  The settling times for this second order system are:  

10%	5%	2%
2.3 / zwn	3 / zwn	4 / zwn
>> Ts10 = 2.3 / (zeta * wn) Ts10 =     34.5086	>> Ts5 = 3 / (zeta * wn) Ts5 =    45.0113	>> Ts2 = 4 / (zeta * wn) Ts2 =    60.0150

The oscillations of this Ferris Wheel will take about 35 seconds to decrease to 10% of their largest magnitude after stopping.

Carnegie Mellon University | University of Michigan

Mechanical Engineering Department