2000
Solutions Manual: Chapter 2
7th Edition
Feedback Control of Dynamic
Systems
Gene F. Franklin
.J. David Powell
.Abbas Emami-Naeini
Assisted by:
H. K. Aghajan
H. Al-Rahmani
P. Coulot
P. Dankoski
S. Everett
R. Fuller
T. Iwata
V. Jones
F. Safai
L. Kobayashi
H-T. Lee
E. Thuriyasena
M. Matsuoka
Chapter 2
Dynamic Models
Problems and Solutions for Section 2.1
1. Write the differential equations for the mechanical systems shown in Fig. 2.41.
For (a) and (b), state whether you think the system will eventually de-
cay so that it has no motion at all, given that there are non-zero initial
conditions for both masses, and give a reason for your answer.
Fig. 2.41 Mechanical systems
Solution:
2001
2002 CHAPTER 2. DYNAMIC MODELS
(a)
Free body diagram for Problem 2.1(a)
m1x1=k1x1b1_x1k2(x1x2)
There is friction affecting the motion of mass 1 which will continue
to take energy out of the system as long as there is any movement of
x1x2
x2
Again, there is friction on mass 2 so there will continue to be a loss
Figure 2.42: Mechanical system for Problem 2.2
2. Write the differential equations for the mechanical systems shown in Fig. 2.42.
State whether you think the system will eventually decay so that it has
no motion at all, given that there are non-zero initial conditions for both
masses, and give a reason for your answer.
Solution:
The key is to draw the Free Body Diagram (FBD) in order to keep the
signs right. To identify the direction of the spring forces on the left side
x1x2
2004 CHAPTER 2. DYNAMIC MODELS
Then the forces are summed on each mass, resulting in
3. Write the equations of motion for the double-pendulum system shown in
Fig. 2.43. Assume the displacement angles of the pendulums are small
enough to ensure that the spring is always horizontal. The pendulum
rods are taken to be massless, of length l, and the springs are attached
3/4 of the way down.
Figure 2.43: Double pendulum
Solution:
2005
3
If we write the moment equilibrium about the pivot point of the left pen-
dulem from the free body diagram,
Or
2006 CHAPTER 2. DYNAMIC MODELS
4. Write the equations of motion of a pendulum consisting of a thin, 2-kg
stick of length lsuspended from a pivot. How long should the rod be in
order for the period to be exactly 1 sec? (The inertia Iof a thin stick
about an endpoint is 1
3ml2. Assume is small enough that sin
=.)
Solution:
Let’s use Eq. (2.14)
Moment about point O.
2007
As we assumed is small,
The frequency only depends on the length of the rod
<Side notes>
5. For the car suspension discussed in Example 2.2, plot the position of the
car and the wheel after the car hits a “unit bump” (i.e., ris a unit step)
using Matlab. Assume that m1= 10 kg,m2= 250 kg,kw= 500;000 N=m,
ks= 10;000 N=m. Find the value of bthat you would prefer if you were
a passenger in the car.
Solution:
The transfer function of the suspension was given in the example in Eq.
(2.12) to be:
2008 CHAPTER 2. DYNAMIC MODELS
(a)
Y(s)
kwb
m1m2(s+ks
b)
% Problem 2.5
6. Write the equations of motion for a body of mass Msuspended from a
fixed point by a spring with a constant k. Carefully define where the
body’s displacement is zero.
Solution:
Some care needs to be taken when the spring is suspended vertically in
2010 CHAPTER 2. DYNAMIC MODELS
The equilibrium value of xincluding the effect of gravity is at x=m
kg
and yrepresents the motion of the mass about that equilibrium point.
7. Automobile manufacturers are contemplating building active suspension
systems. The simplest change is to make shock absorbers with a change-
able damping, b(u1):It is also possible to make a device to be placed in
parallel with the springs that has the ability to supply an equal force, u2;
in opposite directions on the wheel axle and the car body.
2011
(a) Modify the equations of motion in Example 2.2 to include such con-
trol inputs.
(b) Is the resulting system linear?
(c) Is it possible to use the forcer, u2;to completely replace the springs
and shock absorber? Is this a good idea?
Solution:
(a) The FBD shows the addition of the variable force, u2;and shows b
as in the FBD of Fig. 2.5, however, here bis a function of the control
variable, u1:The forces below are drawn in the direction that would
result from a positive displacement of x.
(b) The system is linear with respect to u2because it is additive. But
(c) It is technically possible. However, it would take very high forces
8. In many mechanical positioning systems there is ‡exibility between one
part of the system and another. An example is shown in Figure 2.6
2012 CHAPTER 2. DYNAMIC MODELS
where there is ‡exibility of the solar panels. Figure 2.44 depicts such a
situation, where a force uis applied to the mass Mand another mass
mis connected to it. The coupling between the objects is often modeled
by a spring constant kwith a damping coefficient b, although the actual
situation is usually much more complicated than this.
(a) Write the equations of motion governing this system.
(b) Find the transfer function between the control input, u; and the
output, y:
Figure 2.44: Schematic of a system with ‡exibility
Solution:
(a) The FBD for the system is
or
2013
(b) If we make Laplace Transform of the equations of motion
From Cramer’s Rule,
Finally,
9. Modify the equation of motion for the cruise control in Example 2.1,
Eq(2.4), so that it has a control law; that is, let
u=K(vrv);
where
vr= reference speed
K= constant:
This is a ‘proportional’ control law where the difference between vrand
the actual speed is used as a signal to speed the engine up or slow it down.
Put the equations in the standard state-variable form with vras the input
and vas the state. Assume that m= 1500 kg and b= 70 N s=m;and
2014 CHAPTER 2. DYNAMIC MODELS
find the response for a unit step in vrusing MATLAB. Using trial and
error, find a value of K that you think would result in a control system in
which the actual speed converges as quickly as possible to the reference
speed with no objectional behavior.
Solution:
Rearranging, yields the closed-loop system equations,
2015
We can see that the larger the Kis, the better the performance, with no
2016 CHAPTER 2. DYNAMIC MODELS
10. Determine the dynamic equations for lateral motion of the robot in Fig.
2.45. Assume it has 3 wheels with the front a single, steerable wheel
where you have direct control of the rate of change of the steering angle,
Usteer, with geometry as shown in Fig. 2.46. Assume the robot is going in
approximately a straight line and its angular deviation from that straight
line is very small Also assume that the robot is traveling at a constant
speed, Vo. The dynamic equations relating the lateral velocity of the
center of the robot as a result of commands in Usteer is desired. .
Solution:
This is primarily a problem in kinematics. First, we know that the control
input, Usteer;is the time rate of change of the steering wheel angle, so
2018 CHAPTER 2. DYNAMIC MODELS
Figure 2.46: Model for robot motion
The actual change in the carts lateral position will then be proportional
2019
Problems and Solutions for Section 2.2
11. A first step toward a realistic model of an op amp is given by the equations
below and shown in Fig. 2.47.
Vout =107
s+ 1[V+V]
i+=i= 0
Figure 2.47: Circuit for Problem 11.
Find the transfer function of the simple amplification circuit shown using
this model.
Solution: