# Solution Manual for Feedback Control of Dynamic Systems, 8th Edition

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2000
Solutions Manual: Chapter 2
8th 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
Copyright (c) 2019 Pearson Education
All rights reserved. No part of this publication may be reproduced, stored
in a retrieval system, or transmitted, in any form or by any means, electronic,
mechanical, photocopying, recording, or otherwise, without the prior permission
of the publisher.
Chapter 2
Dynamic Models
Problems and Solutions for Section 2.1
1. Write the diยคerential equations for the mechanical systems shown in Fig. 2.43.
For (a) and (b), 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. Also, for part
(c), answer the question for F=0.
2001
2002
CHAPTER 2. DYNAMIC MODELS
Fig. 2.43 Mechanical systems
Solution:
The key is to draw the Free Body Diagram (FBD) in order to keep the
signs right. For (a), to identify the direction of the spring forces on the
object, let x2 = 0 and โฆxed and increase x1 from 0. Then the k1 spring
will be stretched producing its spring force to the left and the k2 spring
will be compressed producing its spring force to the left also. You can use
the same technique on the damper forces and the other mass.
x2
x1
.
b1x1
m1
k1x1
k2(x1 – x2)
k2(x1 – x2)
m2
Free body diagram for Problem 2.1(a)
(a)
m1 x
โข1
m2 x
โข2
=
=
k1 x1 b1 x_ 1
k2 (x2 x1 )
k2 (x1
k3 (x2
x2 )
y)
There is friction aยคecting the motion of mass 1 which will continue
to take energy out of the system as long as there is any movement of
x1 :Mass 2 is undamped; therefore it will tend to continue oscillating.
However, its motion will drive mass 1 through the spring; therefore,
the entire system will continue to lose energy and will eventually
decay to zero motion for both masses.
k3(x2 – y)
2003
x2
x1
m1
k1x1
k2(x1 – x2)
m2
k2(x1 – x2)
k3x2
.
b1x2
Free body diagram for Problem 2.1(b)
m1 x
โข1
m2 x
โข2
=
=
k1 x1 k2 (x1 x2 )
k2 (x2 x1 ) b1 x_ 2
k3 x2
Again, there is friction on mass 2 so there will continue to be a loss
of energy as long as there is any motion; hence the motion of both
masses will eventually decay to zero.
x2
x1
m1
k1x1
. .
b1(x1 – x2)
. .
b1(x1 – x2)
k2(x1 – x2)
k2(x1 – x2)
m2
Free body diagram for Problem 2.1 (c)
m1 x
โข1
m2 x
โข2
=
=
k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 )
F k2 (x2 x1 ) b1 (x_ 2 x_ 1 )
The situation here is similar to part (a). It is clear that the relative
motion between mass 1 and 2 would decay eventually, but as long
as mass 1 is oscillating, it will drive some relative motion of the two
masses and that will cause energy loss in the damper. So the entire
system will eventually decay to zero.
2. Write the diยคerential equations for the mechanical systems shown in Fig. 2.44.
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.
F
2004
CHAPTER 2. DYNAMIC MODELS
Fig. 2.44 Mechanical system for Problem 2.2
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
object, let x2 = 0 and increase x1 from 0. Then the k1 spring on the left
will be stretched producing its spring force to the left and the k2 spring
will be compressed producing its spring force to the left also. You can use
the same technique on the damper forces and the other mass.
x1
. .
b2(x1 – x2)
x2
. .
b2(x1 – x2)
m2
m1
k1x1
k2(x1 – x2)
Free body diagram for Probelm 2.2
Then the forces are summed on each mass, resulting in
m1 x
โข1
m2 x
โข2
=
=
k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 )
k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) k1 x2
The relative motion between x1 and x2 will decay to zero due to the
damper.
However, the two masses will continue oscillating together
without decay since there is no friction opposing that motion and โกexure
of the end springs is all that is required to maintain the oscillation of the
two masses. However, note that the two end springs have the same spring
constant and the two masses are equal If this had not been true, the two
masses would oscillate with diยคerent frequencies and the damper would
be excited thus taking energy out of the system.
3. Write the equations of motion for the double-pendulum system shown in
Fig. 2.45. Assume the displacement angles of the pendulums are small
enough to ensure that the spring is always horizontal. The pendulum
k1 x2
2005
rods are taken to be massless, of length l, and the springs are attached
3/4 of the way down.
Fig. 2.45 Double pendulum
Solution:
ฮธ1
3
l
4
ฮธ2
k
m
m
3
l sin ฮธ1
4
3
l sin ฮธ 2
4
Deโฆne coordinates
If we write the moment equilibrium about the pivot point of the left pendulem from the free body diagram,
M=
mgl sin 1
3
k l (sin 1
4
3
sin 2 ) cos 1 l = ml2 โข1
4
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CHAPTER 2. DYNAMIC MODELS
ml2 โข1 + mgl sin 1 +
9 2
kl cos 1 (sin 1
16
sin 2 ) = 0
Similary we can write the equation of motion for the right pendulem
3
mgl sin 2 + k l (sin 1
4
3
sin 2 ) cos 2 l = ml2 โข2
4
As we assumed the angles are small, we can approximate using sin 1
1, and cos 2 1. Finally the linearized equations
1 ; sin 2
2 , cos 1
of motion becomes,
9
kl ( 1
16
9
mlโข2 + mg 2 + kl ( 2
16
mlโข1 + mg 1 +
2)
=
0
1)
=
0
Or
โข1 + g 1 + 9 k ( 1
l
16 m
g
โข2 + 2 + 9 k ( 2
l
16 m
2)
=
0
1)
=
0
4. Write the equations of motion of a pendulum consisting of a thin, 2-kg
stick of length l suspended from a pivot. How long should the rod be in
order for the period to be exactly 1 sec? (The inertia I of a thin stick
about an endpoint is 13 ml2 . Assume is small enough that sin = .)
Solution:
Letโs use Eq. (2.14)
M =I ;
2007
l
2
O
ฮธ
mg
Deโฆne coordinates
and forces
Moment about point O.
MO
=
mg
=
1 2โข
ml
3
l
sin = IO โข
2
โข + 3g sin = 0
2l
As we assumed
is small,
โข + 3g = 0
2l
The frequency only depends on the length of the rod
!2 =
3g
2l
s
T
=
2
=2
!
2l
=2
3g
l
=
3g
= 0:3725 m
8 2
2008
CHAPTER 2. DYNAMIC MODELS
Grandfather clocks have a period of 2 sec, i.e., 1 sec for a swing from one
side to the other. This pendulum is shorter because the period is faster.
But if the period had been 2 sec, the pendulum length would have been
1.5 meters, and the clock itself would have been about 2 meters to house
the pendulum and the clock face.
q
2l
(a) Compare the formula for the period, T = 2
3g with the well known
formula for the period
of a point mass hanging with a string with
q
l
length l. T = 2
g.
(b) Important!
In general, Eq. (2.14) is valid only when the reference point for
the moment and the moment of inertia is the mass center of the
body. However, we also can use the formular with a reference point
other than mass center when the point of reference is โฆxed or not
accelerating, as was the case here for point O.
5. For the car suspension discussed in Example 2.2, plot the position of the
car and the wheel after the car hits a โunit bumpโ (i.e., r is a unit step)
using Matlab. Assume that m1 = 10 kg, m2 = 350 kg, kw = 500; 000 N=m,
ks = 10; 000 N=m. Find the value of b that 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:
(a)
kw b
ks
Y (s)
m1 m2 (s + b )
:
= 4
ks
ks
kw
kw ks
R(s)
s + ( mb1 + mb2 )s3 + ( m
+m
+m
)s2 + ( mk1wmb 2 )s + m
1
2
1
1 m2
This transfer function can be put directly into Matlab along with the
numerical values as shown below. Note that b is not the damping ratio, but damping. We need to โฆnd the proper order of magnitude for
b, which can be done by trial and error. What passengers feel is the
position of the car. Some general requirements for the smooth ride
will be, slow response with small overshoot and oscillation. While
the smallest overshoot is with b=5000, the jump in car position happens the fastest with this damping value.
2009
From the โฆgures, b 3000 appears to be the best compromise. There
is too much overshoot for lower values, and the system gets too fast
(and harsh) for larger values.
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CHAPTER 2. DYNAMIC MODELS
% Problem 2.5 Using state space methods
% Can also be done using the transfer function above
clear all, close all
m1 = 10;
m2 = 350;
kw = 500000;
ks = 10000;
Bd = [ 1000 3000 4000 5000];
t = 0:0.01:2;
for i = 1:4
b = Bd(i);
A=[0 1 0 0;-( ks/m1 + kw/m1 ) -b/m1 ks/m1 b/m1;
0 0 0 1; ks/m2 b/m2 -ks/m2 -b/m2 ];
B=[0; kw/m1; 0; 0 ];
C=[ 1 0 0 0; 0 0 1 0 ];
D=0;
y=step(A,B,C,D,1,t);
subplot(2,2,i);
plot( t, y(:,1), โ:โ, t, y(:,2), โ-โ);
legend(โWheelโ,โCarโ);
ttl = sprintf(โResponse with b = %4.1f โ,b );
title(ttl);
end
6. For the quadcopter shown in Figs. 2.13 and 2.14:
(a) Determine the appropriate commands to rotor #s 1, 2, 3, & 4 so a
pure vertical force will be applied to the quadcopter, that is, a force that
will have no eยคect on pitch, roll, or yaw.
(b) Determine the transfer function between Fh , and altitude, h. That is,
โฆnd h(s)=Fh (s).
Solution:
(a) To increase the lifting force, we want to increase the speed of all four
rotors. Since two rotors are rotating CW and two are rotating CCW,
there will be no net torque about the yaw axis. Since rotor #s 1 and 3
are rotating CW, we want to apply a positive torque to those two rotors,
i.e., F
T1 = T3 = +KFh ;
where Fh is the command for an increased lift for motion in the upward
vertical direction and K is the scale factor between torque and lift. Likewise, since rotor #s 2 and 4 are rotating in the CCW direction, we want
tp apply a negative torque to those two rotors, i.e.,
2011
T2 = T4 =
KFh ;
(b) Assuming the mass of the quadrotor is m, the transfer function would
be
h(s)
1
=
Fh (s)
ms2
7. Automobile manufacturers are contemplating building active suspension
systems. The simplest change is to make shock absorbers with a changeable 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.
(a) Modify the equations of motion in Example 2.2 to include such control 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 b is a function of the control
variable, u1 : The forces below are drawn in the direction that would
result from a positive displacement of x.
u2
ks(x-y)
. .
b(x-y)
x
y
m2
m1
kw(x-r)
ks(x-y)
u2
. .
b(x-y)
Free body diagram
m1 x
โข =
m2 yโข =
b (u1 ) (y_ x)
_ + ks (y x) kw (x
ks (y x) b (u1 ) (y_ x)
_ + u2
r)
u2
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CHAPTER 2. DYNAMIC MODELS
(b) The system is linear with respect to u2 because it is additive. But
b is not constant so the system is non-linear with respect to u1 because the control essentially multiplies a state element. So if we add
controllable damping, the system becomes non-linear.
(c) It is technically possible. However, it would take very high forces
and thus a lot of power and is therefore not done. It is a much better solution to modulate the damping coeยข cient by changing oriโฆce
sizes in the shock absorber and/or by changing the spring forces by
increasing or decreasing the pressure in air springs. These features
are now available on some cars… where the driver chooses between
a soft or stiยค ride.
8. In many mechanical positioning systems there is โกexibility between one
part of the system and another. An example is shown in Figure 2.7
where there is โกexibility of the solar panels. Figure 2.46 depicts such a
situation, where a force u is applied to the mass M and another mass
m is connected to it. The coupling between the objects is often modeled
by a spring constant k with a damping coeยข cientb, 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:
Fig. 2.46 Schematic of a system with
โกexibility
Solution:
(a) The FBD for the system is
2013
y
x
k(x – y)
k(x – y)
m
M
. .
b(x – y)
u
. .
b(x – y)
Free body diagrams
which results in the equations
mโข
x =
M yโข =
k (x y) b (x_ y)
_
u + k (x y) + b (x_ y)
_
or
k
b
k
b
x + x_
y
y_
m
m
m
m
b
k
b
k
x
x_ + yโข +
y+
y_
M
M
M
M
x
โข+
=
0
=
1
u
M
(b) If we make Laplace Transform of the equations of motion
s2 X +
k
X
M
k
b
k
b
X + sX
Y
sY
m
m
m
m
b
k
b
sX + s2 Y +
Y +
sY
M
M
M
=
0
=
1
U
M
=
0
U
In matrix form,
ms2 + bs + k
(bs + k)
(bs + k)
M s2 + bs + k
X
Y
From Cramerโs Rule,
ms2 + bs + k 0
(bs + k)
U
2
ms + bs + k
(bs + k)
(bs + k)
M s2 + bs + k
det
Y
=
det
=
Finally,
ms2 + bs + k
(ms2 + bs + k) (M s2 + bs + k)
2U
(bs + k)
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CHAPTER 2. DYNAMIC MODELS
Y
U
=
=
ms2 + bs + k
2
(ms2 + bs + k) (M s2 + bs + k) (bs + k)
ms2 + bs + k
4
mM s + (m + M )bs3 + (M + m)ks2
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(vr
v);
where
vr
K
=
=
reference speed
constant:
This is a โproportionalโcontrol law where the diยคerence between vr and
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 vr as the input
and v as the state. Assume that m = 1500 kg and b = 70 N s= m; and
โฆnd the response for a unit step in vr using Matlab. Using trial and error,
โฆnd 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:
v_ +
substitute in u = K (vr
b
1
v= u
m
m
v)
v_ +
b
1
K
v= u=
(vr
m
m
m
v)
Rearranging, yields the closed-loop system equations,
v_ +
K
K
b
v + v = vr
m
m
m
A block diagram of the scheme is shown below where the car dynamics
are depicted by its transfer function from Eq. 2.7.
2015
Block diagram
The transfer function of the closed-loop system is,
K
V (s)
m
=
b
Vr (s)
s+ m
+K
m
so that the inputs for Matlab are
num
=
K
m
den
=
[1
For K = 100; 500; 1000; 5000 We have,
b
K
+ ]
m m
2016
CHAPTER 2. DYNAMIC MODELS
We can see that the larger the K is, the better the performance, with no
objectionable behaviour for any of the cases. The fact that increasing K
also results in the need for higher acceleration is less obvious from the
plot but it will limit how fast K can be in the real situation because the
engine has only so much poop. Note also that the error with this scheme
gets quite large with the lower values of K. You will โฆnd out how to
eliminate this error in chapter 4 using integral control, which is contained
in all cruise control systems in use today. For this problem, a reasonable
compromise between speed of response and steady state errors would be
K = 1000; where it responds in 5 seconds and the steady state error is
5%.
2017
% Problem 2.9
clear all, close all
% data
m = 1500;
b = 70;
k = [ 100 500 1000 5000 ];
% Overlay the step response
hold on
t=0:0.2:50;
for i=1:length(k)
K=k(i);
num =K/m;
den = [1 b/m+K/m];
sys=tf(num,den);
y = step(sys,t);
plot(t,y)
end
hold oยค
2018
CHAPTER 2. DYNAMIC MODELS
10. Determine the dynamic equations for lateral motion of the robot in Fig.
2.47. 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. 2.48. 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.
Fig. 2.47 Robot for delivery of hospital supplies Source: AP Images
.
Fig. 2.48 Model for robot motion
2019
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
_ s = Usteer
When s is nonzero, the cart will be turning, so that its orientation wrt
the x axis will change at the rate
_ = Vo s :
L
as shown by the diagram below.
Diagram showing turning rate due to
s
T
The actual change in the carts lateral position will then be proportional
to according to
y_ = Vo
as shown below.
2020
CHAPTER 2. DYNAMIC MODELS
Lateral motion as a function of
These linear equations will hold providing
and s stay small enough
that sin ‘ ; and sin s ‘ s :Combining them all, we obtain,
… Vo2
y =
Usteer
L
Note that no dynamics come into play here. It was assumed that the
velocity is constant and the front wheel angle time rate of change is directly
commanded. Therefore, there was no need to invoke Eqs (2.1) or (2.10).
As you will see in future chapters, feedback control of such a system with
a triple integration is tricky and needs signiโฆcant damping in the feedback
path to achieve stability.
11. Determine the pitch, yaw, and roll control equations for the hexacopter
shown in Fig. 2.49 that are similar to those for the quadcopter given in
Eqs. (2.18) to (2.20).
Fig. 2.49 Hexacopter
2021
Assume that rotor #1 is in the direction of โกight and the remaining rotors
are numbered CW from that rotor. In other words, rotors #1 and #4
will determine the pitch motion. Rotor #s 2, 3, 5, & 6 will determine
roll motion. Pitch, roll and yaw motions are deโฆned by the coordinate
system shown in Fig. 2.14 in Example 2.5. In addition to developing the
equations for the 3 degrees of freedom in terms of how the six rotor motors
should be commanded (similar to thoseforthe quadrotorinEqs. (2.18)โ
(2.20)), it will also be necessary to decide which rotors are turning CW
and which ones are turning CCW. The direction of rotation for the rotors
needs to be selected so there is no net torque about the vertical axis; that
is, the hexicopter will have no tendancy for yaw rotation in steadystate.
Furthermore, a control action to aยคect pitch should have no eยคect on yaw
or roll. Likewise, a control action for roll should have no eยคect on pitch
or yaw, and a control action for yaw should have no eยคect on pitch or
roll. In other words, the control actions should produce no cross-coupling
between pitch, roll, and yaw just as was the case for the quadcopter in
Example 2.5.
Solution: For starters, the instructions above give us the following definitions
Hexacopter coordinate system deโฆnition
To obtain some symmetry in the rotations, letโs assign rotor #s 1, 3, & 5
to rotate CW, while rotor #s 2, 4, & 6 rotate CCW.
Now letโs start out by examining the yaw torque eยคects. If we assume
the same control action as we found for the quadrotor in Eq. (2.20), we
would apply a negative torque on all six rotors. This will clearly produce
a positive yawing torque on the hexacopter with no net change in vertical
2022
CHAPTER 2. DYNAMIC MODELS
lift for the same reasons given on pages 38 and 39.
However, it not
necessarily produce zero torque about the pitch and roll axes. So letโs
check that and adjust things if required.
Unlike the quadrotor where
the two rotors on the x-axis were both turning CW, rotors1 and 4 are
turning in the opposite direction. So in this case, giving those two rotors
an equal negative torque will speed one up rotor 4 and slow down,rotor 1,
thus producing a negative torque about the y-axis. However, note that
rotors 2 and 6 will speed up, while rotors 3 and 5 will slow down, thus
producing a positive torque.about the y-axis. These two torques cancel
each other because the angle between rotors 2, 3, 5 & 6 and the y-axis are
all 30o between the arms and the y-axis. Since the sine of 30o = 1/2, that
means that the torque about the y-axis from rotors 2, 3, 5 & 6 will oยคset
the torque from rotors 1 & 4. For no roll moment, we require a net zero
torque about the x-axis. Rotors 1 & 4 are on the x-axis, so no torque
from them. Rotors 2 & 6 are both CCW, so they produce no net torque
about the x-axis, and the same applies for rotors 3 & 5; hence when we
apply an equal torque on all six rotors in the same direction along the
z-axis, we obtain a yaw torque with no eยคect on pitch or roll or a vertical
force. So, the control using:
Yaw control:
T1 = T2 = T3 = T4 = T5 = T6 =
will produce motion in yaw, with no eยคect on pitch, roll,
motion.
T ;
or vertical
For pitch control, we want an increase in the speed and lift of the
CW rotor 1 (ie a positive T1 ) and a decrease in the speed of the CCW
rotor 4, which requires a positive T4 ; so T1 = T4 = +T . But note
that this action will produce a yawing torque, which.can be canceled if we
apply an equal and opposite yawing torque from rotors 2, 3, 5, 6. This
accomplished by T2 = T3 = T5 = T6 = T =2: In fact, this additional
control will also add to the torque about the pitch (y-axis) by 50% while
having no eยคect on roll. Thus the control using:
Pitch control:
T1 = 2 T2 = 2 T3 =
T4 = 2 T5 = 2 T6 =
2
T ;
3
will produce motion in pitch, with no eยคect on yaw, roll, or vertical motion.
For roll control, by increasing the speed of rotors 5 & 6, and decreasing
the speed of rotors 2 & 3, we will obtain a positive roll torque about the
x-axis. No change is needed for rotors 1 & 4.
To accomplish these
speed changes, we need to apply a positive torque to the CW rotor 5 and
a negative torque to the CCW rotor 6. Likewise, we need a positive
torque to the CCW rotor 2 and a negative torque to the CW rotor 3.
2023
Since the torques applied are half positive and half negative, there will be
no yaw torque. Furthermore, there will be no pitch torque of change in
overall lift with these torque applications. However, the moment arm is
not 1 arm length; rather it is cos(30o ), so to keep the scale factor for the
two arms consistent with the pitch axis, we need to divide by 2 cos(30o ):
Therefore, the control using:
Roll control:
T2 =
T3 = T5 =
T6 = T =(2 cos(30o ));
will produce motion in roll, with no eยคect on pitch, yaw, or vertical motion.
12. In most cases, quadcopters have a camera mounted that does not swivel
in the x y plane and its direction of view is oriented at 45o to the arms
supporting the rotors. Therefore, these drones typically โกy in a direction
that is aligned with the camera rather than along an axis containing two
of the rotors. To simplify the โกight dynamics, the x-direction of the
coordinate system is aligned with the camera direction. Based on the
coordinate deโฆnitions for the axes in Fig. 2.14, assume the x-axis lies half
way between rotors # 1 and 2 and determine the rotor commands for the
four rotors that would accomplish independent motion for pitch, roll, and
yaw.
Solution:
The orientation of the coordinate system and arrangement of the rotors is
2024
CHAPTER 2. DYNAMIC MODELS
Coorinate system and rotor arrangement
It should be clear from the discussion on pages 38 and 39 of the book (plus
the discussion here for Problem 2.11) that the following commands to the
rotors produces the desired independent motion:
Roll control:
Pitch control:
Yaw control:
T1 = T2 =
T3 =
T4 = T ;
T1 =
T2 =
T3 = T4 = T ;
T1 = T2 = T3 = T4 = T :
Problems and Solutions for Section 2.2
13. A โฆrst step toward a realistic model of an op amp is given by the equations
below and shown in Fig. 2.50.
Vout
=
i+
=
107
[V+
s+1
i =0
V ]
2025
Fig. 2.50 Circuit for Problem 2.13
Find the transfer function of the simple ampliโฆcation circuit shown using
this model.
Solution:
As i = 0,
(a)
Vin V
Rin
V =
Vout
=
=
=
=
V
Vout
Rf
Rf
Rin
Vin +
Vout
Rin + Rf
Rin + Rf
107
[V+ V ]
s+1
107
Rf
V+
Vin
s+1
Rin + Rf
107
s+1
Rin
Vout
Rin + Rf
Rf
Rin
Vin +
Vout
Rin + Rf
Rin + Rf
R
f
107 Rin +R
Vout
f
=
in
Vin
s + 1 + 107 R R+R
in
f
14. Show that the op amp connection shown in Fig. 2.51 results in Vo = Vin
if the op amp is ideal. Give the transfer function if the op amp has the
non-ideal transfer function of Problem 2.13.
2026
CHAPTER 2. DYNAMIC MODELS
Fig. 2.51 Circuit for Problem 2.14
Solution:
Ideal case:
Vin
V+
V
=
=
=
V+
V
Vout
Non-ideal case:
Vin = V+ ; V = Vout
but,
V+ 6= V
instead,
Vout
=
=
107
[V+
s+1
107
[Vin
s+1
V ]
Vout ]
so,
107
Vout
107
107
= s+1107 =
=
7
Vin
s + 1 + 10
s + 107
1 + s+1
2027
15. A common connection for a motor power ampliโฆer is shown in Fig. 2.52.
The idea is to have the motor current follow the input voltage and the
connection is called a current ampliโฆer. Assume that the sense resistor,
Rs is very small compared with the feedback resistor, R and โฆnd the
transfer function from Vin to Ia : Also show the transfer function when
Rf = 1:
Node A
Node B
Fig. 2.52 Op Amp circuit for Problem 2.15 with nodes
marked.
Solution:
At node A,
Vin 0 Vout 0 VB 0
=0
+
+
Rin
Rf
R
At node B, with Rs
Ia +
(201)
R
0
VB
0 VB
+
R
Rs
=
VB
=
VB
0
(202)
RRs
Ia
R + Rs
R s Ia
The dynamics of the motor is modeled with negligible inductance as
Jm โขm + b _ m
Jm s + b
=
=
Kt I a
Kt I a
(203)
2028
CHAPTER 2. DYNAMIC MODELS
At the output, from Eq. (202). Eq. (203) and the motor equation Va =
I a R a + Ke s
Vo
=
I a R s + Va
=
I a R s + I a R a + Ke
Kt I a
Jm s + b
Substituting this into Eq.(201)
Vin
1
Kt I a
Ia R s
+
I a R s + I a R a + Ke
+
=0
Rin
Rf
Jm s + b
R
This expression shows that, in the steady state when s ! 0; the current
is proportional to the input voltage.
If fact, the current ampliโฆer normally has no feedback from the output
voltage, in which case Rf ! 1 and we have simply
Ia
=
Vin
R
Rin Rs
16. An op amp connection with feedback to both the negative and the positive
terminals is shown in Fig 2.53. If the op amp has the non-ideal transfer
function given in Problem 13, give the maximum value possible for the
r
positive feedback ratio, P =
in terms of the negative feedback
r+R
Rin
ratio,N =
for the circuit to remain stable.
Rin + Rf
Fig. 2.53 Op Amp circuit for Problem 2.16
2029
Solution:
Vin V
Vout V
+
Rin
Rf
Vout V+
0 V+
+
R
r
V
0
=
0
Rf
Rin
Vin +
Vout
Rin + Rf
Rin + Rf
= (1 N ) Vin + N Vout
r
Vout = P Vout
=
r+R
=
V+
Vout
=
=
=
107
[V+ V ]
s+1
107
[P Vout (1
s+1
Vout
Vin
=
=
=
0
P
<
T 1 > T 2 > T 3
and, with a same cold air โกow into every room, the ones with some sun
load will be hotest.
Letโs redeโฆnce the resistances
Ro = resistance to heat ow through one unit of outer wall
Ri = resistance to heat ow through one unit of inner wall
Room type 1:
2050
CHAPTER 2. DYNAMIC MODELS
T_1
=
=
qout
=
qin
=
2
(T1
Ri
2
(To
Ro
1
(qin qout )
C
1 2
(To T1 )
C Ro
T2 ) + q
T1 )
2
(T1
Ri
T2 )
2
(T1
Ri
T2 )
q
Room type 2:
1
T_2 =
C
qin
=
qout
=
1
(To
Ro
1
(To
Ro
1
(T2
Ri
T2 ) +
T2 ) +
T3 ) + q
2
(T1
Ri
T2 )
1
(T2
Ri
T3 )
q
Room type 3:
qin
=
qout
=
1
T_3 =
C
4
(T2
Ri
q
4
(T2
Ri
T3 )
T3 )
q
27. For the two-tank โกuid-โกow system shown in Fig. 2.59, โฆnd the diยคerential
equations relating the โกow into the โฆrst tank to the โกow out of the second
tank.
2051
Fig. 2.60 Two-tank โกuid-โกow system for Problem 27
Solution:
This is a variation on the problem solved in Example 2.21 and the deโฆnitions of terms is taken from that. From the relation between the height
of the water and mass โกow rate, the continuity equations are
m
_1
m
_2
=
=
A1 h_ 1 = win w
A2 h_ 2 = w wout
Also from the relation between the pressure and outgoing mass โกow rate,
w
=
wout
=
1
1
( gh1 ) 2
R1
1
1
( gh2 ) 2
R2
Finally,
h_ 1
=
h_ 2
=
1
1
1
( gh1 ) 2 +
win
A1 R 1
A1
1
1
1
1
( gh1 ) 2
( gh2 ) 2 :
A2 R 1
A2 R2
28. A laboratory experiment in the โกow of water through two tanks is sketched
in Fig. 2.61. Assume that Eq. (2.96) describes โกow through the equal-sized
holes at points A, B, or C.
(a) With holes at B and C but none at A, write the equations of motion
for this system in terms of h1 and h2 . Assume that when h2 = 15 cm,
the outโกow is 200 g/min.
(b) At h1 = 30 cm and h2 = 10 cm, compute a linearized model and the
transfer function from pump โกow (in cubic centimeters per minute)
to h2 .
2052
CHAPTER 2. DYNAMIC MODELS
(c) Repeat parts (a) and (b) assuming hole B is closed and hole A is
open. Assume that h3 = 20 cm, h1 > 20 cm, and h2 < 20:cm.
Fig. 2.61 Two-tank โกuid-โกow system for Problem 28
Solution:
(a) Following the solution of Example 2.21, and assuming the area of
both tanks is A; the values given for the heights ensure that the
water will โกow according to
WB
Win
WB
=
WC
=
WC
WB
=
=
1
[ g (h1
R
1
1
[ gh2 ] 2
R
Ah_ 2
Ah_ 1
1
h2 )] 2
From the outโกow information given, we can compute the oriโฆce resistance, R; noting that for water,
= 1 gram/cc and g = 981
cm/sec2 ' 1000 cm/sec2 :
WC
=
R
=
=
1p
1p
gh2 =
g 15 cm
200 g= mn =
R p
R
p
1 g= cm3 1000 cm= s2
g 10 cm
=
200 g= mn
200 g=60 s
s
1
1
122:5
g cm2 s2
60
= 36:7 g 2 cm 2
200
cm3 s2 g2
15 cm
2053
(b) The nonlinear equations from above are
h_ 1
=
h_ 2
=
1
1 p
g (h1 h2 ) +
Win
AR
A
1 p
1 p
g (h1 h2 )
gh2
AR
AR
The square root functions need to be linearized about the nominal
heights. In general the square root function can be linearized as
below
p
x0 + x
=
s
=
p
x0 1 +
x
x0
x0 1 +
1 x
2 x0
So letโs assume that h1 = h10 + h1 and h2 = h20 + h2 where h10 = 30
cm and h20 = 10 cm. And for round numbers, letโs assume the area
of each tank A = 100 cm2 : The equations above then reduce to
h_ 1
=
h_ 2
=
p
1
1
(1)(1000) (30 + h1 10
h2 ) +
Win
(1)(100)(36:7)
(1)(100)
p
p
1
1
(1)(1000) (30 + h1 10
h2 )
(1)(1000)(10 + h2 )
(1)(100)(36:7)
(1)(100)(36:7)
which, with the square root approximations, is equivalent to,
p
2
1
1
1
h_ 1 =
(1 +
h1
h2 ) +
Win
36:7
40
40
100
p
2
1
1
1
1
(1 +
h1
h2 )
(1 +
h2 )
h_ 2 =
36:7
40
40
36:7
20
p
10
The nominal inโกow Wnom = 3:67
2cc/sec is required in order for the
system to be in equilibrium, as can be seen from the โฆrst equation.
So we will deโฆne the total inโกow to be Win = Wnom + W: Including
the nominal inโกow, the equations become
p
2
1
_h1 =
( h1
h2 ) +
W
1468
100
p
p
p
1
2
2
2 1
h1 + (
) h2 +
h_ 2 =
1468
1468 734
36:7
However, holding the nominal โกow rate maintains h1 at equilibrium,
but h2 will not stay at equilibrium. Instead, there will be a constant term increasing h2 : Thus the standard transfer function will
not result.
2054
CHAPTER 2. DYNAMIC MODELS
(c) With hole B closed and hole A open, the relevant relations are
Win
WA
h_ 1
=
h_ 2
=
WA
=
WA
=
WC
=
WC
=
Ah_ 1
1p
g(h1
R
Ah_ 2
1p
gh2
R
h3 )
1 p
1
g(h1 h3 ) +
Win
AR
A
1 p
1 p
g(h1 h3 )
gh2
AR
AR
For the value of R, we will use the same calculation from part (a),
since that value has not changed with diยคerent hole openings, so we
still have
122:5
R=
60
200
s
g cm2 s2
= 36:7 g
cm3 s2 g2
1
2
cm
1
2
. With the same deโฆnitions for the perturbed quantities as for part
(b), plus h3 = 20 cm,.we obtain
h_ 1
=
h_ 2
=
p
1
1
Win
(1)(1000)(30 + h1 20) +
(1)(100)(36:7)
(1)(100)
p
1
(1)(1000)(30 + h1 20)
(1)(100)(36:7)
p
1
(1)(1000)(10 + h2 )
(1)(100)(36:7)
which, with the linearization carried out, reduces to
h_ 1
=
h_ 2
=
1
1
1
1
(1 +
h1 ) +
Wnom +
W
(36:7)
20
(100)
(100)
1
1
1
1
(1 +
h1 )
(1 +
h2 )
(36:7)
20
(36:7)
20
10
and with the nominal โกow rate of Win = 3:67
removed
h_ 1
=
h_ 2
=
1
1
h1 +
W
734
100
1
1
h1
h2
734
734
2055
Taking the Laplace transform of these two equations, and solving for
the desired transfer function (in cc/sec) yields
H2 (s)
1
0:01
:
=
W (s)
734 (s + 1=734)2
which becomes, with the inโกow in grams/min,
H2 (s)
0:000817
1 (0:01)(60)
=:
=
2
W (s)
734 (s + 1=734)
(s + 1=734)2
29. The equations for heating a house are given by Eqs. (2.81) and (2.82)
and, in a particular case can be written with time in hours as
C
dTh
= Ku
dt
Th
To
R
where
(a) C is the Thermal capacity of the house, BT U=o F
(b) Th is the temperature in the house, o F
(c) To is the temperature outside the house, o F
(d) K is the heat rating of the furnace, = 90; 000 BT U=hour
(e) R is the thermal resistance, o F per BT U=hour
(f) u is the furnace switch, =1 if the furnace is on and =0 if the furnace
is oยค.
It is measured that, with the outside temperature at 32 o F and the house
at 60 o F , the furnace raises the temperature 2 o F in 6 minutes (0.1
hour). With the furnace oยค, the house temperature falls 2 o F in 40
minutes. What are the values of C and R for the house?
Solution:
For the โฆrst case, the furnace is on which means u = 1.
C
dTh
dt
=
K
T_h
=
K
C
1
(Th To )
R
1
(Th To )
RC
and with the furnace oยค,
T_h =
1
(Th
RC
To )
In both cases, it is a โฆrst order system and thus the solutions involve
exponentials in time. The approximate answer can be obtained by simply
2056
CHAPTER 2. DYNAMIC MODELS
looking at the slope of the exponential at the outset. This will be fairly
accurate because the temperature is only changing by 2 degrees and this
represents a small fraction of the 30 degree temperature diยคerence. Letโs
solve the equation for the furnace oยค โฆrst
Th
=
t
1
(Th
RC
To )
plugging in the numbers available, the temperature falls 2 degrees in 2/3
hr, we have
1
2
=
(60 32)
2=3
RC
which means that
RC = 28=3
For the second case, the furnace is turned on which means
K
Th
=
t
C
1
(Th
RC
To )
and plugging in the numbers yields
2
90; 000
=
0:1
C
1
(60
28=3
32)
and we have
C
=
R
=
90; 000
= 3913
23
RC
28=3
=
= 0:00239
C
3913

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