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Solutions Manual

6th 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 1

An Overview and Brief

History of Feedback Control

1.1 Problems and Solutions

1. Draw a component block diagram for each of the following feedback control systems.

(a) The manual steering system of an automobile (b) Drebbelís incubator

(c) The water level controlled by a áoat and valve

(d) Wattís steam engine with áy-ball governor In each case, indicate the location of the elements listed below and give the units associated with each signal.

 the process  the process output signal  the sensor  the actuator  the actuator output signal  The reference signal

Notice that in a number of cases the same physical device may per- form more than one of these functions. Solution:

(a) A manual steering system for an automobile:

102 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL

(b) Drebbelís incubator:

(c) Water level regulator:

(d) Fly-ball governor:

1.1. PROBLEMS AND SOLUTIONS 103

(e) Automatic steering of a ship:

(f) A public address system:

2. Identify the physical principles and describe the operation of the thermo- stat in your home or o¢ ce. Solution: A thermostat is a device for maintaining a temperature constant at a desired value. It is equipped with a temperature sensor which detects deviation from the desired value, determines whether the temperature setting is exceeded or not, and transmits the information to a furnace or air conditioner so that the termperature in the room is brought back

104 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL

T h i c k s t o c k C o n s i s t e n c y m e t e r

C o n t r o l l e r W h i t e w a t e r

C V

S c ht o e r s a t g e R e f i n e r M c h ae cs ht i n e

W i r e

M o i s t u r e m e t e r

S c r e e n s a n d c l e a n e r s

H e a d b o x D r y i n g s e c t i o n

P r e s s e s R e e l

Figure 1.1: A paper making machine From Karl Astrom, (1970, page 192) reprinted with permission.

to the desired setting. Examples: Tubes Ölled with liquid mercury are attached to a bimetallic strip which tilt the tube and cause the mer- cury to slide over electrical contacts. A bimetallic strip consists of two strips of metal bonded together, each of a di§erent expansion coe¢ cient so that temperature changes bend the metal. In some cases, the bending of bimetallic strips simply cause electrical contacts to open or close di- rectly. In most cases today, temperature is sensed electronically using,for example, a thermistor, a resistor whose resistance changes with tempera- ture. Modern computer-based thermostats are programmable, sense the current from the thermistor and convert that to a digital signal.

3. A machine for making paper is diagrammed in Fig. 1.12. There are two main parameters under feedback control: the density of Öbers as controlled by the consistency of the thick stock that áows from the headbox onto the wire, and the moisture content of the Önal product that comes out of the dryers. Stock from the machine chest is diluted by white water returning from under the wire as controlled by a control valve (CV). A meter supplies a reading of the consistency. At the ìdry endî of the machine, there is a moisture sensor. Draw a signal graph and identify the seven components listed in Problem 1 for

(a) control of consistency

(b) control of moisture Solution:

(a) Control of paper machine consistency:

1.1. PROBLEMS AND SOLUTIONS 105

(b) Control of paper machine moisture:

4. Many variables in the human body are under feedback control. For each of the following controlled variables, draw a graph showing the process being controlled, the sensor that measures the variable, the actuator that causes it to increase and/or decrease, the information path that completes the feedback path, and the disturbances that upset the variable. You may need to consult an encyclopedia or textbook on human physiology for information on this problem.

(a) blood pressure

(b) blood sugar concentration

(c) heart rate

(d) eye-pointing angle

(e) eye-pupil diameter Solution: Feedback control in human body:

106 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL

Variable Sensor Actuator Inform ation path Disturbances a) Blood pressure -Arterial -Cardiac output -A§erent nerve -Bleeding baroreceptors -Arteriolar/venous Öb ers -Drugs dilation -Stress,Pain b) Blood sugar -Pancreas -Pancreas secreting -Blood áow to -Diet concentration insulin pancreas -Exercise (Glucose) c) Heart rate -Diastolic volum e -Electrical stimulation -M echanical draw -Horm one release sensors of sino-atrial node of blood from heart -Exercise -Cardiac sym pathetic and cardiac muscle -Circulating nerves epinephrine d) Eye p ointing -Optic nerve -Extraocular muscles -Cranial innervation -Head m ovem ent angle -Im age detection -M uscle twitch e) Pupil diam eter -Rods -Pupillary sphincter -Autonom ous -Ambient light muscles system -Drugs f ) Blood calcium -Parathyroid gland -Ca from b ones to blood - Parathorm one -Ca need in b one level detectors -Gastrointestinal horm one a§ecting -Drugs absorption e§ector sites

5. Draw a graph of the components for temperature control in a refrigerator or automobile air-conditioning system. Solution:

This is the simplest possible system. Modern cases include computer control as described in later chapters.

6. Draw a graph of the components for an elevator-position control. Indi- cate how you would measure the position of the elevator car. Consider a combined coarse and Öne measurement system. What accuracies do you suggest for each sensor? Your system should be able to correct for the fact that in elevators for tall buildings there is signiÖcant cable stretch as a function of cab load. Solution: A coarse measurement can be obtained by an electroswitch located before the desired áoor level. When touched, the controller reduces the motor speed. A ìÖneî sensor can then be used to bring the elevator precisely to the áoor level. With a sensor such as the one depicted in the Ögure, a linear control loop can be created (as opposed to the on-o§ type of the coarse control).Accuracy required for the course switch is around 5 cm; for the Öne áoor alignment, an accuracy of about 2 mm is desirable to eliminate any noticeable step for those entering or exiting the elevator.

1.1. PROBLEMS AND SOLUTIONS 107

7. Feedback control requires being able to sense the variable being controlled. Because electrical signals can be transmitted, ampliÖed, and processed easily, often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. Describe a sensor that would give an electrical output proportional to:

(a) temperature (b) pressure (c) liquid level (d) áow of liquid along a pipe (or blood along an artery) force (e) linear position (f) rotational position (g) linear velocity (h) rotational speed (i) translational acceleration (j) torque Solution: Sensors for feedback control systems with electrical output. Exam- ples

(a) Temperature: Thermistor- temperature sensitive resistor with resis- tance change proportional to temperature; Thermocouple; Thyrister. Modern thermostats are computer controlled and programmable. (b) Pressure: Strain sensitive resistor mounted on a diaphragm which bends due to changing pressure

108 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL

(c) Liquid level: Float connected to potentiometer. If liquid is conductive the impedance change of a rod immersed in the liquid may indicate the liquid level.

(d) Flow of liquid along a pipe: A turbine actuated by the áow with a magnet to trigger an external counting circuit. Hall e§ect produces an electronic output in response to magnetic Öeld changes. Another way: Measure pressure di§erence from venturi into pressure sensor as in Ögure; Flowmeter. For blood áow, an ultrasound device like a SONAR can be used.

(e) Position. When direct mechanical interaction is possible and for ìsmallî dis- placements, the same ideas may be used. For example a potentiome- ter may be used to measure position of a mass in an accelerator (h). However in many cases such as the position of an aircraft, the task is much more complicated and measurement cannot be made directly. Calculation must be carried out based on other measurements, for example optical or electromagnetic direction measurements to several known references (stars,transmitting antennas ...); LVDT for linear, RVDT for rotational. (f) Rotational position. The most common traditional device is a pote- niometer. Also common are magnetic machines in shich a rotating magnet produces a variable output based on its angle. (g) Linear velocity. For a vehicle, a RADAR can measure linear velocity. In other cases, a rack-and-pinion can be used to translate linear to rotational motion and an electric motor(tachometer) used to measure the speed. (h) Speed: Any toothed wheel or gear on a rotating part may be used to trigger a magnetic Öeld change which can be used to trigger an elec- trical counting circuit by use of a Hall e§ect (magnetic to electrical) sensor. The pulses can then be counted over a set time interval to produce angular velocity: Rate gyro; Tachometer

1.1. PROBLEMS AND SOLUTIONS 109

(i) Acceleration: A mass movement restrained by a spring measured by a potentiometer. A piezoelectric material may be used instead (a ma- terial that produces electrical current with intensity proportional to acceleration). In modern airbags, an integrated circuit chip contains a tiny lever and íproof massíwhose motion is measured generating a voltage proportional to acceleration.

(j) Force, torque: A dynamometer based on spring or beam deáections, which may be measured by a potentiometer or a strain-gauge.

8. Each of the variables listed in Problem 7 can be brought under feedback control. Describe an actuator that could accept an electrical input and be used to control the variables listed. Give the units of the actuator output signal. Solution:

(a) Resistor with voltage applied to it ormercury arc lamp to generate heat for small devices. a furnace for a building.. (b) Pump: Pumping air in or out of a chamberto generate pressure. Else, a ítorque motoríproduces force.. (c) Valve and pump: forcing liquid in or out of the container. (d) A valve is nromally used to control áow. (e) Electric motor (f) Electric motor (g) Electric motor (h) Electric motor (i) Translational acceleration is usually controlled by a motor or engine to provide force on the vehicle or other object. (j) Torque motor. In this motor the torque is directly proportional to the input (current).

Chapter 2

Dynamic Models

Problems and Solutions for Section 2.

1. Write the di§erential equations for the mechanical systems shown in Fig. 2.39. 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.

Figure 2.39: Mechanical systems

Solution:

2004 CHAPTER 2. DYNAMIC MODELS

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 x 2 = 0 and Öxed and increase x 1 from 0. Then the k 1 spring will be stretched producing its spring force to the left and the k 2 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.

(a)

Free body diagram for Problem 2.1(a)

m 1 x 1 = k 1 x 1 b 1 x_ 1 k 2 (x 1 x 2 ) m 2 x 2 = k 2 (x 2 x 1 ) k 3 (x 2 y) b 2 x_ 2

There is friction a§ecting the motion of both masses; therefore the system will decay to zero motion for both masses.

Free body diagram for Problem 2.1(b)

m 1 x 1 = k 1 x 1 k 2 (x 1 x 2 ) b 1 x_ 1 m 2 x 2 = k 2 (x 2 x 1 ) k 3 x 2

Although friction only a§ects the motion of the left mass directly, continuing motion of the right mass will excite the left mass, and that interaction will continue until all motion damps out.

Figure 2.40: Mechanical system for Problem 2.

m 1 m^2

x 1 x^2

k (x - x ) 2 1 2 k (x - x ) 2 1 2

k x 1 1

b (x - x ) 1 1 2 b (x - x )^1 1

....

F

Free body diagram for Problem 2.1(c)

m 1 x 1 = k 1 x 1 k 2 (x 1 x 2 ) b 1 ( x 1 x 2 ) m 2 x 2 = F k 2 (x 2 x 1 ) b 1 ( x 2 x 1 )

2. Write the di§erential equations for the mechanical systems shown in Fig. 2.40. 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 object, let x 2 = 0 and increase x 1 from 0. Then the k 1 spring on the left will be stretched producing its spring force to the left and the k 2 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.

m 1 m 2

x 1 x 2

x 2

k (x - x ) 2 1 2

k x 1 1 k 1

b (x - x ) 2 1 2 b (x - x )^2 1

....

Free body diagram for Problem 2.

2006 CHAPTER 2. DYNAMIC MODELS

Then the forces are summed on each mass, resulting in

m 1 x 1 = k 1 x 1 k 2 (x 1 x 2 ) b 1 ( x 1 x 2 ) m 2 x 2 = k 2 (x 1 x 2 ) b 1 ( x 1 x 2 ) k 1 x 2

The relative motion between x 1 and x 2 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 no áex- ure of the end springs is all that is required to maintain the oscillation of the two masses.

3. Write the equations of motion for the double-pendulum system shown in Fig. 2.41. 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 2.41: Double pendulum

Solution:

θ 1 θ 2

k

sin 1

l θ sin 2

l θ

m m

l

DeÖne coordinates

If we write the moment equilibrium about the pivot point of the left pen- dulem from the free body diagram,

M = mgl sin  1 k

l (sin  1 sin  2 ) cos  1

l = ml^2  1

ml^2  1 + mgl sin  1 +

kl^2 cos  1 (sin  1 sin  2 ) = 0

Similary we can write the equation of motion for the right pendulem

mgl sin  2 + k

l (sin  1 sin  2 ) cos  2

l = ml^2  2

As we assumed the angles are small, we can approximate using sin  1   1 ; sin  2   2 , cos  1  1 , and cos  2  1. Finally the linearized equations of motion becomes,

ml 1 + mg 1 +

kl ( 1  2 ) = 0

ml 2 + mg 2 +

kl ( 2  1 ) = 0

Or

2008 CHAPTER 2. DYNAMIC MODELS

 1 + g l

 1 +

k m

( 1  2 ) = 0

 2 + g l

 2 +

k m

( 2  1 ) = 0

4. Write the equations of motion of a pendulum consisting of a thin, 4-kg stick of length l suspended from a pivot. How long should the rod be in order for the period to be exactly 2 secs? (The inertia I of a thin stick about an endpoint is 13 ml^2. Assume  is small enough that sin  = .) Solution: Letís use Eq. (2.14)

M = I ;

mg

θ

O^ l

DeÖne coordinates and forces

Moment about point O.

MO = mg 

l 2 sin  = IO

=

ml^2 

 +^3 g 2 l

sin  = 0

As we assumed  is small,

 +^3 g 2 l

 = 0

The frequency only depends on the length of the rod

!^2 =

3 g 2 l

T =

2 

!

= 2

s 2 l 3 g

= 2

l = 3 g 2 ^2

= 1:49 m

(a) Compare the formula for the period, T = 2

q 2 l 3 g with the well known formula for the period of a point mass hanging with a string with length l. T = 2

q l 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 Öxed 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 ìunit bumpî (i.e., r is a unit step) using MATLAB. Assume that m 1 = 10 kg, m 2 = 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)

Y (s) R(s)

=

kw b m 1 m 2 (s^ +^

ks b ) s^4 + ( (^) mb 1 + (^) mb 2 )s^3 + ( (^) mks 1 + (^) mks 2 + (^) mkw 1 )s^2 + ( (^) mk 1 wm^ b 2 )s + (^) mkw 1 mks 2

:

This transfer function can be put directly into Matlab along with the numerical values as shown below. Note that b is not the damping

2010 CHAPTER 2. DYNAMIC MODELS

ratio, but damping. We need to Önd 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.

From the Ögures, b  3000 would be acceptable. There is too much overshoot for lower values, and the system gets too fast (and harsh) for larger values.

% Problem 2. clear all, close all

m1 = 10; m2 = 350; kw = 500000; ks = 10000; B = [ 1000 2000 3000 4000 ]; t = 0:0.01:2;

for i = 1: b = B(i); num = kwb/(m1m2)[1 ks/b]; den =[1 (b/m1+b/m2) (ks/m1+ks/m2+kw/m1) (kwb/(m1m2) kwks/(m1*m2)]; sys=tf(num,den); y = step( sys, t );

subplot(2,2,i); plot( t, y(:,1), í:í, t, y(:,2), í-í); legend(íWheelí,íCarí); ttl = sprintf(íResponse with b = %4.1f í,b ); title(ttl); end

6. Write the equations of motion for a body of mass M suspended from a Öxed point by a spring with a constant k. Carefully deÖne where the bodyís displacement is zero.

Solution:

Some care needs to be taken when the spring is suspended vertically in the presence of the gravity. We deÖne x = 0 to be when the spring is unstretched with no mass attached as in (a). The static situation in (b) results from a balance between the gravity force and the spring.

From the free body diagram in (b), the dynamic equation results

mx = kx mg:

We can manipulate the equation

mx = k



x +

m k g



;

so if we replace x using y = x + mk g,

my = ky my + ky = 0

The equilibrium value of x including the e§ect of gravity is at x = mk g and y represents the motion of the mass about that equilibrium point. An alternate solution method, which is applicable for any problem involving vertical spring motion, is to deÖne the motion to be with respect to the static equilibrium point of the springs including the e§ect of gravity, and then to proceed as if no gravity was present. In this problem, we would deÖne y to be the motion with respect to the equilibrium point, then the FBD in (c) would result directly in

my = ky:

7. Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a change- able damping, b(u 1 ): 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, u 2 ; in opposite directions on the wheel axle and the car body.

2012 CHAPTER 2. DYNAMIC MODELS

(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, u 2 ; to completely replace the springs and shock absorber? Is this a good idea?

Solution:

(a) The FBD shows the addition of the variable force, u 2 ; and shows b as in the FBD of Fig. 2.5, however, here b is a function of the control variable, u 1 : The forces below are drawn in the direction that would result from a positive displacement of x.

Free body diagram

m 1 x = b (u 1 ) ( y x) + ks (y x) kw (x r) u 2 m 2 y = ks (y x) b (u 1 ) ( y x) + u 2

(b) The system is linear with respect to u 2 because it is additive. But b is not constant so the system is non-linear with respect to u 1 be- cause 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 bet- ter solution to modulate the damping coe¢ cient by changing oriÖce 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 sti§ ride.

8. 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 = reference speed K = constant:

This is a ëproportionalícontrol law where the di§erence 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 = 1000 kg and b = 50 N
s= m; and Önd the response for a unit step in vr using MATLAB. Using trial and error, Önd 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 _ +

b m

v =

m

u

substitute in u = K (vr v)

v_ +

b m

v =

m

u =

K

m

(vr v)

Rearranging, yields the closed-loop system equations,

v _ + b m

v +

K

m

v =

K

m

vr

A block diagram of the scheme is shown below where the car dynamics are depicted by its transfer function from Eq. 2.7.

m

b s

m

1 Σ (^) K

vr u v

−

Block diagram

The transfer function of the closed-loop system is,

V (s) Vr (s)

=

K m s + (^) mb + Km

so that the inputs for Matlab are

2014 CHAPTER 2. DYNAMIC MODELS

num =

K

m den = [1

b m

+

K

m

]

For K = 100; 500 ; 1000 ; 5000 We have,

Time (sec.)

Amplitude

Step Response

0 5 10 15 20 25 30 35 40 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

From: U(1)

To: Y(1)^ K=100

K=500 K=1000

K=5000

Time responses

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 Önd 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 is 5 seconds and the steady state error is 5%.

% Problem 2.8 clear all, close all

% data m = 1000; b = 50; k = [ 100 500 1000 5000 ];

% Overlay the step response hold on for i=1:length(k) K=k(i); num =K/m; den = [1 b/m+K/m]; step( num, den); end

9. In many mechanical positioning systems there is áexibility between one part of the system and another. An example is shown in Figure 2.6 where there is áexibility of the solar panels. Figure 2.42 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 coe¢ cient 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 2.42: Schematic of a system with áexibility

Solution:

(a) The FBD for the system is

2016 CHAPTER 2. DYNAMIC MODELS

Free body diagrams

which results in the equations

mx = k (x y) b ( x y) M y = u + k (x y) + b ( x y)

or

x  +

k m

x +

b m

x_

k m

y

b m

y_ = 0

k M

x b M

x_ + y + k M

y + b M

y_ =

M

u

(b) If we make Laplace Transform of the equations of motion

s^2 X +

k m

X +

b m

sX

k m

Y

b m

sY = 0

k M

X

b M

sX + s^2 Y +

k M

Y +

b M

sY =

M

U

In matrix form,  ms^2 + bs + k (bs + k) (bs + k) M s^2 + bs + k

 

X

Y



=



0

U



From Cramerís Rule,

Y =

det



ms^2 + bs + k 0 (bs + k) U



det



ms^2 + bs + k (bs + k) (bs + k) M s^2 + bs + k



=

ms^2 + bs + k (ms^2 + bs + k) (M s^2 + bs + k) (bs + k)^2

U

Finally,

Y

U

=

ms^2 + bs + k (ms^2 + bs + k) (M s^2 + bs + k) (bs + k)^2

=

ms^2 + bs + k mM s^4 + (m + M )bs^3 + (M + m)ks^2