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

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    100 Solutions Manual: Chapter 1 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

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    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) Drebbelí s incubator (c) The water level controlled by a á oat and valve (d) Wattí s steam engine with á y-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 desired output signal  the sensor  the actuator  the actuator output signal  the controller  the controller output signal  the reference signal  the error 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: 101

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    102 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL (b) Drebbelí s incubator: (c) Water level regulator: (d) Fly-ball governor:

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    1.1. PROBLEMS AND SOLUTIONS 103 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 temperature in the room is brought back to the desired setting. Examples: Tubes Ölled with liquid mercury are attached to a bimetallic strip which tilt the tube and cause the mercury 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 temper- ature changes bend the metal. In some cases, the bending of bimetallic strips simply cause electrical contacts to open or close directly. In most cases today, temperature is sensed electronically using,for example, a ther- mistor, a resistor whose resistance changes with temperature. Modern computer-based thermostats are programmable, sense the current from the thermistor and convert that to a digital signal. Fig 1.12 A Paper Making Machine

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    104 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL 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 á ows 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: (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.

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    1.1. PROBLEMS AND SOLUTIONS 105 (a) blood pressure (b) blood sugar concentration (c) heart rate (d) eye-pointing angle (e) eye-pupil diameter Solution: Feedback control in human body: 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 ones level detectors -Gastrointestinal horm one a§ecting -Drugs absorption e§ector sites 5. 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 á oor level. When touched, the controller reduces the motor speed. A ìÖneî sensor can then be used to bring the elevator precisely to the á oor 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 á oor alignment, an accuracy of about 2 mm is desirable to eliminate any noticeable step for those entering or exiting the elevator.

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    106 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL 6. 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) á ow 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; Thyristor. Modern thermostats are computer controlled and programmable. (b) Pressure: Strain sensitive resistor mounted on a diaphragm which bends due to changing pressure

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    1.1. PROBLEMS AND SOLUTIONS 107 (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 á ow 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 á ow, 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 poten- tiometer. Also common are magnetic machines in which 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

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    108 CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL (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 í proof massíwhose motion is measured generating a voltage proportional to acceleration. (j) Force, torque: A dynamometer based on spring or beam deá ections, which may be measured by a potentiometer or a strain-gauge. 7. Each of the variables listed in Problem 6 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 or mercury arc lamp to generate heat for small devices. a furnace for a building.. (b) Pump: Pumping air in or out of a chamber to generate pressure. Else, a í torque motoríproduces force.. (c) Valve and pump: forcing liquid in or out of the container. (d) A valve is normally used to control á ow. (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). 8. Feedback in Biology (a) Negative Feedback in Biology : When a person is under long term stress (say a couple of weeks before an exam!), hypothalamus (in the brain) se- cretes a hormone called CRF (Corticotrophin Releasing Factor) which binds to a receptor in the pituitary gland stimulating it to produce ACTH

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    1.1. PROBLEMS AND SOLUTIONS 109 (Adrenocorticotropic hormone), which in turn stimulates the adrenal cor- tex (outer part of the adrenal glands) to release the stress hormone Glu- cocorticoids (GC). This in turn shuts down (turns o§ the stress response) for both CRF and ACTH production by negative feedback via the blood- stream until GC returns to its normal level. Draw a block diagram of this closed-loop system. (b) Positive Feedback in Biology : This happens in some unique circum- stances. Consider the birth process of a baby. Pressure from the head of the baby going through the birth canal causes contractions via secre- tion of a hormone called Oxytocin which causes more pressure which in turn intensiÖes contractions. Once the baby is born, the system goes back to normal (negative feedback). Draw a block diagram of this closed-loop system. Solution: (a) Negative Feedback in Biology - Stress Stress induced negative feedback (b) Positive Feedback in Biology - Child birth Child birth induced positive feedback

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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 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.

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    Chapter 2 Dynamic Models Problems and Solutions for Section 2.1 1. Write the di§erential 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 condi- tions for both masses, and give a reason for your answer. Also, for part (c), answer the question for F=0. 2001

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    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 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) m 1 m 2 x 1 x 2 k (x -x ) 2 1 2 k (x -x ) 2 1 2 k (x 3 2 - y) k x 1 1 b x 1 1 . 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 ) There is friction a§ecting the motion of mass 1 which will continue to take energy out of the system as long as there is any movement of x 1 : 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.

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    2003 m 1 m 2 x 1 x 2 x 2 k (x -x ) 2 1 2 k (x -x ) 2 1 2 k x 1 1 b x 1 2 . x 2 x 2 k 3 Free body diagram for Problem 2.1(b) m 1  x 1 = k 1 x 1 k 2 ( x 1 x 2 ) m 2  x 2 = k 2 ( x 2 x 1 ) b 1 _ x 2 k 3 x 2 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. 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 2 . . . . 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 ) 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 di§erential 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.

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    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 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 2 . . . . Free body diagram for Probelm 2.2 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 á exure 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 di§erent 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

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