Главная
Instructor's Solutions Manual for Modern Control Systems, 12th Edition
Instructor's Solutions Manual for Modern Control Systems, 12th Edition
Richard C. Dorf, Robert H. Bishop
0 /
0
Насколько Вам понравилась эта книга?
Какого качества скаченный файл?
Скачайте книгу, чтобы оценить ее качество
Какого качества скаченные файлы?
Категории:
Язык:
english
Страницы:
754
ISBN 10:
7801360249
Файл:
PDF, 4,90 MB
Ваши теги:
Скачать (pdf, 4,90 MB)
 Открыть в браузере
 Checking other formats...
 Конвертировать в EPUB
 Конвертировать в FB2
 Конвертировать в MOBI
 Конвертировать в TXT
 Конвертировать в RTF
 Конвертированный файл может отличаться от оригинала. При возможности лучше скачивать файл в оригинальном формате.
 Пожалуйста, сначала войдите в свой аккаунт

Нужна помощь? Пожалуйста, ознакомьтесь с инструкцией как отправить книгу на Kindle
В течение 15 минут файл будет доставлен на Ваш email.
В течение 15 минут файл будет доставлен на Ваш kindle.
Примечание: Вам необходимо верифицировать каждую книгу, которую Вы отправляете на Kindle. Проверьте свой почтовый ящик на наличие письма с подтверждением от Amazon Kindle Support.
Примечание: Вам необходимо верифицировать каждую книгу, которую Вы отправляете на Kindle. Проверьте свой почтовый ящик на наличие письма с подтверждением от Amazon Kindle Support.
Возможно Вас заинтересует Powered by Rec2Me
Ключевые слова
upper saddle river^{1507}
upper saddle^{1507}
saddle river^{1507}
pearson education^{1506}
transmission^{757}
permissions department^{753}
rights and permissions^{753}
prohibited reproduction^{753}
photocopying^{753}
retrieval system^{753}
likewise^{753}
sec^{739}
loop^{511}
rad^{398}
root locus^{387}
plot^{365}
axis^{359}
frequency^{347}
bode^{333}
loop transfer^{269}
feedback^{266}
loop transfer function^{260}
characteristic equation^{228}
controller^{226}
deg^{217}
yields^{209}
diagram^{187}
input^{185}
bode plot^{184}
sys^{183}
overshoot^{179}
amplitude^{173}
real axis^{170}
stability^{145}
phase margin^{144}
disturbance^{140}
lim^{138}
magnitude^{131}
ess^{127}
poles^{114}
percent overshoot^{110}
imag axis^{110}
select^{110}
solving^{108}
locus is shown^{105}
feedback control^{101}
bode diagram^{100}
settling time^{99}
loop system^{97}
real axis figure^{96}
axis figure^{96}
state variable^{93}
num^{90}
compensator^{89}
matrix^{88}
det^{88}
pole^{87}
dashed^{83}
exercises^{82}
plot is shown^{80}
Связанные Подборки
0 comments
Вы можете оставить отзыв о книге и поделиться своим опытом. Другим читателям будет интересно узнать Ваше мнение о прочитанных книгах. Независимо от того, пришлась ли Вам книга по душе или нет, если Вы честно и подробно расскажете об этом, люди смогут найти для себя новые книги, которые их заинтересуют.
1

2

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. MODERN CONTROL SYSTEMS SOLUTION MANUAL Richard C. Dorf Robert H. Bishop University of California, Davis Marquette University A companion to MODERN CONTROL SYSTEMS TWELFTH EDITION Richard C. Dorf Robert H. Bishop Prentice Hall Upper Saddle River Boston Columbus San Francisco New York Indianapolis London Toronto Sydney Singapore Tokyo Montreal Dubai Madrid Hong Kong Mexico City Munich Paris Amsterdam Cape Town Educator Home  eLearning & Assessment  Support/Contact Us  Find your rep  Exam copy bookbag Instructor's Solutions Manual for Modern Control Systems, 12/E Richard C. Dorf, University of California, Davis Robert H. Bishop, University of Texas at Austin ISBN10: 013602498X ISBN13: 9780136024989 Publisher: Prentice Hall Copyright: 2011 Format: Online Supplement Published: 08/16/2010 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage ; in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. P R E F A C E In each chapter, there are five problem types: Exercises Problems Advanced Problems Design Problems/Continuous Design Problem Computer Problems In total, there are over 1000 problems. The abundance of problems of increasing complexity gives students confidence in their problemsolving ability as they work their way from the exercises to the design and computerbased problems. It is assumed that instructors (and students) have access to MATLAB and the Control System Toolbox or to LabVIEW and the MathScript RT Module. All of the computer solutions in this Solution Manual were developed and tested on an Apple MacBook Pro platform using MATLAB 7.6 Release 2008a and the Control System Toolbox Version 8.1 and LabVIEW 2009. It is not possible to verify each solution on all the available computer platforms that are compatible with MATLAB and LabVIEW MathScript RT Module. Please forward any incompatibilities you encounter with the scripts to Prof. Bishop at the email address given below. The authors and the staff at Prentice Hall would like to establish an open line of communication with the instructors using Modern Control Systems. We encourage you to contact Prentice Hall with comments and suggestions for this and future editions. Robert H. Bishop rhbishop@marquette.edu iii © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. T A B L E  O F  C O N T E N T S 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. iv Introduction to Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Mathematical Models of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 State Variable Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Feedback Control System Characteristics . . . . . . . . . . . . . . . . . . . . . . . 133 The Performance of Feedback Control Systems . . . . . . . . . . . . . . . . . 177 The Stability of Linear Feedback Systems . . . . . . . . . . . . . . . . . . . . . . 234 The Root Locus Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 Frequency Response Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Stability in the Frequency Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 The Design of Feedback Control Systems . . . . . . . . . . . . . . . . . . . . . . . 519 The Design of State Variable Feedback Systems . . . . . . . . . . . . . . . . 600 Robust Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659 Digital Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. C H A P T E R 1 Introduction to Control Systems There are, in general, no unique solutions to the following exercises and problems. Other equally valid block diagrams may be submitted by the student. Exercises E1.1 A microprocessor controlled laser system: Controller Desired power output Error  Microprocessor Current i(t) Laser Power Sensor power A driver controlled cruise control system: Controller Process Foot pedal Desired speed Power out Measurement Measured E1.2 Process  Driver Car and Engine Actual auto speed Measurement Visual indication of speed E1.3 Speedometer Although the principle of conservation of momentum explains much of the process of flycasting, there does not exist a comprehensive scientific explanation of how a flyfisher uses the small backward and forward motion of the fly rod to cast an almost weightless fly lure long distances (the 1 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 2 CHAPTER 1 Introduction to Control Systems current worldrecord is 236 ft). The fly lure is attached to a short invisible leader about 15ft long, which is in turn attached to a longer and thicker Dacron line. The objective is cast the fly lure to a distant spot with deadeye accuracy so that the thicker part of the line touches the water first and then the fly gently settles on the water just as an insect might. Flyfisher Desired position of the fly Controller  Wind disturbance Mind and body of the flyfisher Process Rod, line, and cast Actual position of the fly Measurement Visual indication of the position of the fly E1.4 Vision of the flyfisher An autofocus camera control system: Oneway trip time for the beam Conversion factor (speed of light or sound) K1 Beam Emitter/ Receiver Beam return Distance to subject Subject Lens focusing motor Lens © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 3 Exercises E1.5 Tacking a sailboat as the wind shifts: Error Desired sailboat direction  Controller Actuators Sailor Rudder and sail adjustment Wind Process Sailboat Actual sailboat direction Measurement Measured sailboat direction Gyro compass E1.6 An automated highway control system merging two lanes of traffic: Controller Error Desired gap  Embedded computer Actuators Brakes, gas or steering Process Active vehicle Actual gap Measurement Measured gap Radar E1.7 Using the speedometer, the driver calculates the difference between the measured speed and the desired speed. The driver throotle knob or the brakes as necessary to adjust the speed. If the current speed is not too much over the desired speed, the driver may let friction and gravity slow the motorcycle down. Controller Desired speed Error  Driver Actuators Throttle or brakes Measurement Visual indication of speed Speedometer Process Motorcycle Actual motorcycle speed © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 4 CHAPTER 1 E1.8 Introduction to Control Systems Human biofeedback control system: Controller Desired body temp Process Hypothalumus  Message to blood vessels Actual body temp Human body Measurement Visual indication of body temperature E1.9 TV display Body sensor Eenabled aircraft with groundbased flight path control: Corrections to the flight path Desired Flight Path  Controller Aircraft Gc(s) G(s) Flight Path Health Parameters Meteorological data Location and speed Optimal flight path GroundBased Computer Network Optimal flight path Meteorological data Desired Flight Path E1.10 Specified Flight Trajectory Health Parameters Corrections to the flight path Gc(s) G(s) Controller Aircraft Location and speed Flight Path Unmanned aerial vehicle used for crop monitoring in an autonomous mode: Trajectory error  Controller UAV Gc(s) G(s) Flight Trajectory Sensor Location with respect to the ground Map Correlation Algorithm Ground photo Camera © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 5 Exercises E1.11 An inverted pendulum control system using an optical encoder to measure the angle of the pendulum and a motor producing a control torque: Actuator Voltage Error Desired angle  Controller Process Torque Motor Pendulum Angle Measurement Measured angle E1.12 In the video game, the player can serve as both the controller and the sensor. The objective of the game might be to drive a car along a prescribed path. The player controls the car trajectory using the joystick using the visual queues from the game displayed on the computer monitor. Controller Desired game objective Optical encoder Error  Player Actuator Joystick Measurement Player (eyesight, tactile, etc.) Process Video game Game objective © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 6 CHAPTER 1 Introduction to Control Systems Problems P1.1 Desired temperature set by the driver An automobile interior cabin temperature control system block diagram: Error  Controller Process Thermostat and air conditioning unit Automobile cabin Automobile cabin temperature Measurement Measured temperature P1.2 Temperature sensor A human operator controlled valve system: Controller Process Error * Desired fluid output *  Tank Valve Fluid output Measurement Visual indication of fluid output * Meter * = operator functions P1.3 A chemical composition control block diagram: Controller Process Error Desired chemical composition  Mixer tube Valve Measurement Measured chemical composition Infrared analyzer Chemical composition © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 7 Problems P1.4 A nuclear reactor control block diagram: Controller Process Error Desired power level Reactor and rods Motor and amplifier  Output power level Measurement Measured chemical composition P1.5 A light seeking control system to track the sun: Measurement Light source Dual Photocells P1.6 Ionization chamber Controller Ligh intensity Trajectory Planner Desired carriage position Controller  Motor, carriage, and gears K Photocell carriage position If you assume that increasing worker’s wages results in increased prices, then by delaying or falsifying costofliving data you could reduce or eliminate the pressure to increase worker’s wages, thus stabilizing prices. This would work only if there were no other factors forcing the costofliving up. Government price and wage economic guidelines would take the place of additional “controllers” in the block diagram, as shown in the block diagram. Controller Process Marketbased prices Initial wages Process Motor inputs Error  Industry Government price guidelines Controller Wage increases Government wage guidelines Costofliving K1 Prices © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 8 CHAPTER 1 P1.7 Introduction to Control Systems Assume that the cannon fires initially at exactly 5:00 p.m.. We have a positive feedback system. Denote by ∆t the time lost per day, and the net time error by ET . Then the follwoing relationships hold: ∆t = 4/3 min. + 3 min. = 13/3 min. and ET = 12 days × 13/3 min./day . Therefore, the net time error after 15 days is ET = 52 min. P1.8 The studentteacher learning process: Process Controller Lectures Error Desired knowledge  Teacher Knowledge Student Measurement Exams Measured knowledge P1.9 A human arm control system: Process Controller u Desired arm location e y s Brain Nerve signals z Measurement Visual indication of arm location Pressure Eyes and pressure receptors Arm & muscles d Arm location © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 9 Problems P1.10 An aircraft flight path control system using GPS: Controller Desired flight path from air traffic controllers Actuators Computer Autopilot Error  Process Ailerons, elevators, rudder, and engine power Flight path Aircraft Measurement Measured flight path P1.11 The accuracy of the clock is dependent upon a constant flow from the orifice; the flow is dependent upon the height of the water in the float tank. The height of the water is controlled by the float. The control system controls only the height of the water. Any errors due to enlargement of the orifice or evaporation of the water in the lower tank is not accounted for. The control system can be seen as: Desired height of the water in float tank P1.12 Global Positioning System  Controller Process Float level Flow from upper tank to float tank Actual height Assume that the turret and fantail are at 90◦ , if θw 6= θF 90◦ . The fantail operates on the error signal θw  θT , and as the fantail turns, it drives the turret to turn. y Wind qW = Wind angle qF = Fantail angle qT = Turret angle Controller * qW qF qT qW * Turret x  Process Torque Error Fantail Fantail Gears & turret qT © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 10 CHAPTER 1 P1.13 Introduction to Control Systems This scheme assumes the person adjusts the hot water for temperature control, and then adjusts the cold water for flow rate control. Controller Error Desired water temperature Process Hot water system Valve adjust  Hot water Actual water temperature and flow rate Desired water flow rate Cold water system Valve adjust  Cold water Measurement Measured water flow Measured water temperature P1.14 Human: visual and touch If the rewards in a specific trade is greater than the average reward, there is a positive influx of workers, since q(t) = f1 (c(t) − r(t)). If an influx of workers occurs, then reward in specific trade decreases, since c(t) = −f2 (q(t)). Controller Average rewards r(t) P1.15 Desired Fuel Pressure Error  f1(c(t)r(t)) Process q(t)  f2(q(t)) Total of rewards c(t) A computer controlled fuel injection system:  Controller Process Electronic Control Unit High Pressure Fuel Supply Pump and Electronic Fuel Injectors Measurement Measured fuel pressure Fuel Pressure Sensor Fuel Pressure © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 11 Problems P1.16 With the onset of a fever, the body thermostat is turned up. The body adjusts by shivering and less blood flows to the skin surface. Aspirin acts to lowers the thermal setpoint in the brain. Controller Desired temperature or setpoint from body thermostat in the brain Process Adjustments within the body  Body temperature Body Measurement Measured body temperature Internal sensor P1.17 Hitting a baseball is arguably one of the most difficult feats in all of sports. Given that pitchers may throw the ball at speeds of 90 mph (or higher!), batters have only about 0.1 second to make the decision to swing—with bat speeds aproaching 90 mph. The key to hitting a baseball a long distance is to make contact with the ball with a high bat velocity. This is more important than the bat’s weight, which is usually around 33 ounces (compared to Ty Cobb’s bat which was 41 ounces!). Since the pitcher can throw a variety of pitches (fast ball, curve ball, slider, etc.), a batter must decide if the ball is going to enter the strike zone and if possible, decide the type of pitch. The batter uses his/her vision as the sensor in the feedback loop. A high degree of eyehand coordination is key to success—that is, an accurate feedback control system. P1.18 Define the following variables: p = output pressure, fs = spring force = Kx, fd = diaphragm force = Ap, and fv = valve force = fs  fd . The motion of the valve is described by ÿ = fv /m where m is the valve mass. The output pressure is proportional to the valve displacement, thus p = cy , where c is the constant of proportionality. Constant of proportionality Spring Screw displacement x(t) K fs  Valve position fv Valve c y Diaphragm area fd A Output pressure p(t) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 12 CHAPTER 1 P1.19 Introduction to Control Systems A control system to keep a car at a given relative position offset from a lead car: Throttle Position of follower Follower car Actuator u  Controller Relative position  Position of lead Lead car Fuel throttle (fuel) Video camera & processing algorithms Reference photo Desired relative position P1.20 A control system for a highperformance car with an adjustable wing: Desired road adhesion  Process Actuator Controller Computer Adjustable wing Road conditions Race Car Road adhesion Measurement Measured road adhesion P1.21 K Tire internal strain gauges A control system for a twinlift helicopter system: Measurement Measured separation distance Desired separation distance  Controller Process Separation distance Pilot Desired altitude Radar Helicopter Altitude Measurement Measured altitude Altimeter © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 13 Problems P1.22 The desired building deflection would not necessarily be zero. Rather it would be prescribed so that the building is allowed moderate movement up to a point, and then active control is applied if the movement is larger than some predetermined amount. Process Controller Desired deflection Hydraulic stiffeners  Building Deflection Measurement Measured deflection P1.23 Strain gauges on truss structure K The humanlike face of the robot might have microactuators placed at strategic points on the interior of the malleable facial structure. Cooperative control of the microactuators would then enable the robot to achieve various facial expressions. Controller Process Error Desired actuator position  Voltage Electromechanical actuator Amplifier Actuator position Measurement Position sensor Measured position P1.24 We might envision a sensor embedded in a “gutter” at the base of the windshield which measures water levels—higher water levels corresponds to higher intensity rain. This information would be used to modulate the wiper blade speed. Process Controller Desired wiper speed Wiper blade and motor Electronic Control Unit  Measurement K Measured water level Water depth sensor Wiper blade speed © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 14 CHAPTER 1 Introduction to Control Systems A feedback control system for the space traffic control: P1.25 Controller Error Desired orbit position Control law  Actuator Jet commands Process Applied forces Reaction control jets Satellite Actual orbit position Measurement Measured orbit position Radar or GPS Earthbased control of a microrover to point the camera: P1.26 Microrover Camera position command Receiver/ Transmitter Controller G(s) Gc(s) Rover position Camera Camera Position m Ca ap er Sensor ea iti os M Measured camera position on d re su d an m m co ap er m ca on iti os P1.27 Desired Charge Level Control of a methanol fuel cell:  Controller Recharging System Gc(s) GR(s) Methanol water solution G(s) Sensor Measured charge level Fuel Cell H(s) Charge Level © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 15 Advanced Problems Advanced Problems AP1.1 Control of a robotic microsurgical device: Microsurgical robotic manipulator Controller Desired Endeffector Position  G(s) Gc(s) Endeffector Position Sensor H(s) AP1.2 An advanced wind energy system viewed as a mechatronic system: AERODYNAMIC DESIGN STRUCTURAL DESIGN OF THE TOWER ELECTRICAL AND POWER SYSTEMS SENSORS Rotor rotational sensor Wind speed and direction sensor ACTUATORS Motors for manipulatiing the propeller pitch Physical System Modeling CONTROL SYSTEM DESIGN AND ANALYSIS ELECTRICAL SYSTEM DESIGN AND ANALYSIS POWER GENERATION AND STORAGE Sensors and Actuators WIND ENERGY SYSTEM Software and Data Acquisition CONTROLLER ALGORITHMS DATA ACQUISTION: WIND SPEED AND DIRECTION ROTOR ANGULAR SPEED PROPELLOR PITCH ANGLE AP1.3 Signals and Systems Computers and Logic Systems COMPUTER EQUIPMENT FOR CONTROLLING THE SYSTEM SAFETY MONITORING SYSTEMS The automatic parallel parking system might use multiple ultrasound sensors to measure distances to the parked automobiles and the curb. The sensor measurements would be processed by an onboard computer to determine the steering wheel, accelerator, and brake inputs to avoid collision and to properly align the vehicle in the desired space. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 16 CHAPTER 1 Introduction to Control Systems Even though the sensors may accurately measure the distance between the two parked vehicles, there will be a problem if the available space is not big enough to accommodate the parking car. Controller Desired automobile position Error Actuators Onboard computer  Steering wheel, accelerator, and brake Process Actual automobile position Automobile Measurement Position of automobile relative to parked cars and curb Ultrasound There are various control methods that can be considered, including placing the controller in the feedforward loop (as in Figure 1.3). The adaptive optics block diagram below shows the controller in the feedback loop, as an alternative control system architecture. AP1.4 Process Astronomical object Uncompensated image Astronomical telescope mirror Compensated image Measurement Wavefront reconstructor Wavefront corrector Wavefront sensor Actuator & controller AP1.5 Desired floor Error  The control system might have an inner loop for controlling the acceleration and an outer loop to reach the desired floor level precisely. Controller #2 Outer Loop Desired acceleration Error  Controller #1 Elevator motor, cables, etc. Inner Loop Measured acceleration Acceleration Measurement Elevator Floor © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 17 Advanced Problems An obstacle avoidance control system would keep the robotic vacuum cleaner from colliding with furniture but it would not necessarily put the vacuum cleaner on an optimal path to reach the entire floor. This would require another sensor to measure position in the room, a digital map of the room layout, and a control system in the outer loop. AP1.6 Process Desired distance from obstacles Error  Controller Measured distance from obstacle Motors, wheels, etc. Infrared sensors Robotic vacuum cleaner Distance from obstacles © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 18 CHAPTER 1 Introduction to Control Systems Design Problems CDP1.1 The machine tool with the movable table in a feedback control configuration: Controller Error Desired position x Amplifier  Actuator Process Machine tool with table Positioning motor Actual position x Measurement Position sensor Measured position DP1.1 Use the stereo system and amplifiers to cancel out the noise by emitting signals 180◦ out of phase with the noise. Process Controller Noise signal Desired noise = 0 Shift phase by 180 deg  Machine tool with table Positioning motor Noise in cabin Measurement Microphone DP1.2 Desired speed of auto set by driver 1/K An automobile cruise control system: Controller Desired shaft speed  Electric motor Process Automobile and engine Valve Measurement Measured shaft speed Shaft speed sensor Drive shaf t speed K Actual speed of auto © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 19 Design Problems DP1.3 An automoted cow milking system: Measurement Cow location Vision system Motor and gears  Desired cup location Process Actuator Controller Location of cup Robot arm and cup gripper Cow and milker Milk Measurement Vision system Measured cup location DP1.4 A feedback control system for a robot welder: Controller Desired position Process Computer and amplifier Error  Voltage Motor and arm Weld top position Measurement Vision camera Measured position DP1.5 A control system for one wheel of a traction control system: Antislip controller Engine torque +  Wheel dynamics +  Wheel speed Sensor + Actual slip 1/Rw Vehicle dynamics Brake torque + Vehicle speed Antiskid controller Rw = Radius of wheel Sensor Measured slip © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 20 CHAPTER 1 Introduction to Control Systems A vibration damping system for the Hubble Space Telescope: DP1.6 Controller Desired jitter = 0 Error Computer  Actuators Gyro and reaction wheels Process Signal to cancel the jitter Spacecraft dynamics Jitter of vibration Measurement Measurement of 0.05 Hz jitter DP1.7 A control system for a nanorobot: Controller Desired nanorobot position Rate gyro sensor Error  Biocomputer Actuators Plane surfaces and propellers Process Nanorobot Actual nanorobot position Measurement External beacons Many concepts from underwater robotics can be applied to nanorobotics within the bloodstream. For example, plane surfaces and propellers can provide the required actuation with screw drives providing the propulsion. The nanorobots can use signals from beacons located outside the skin as sensors to determine their position. The nanorobots use energy from the chemical reaction of oxygen and glucose available in the human body. The control system requires a biocomputer–an innovation that is not yet available. For further reading, see A. Cavalcanti, L. Rosen, L. C. Kretly, M. Rosenfeld, and S. Einav, “Nanorobotic Challenges n Biomedical Application, Design, and Control,” IEEE ICECS Intl Conf. on Electronics, Circuits and Systems, TelAviv, Israel, December 2004. DP1.8 The feedback control system might use gyros and/or accelerometers to measure angle change and assuming the HTV was originally in the vertical position, the feedback would retain the vertical position using commands to motors and other actuators that produced torques and could move the HTV forward and backward. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 21 Design Problems Process Desired angle from vertical (0o) Error  Controller Measured angle from vertical Motors, wheels, etc. Gyros & accelerometers HTV Angle from vertical © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. C H A P T E R 2 Mathematical Models of Systems Exercises E2.1 We have for the openloop y = r2 and for the closedloop e = r − y and y = e2 . So, e = r − e2 and e2 + e − r = 0 . 16 14 12 y 10 8 openloop 6 4 closedloop 2 0 0 0.5 1 1.5 2 r 2.5 3 3.5 4 FIGURE E2.1 Plot of openloop versus closedloop. For example, if r = 1, then e2 + e − 1 = 0 implies that e = 0.618. Thus, y = 0.382. A plot y versus r is shown in Figure E2.1. 22 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 23 Exercises E2.2 Define f (T ) = R = R0 e−0.1T and ∆R = f (T ) − f (T0 ) , ∆T = T − T0 . Then, ∆R = f (T ) − f (T0 ) = ∂f ∂T T =T0 =20◦ ∆T + · · · where ∂f ∂T T =T0 =20◦ = −0.1R0 e−0.1T0 = −135, when R0 = 10, 000Ω. Thus, the linear approximation is computed by considering only the firstorder terms in the Taylor series expansion, and is given by ∆R = −135∆T . The spring constant for the equilibrium point is found graphically by estimating the slope of a line tangent to the force versus displacement curve at the point y = 0.5cm, see Figure E2.3. The slope of the line is K ≈ 1. 2 1.5 Spring breaks 1 0.5 0 Force (n) E2.3 0.5 1 1.5 2 2.5 3 2 Spring compresses 1.5 1 0.5 0 0.5 1 y=Displacement (cm) FIGURE E2.3 Spring force as a function of displacement. 1.5 2 2.5 3 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 24 CHAPTER 2 E2.4 Mathematical Models of Systems Since R(s) = 1 s we have Y (s) = 4(s + 50) . s(s + 20)(s + 10) The partial fraction expansion of Y (s) is given by Y (s) = A1 A2 A3 + + s s + 20 s + 10 where A1 = 1 , A2 = 0.6 and A3 = −1.6 . Using the Laplace transform table, we find that y(t) = 1 + 0.6e−20t − 1.6e−10t . The final value is computed using the final value theorem: 4(s + 50) lim y(t) = lim s =1. 2 t→∞ s→0 s(s + 30s + 200) E2.5 The circuit diagram is shown in Figure E2.5. R2 v+ A + vin  FIGURE E2.5 Noninverting opamp circuit. With an ideal opamp, we have vo = A(vin − v − ), + v0  R1 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 25 Exercises where A is very large. We have the relationship R1 vo . R1 + R2 v− = Therefore, vo = A(vin − R1 vo ), R1 + R2 and solving for vo yields vo = A 1+ AR1 R1 +R2 1 Since A ≫ 1, it follows that 1 + RAR ≈ 1 +R2 vo simplifies to vo = E2.6 vin . AR1 R1 +R2 . Then the expression for R1 + R2 vin . R1 Given y = f (x) = ex and the operating point xo = 1, we have the linear approximation y = f (x) = f (xo ) + ∂f ∂x x=xo (x − xo ) + · · · where df dx f (xo ) = e, = e, x=xo =1 and x − xo = x − 1. Therefore, we obtain the linear approximation y = ex. E2.7 The block diagram is shown in Figure E2.7. R(s) Ea(s) + G1(s) G2(s)  H(s) FIGURE E2.7 Block diagram model. I(s) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 26 CHAPTER 2 Mathematical Models of Systems Starting at the output we obtain I(s) = G1 (s)G2 (s)E(s). But E(s) = R(s) − H(s)I(s), so I(s) = G1 (s)G2 (s) [R(s) − H(s)I(s)] . Solving for I(s) yields the closedloop transfer function G1 (s)G2 (s) I(s) = . R(s) 1 + G1 (s)G2 (s)H(s) E2.8 The block diagram is shown in Figure E2.8. H2(s)  R(s) K  E(s)  G1(s) W(s)  A(s) G2(s) Z(s) 1 s Y(s) H3(s) H1(s) FIGURE E2.8 Block diagram model. Starting at the output we obtain Y (s) = 1 1 Z(s) = G2 (s)A(s). s s But A(s) = G1 (s) [−H2 (s)Z(s) − H3 (s)A(s) + W (s)] and Z(s) = sY (s), so 1 Y (s) = −G1 (s)G2 (s)H2 (s)Y (s) − G1 (s)H3 (s)Y (s) + G1 (s)G2 (s)W (s). s Substituting W (s) = KE(s) − H1 (s)Z(s) into the above equation yields Y (s) = −G1 (s)G2 (s)H2 (s)Y (s) − G1 (s)H3 (s)Y (s) 1 + G1 (s)G2 (s) [KE(s) − H1 (s)Z(s)] s © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 27 Exercises and with E(s) = R(s) − Y (s) and Z(s) = sY (s) this reduces to Y (s) = [−G1 (s)G2 (s) (H2 (s) + H1 (s)) − G1 (s)H3 (s) 1 1 − G1 (s)G2 (s)K]Y (s) + G1 (s)G2 (s)KR(s). s s Solving for Y (s) yields the transfer function Y (s) = T (s)R(s), where T (s) = E2.9 KG1 (s)G2 (s)/s . 1 + G1 (s)G2 (s) [(H2 (s) + H1 (s)] + G1 (s)H3 (s) + KG1 (s)G2 (s)/s From Figure E2.9, we observe that Ff (s) = G2 (s)U (s) and FR (s) = G3 (s)U (s) . Then, solving for U (s) yields U (s) = 1 Ff (s) G2 (s) FR (s) = G3 (s) U (s) . G2 (s) and it follows that Again, considering the block diagram in Figure E2.9 we determine Ff (s) = G1 (s)G2 (s)[R(s) − H2 (s)Ff (s) − H2 (s)FR (s)] . But, from the previous result, we substitute for FR (s) resulting in Ff (s) = G1 (s)G2 (s)R(s)−G1 (s)G2 (s)H2 (s)Ff (s)−G1 (s)H2 (s)G3 (s)Ff (s) . Solving for Ff (s) yields G1 (s)G2 (s) Ff (s) = R(s) . 1 + G1 (s)G2 (s)H2 (s) + G1 (s)G3 (s)H2 (s) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 28 CHAPTER 2 Mathematical Models of Systems H2(s) +  R(s) U(s) G2(s) Ff (s) U(s) G3(s) FR(s) G1(s)  H2(s) FIGURE E2.9 Block diagram model. E2.10 The shock absorber block diagram is shown in Figure E2.10. The closedloop transfer function model is T (s) = Gc (s)Gp (s)G(s) . 1 + H(s)Gc (s)Gp (s)G(s) Controller Gear Motor Plunger and Piston System Gc(s) Gp(s) G(s) + R(s) Desired piston travel  Y(s) Piston travel Sensor H(s) Piston travel measurement FIGURE E2.10 Shock absorber block diagram. E2.11 Let f denote the spring force (n) and x denote the deflection (m). Then K= ∆f . ∆x Computing the slope from the graph yields: (a) xo = −0.14m → K = ∆f /∆x = 10 n / 0.04 m = 250 n/m (b) xo = 0m → K = ∆f /∆x = 10 n / 0.05 m = 200 n/m (c) xo = 0.35m → K = ∆f /∆x = 3n / 0.05 m = 60 n/m © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 29 Exercises E2.12 The signal flow graph is shown in Fig. E2.12. Find Y (s) when R(s) = 0. K Td(s) 1 1 K2 G(s) Y (s) 1 FIGURE E2.12 Signal flow graph. The transfer function from Td (s) to Y (s) is Y (s) = G(s)(1 − K1 K2 )Td (s) G(s)Td (s) − K1 K2 G(s)Td (s) = . 1 − (−K2 G(s)) 1 + K2 G(s) If we set K1 K2 = 1 , then Y (s) = 0 for any Td (s). E2.13 The transfer function from R(s), Td (s), and N (s) to Y (s) is K K 1 R(s)+ 2 Td (s)− 2 N (s) Y (s) = 2 s + 10s + K s + 10s + K s + 10s + K Therefore, we find that Y (s)/Td (s) = E2.14 s2 1 + 10s + K and Y (s)/N (s) = − s2 K + 10s + K Since we want to compute the transfer function from R2 (s) to Y1 (s), we can assume that R1 = 0 (application of the principle of superposition). Then, starting at the output Y1 (s) we obtain Y1 (s) = G3 (s) [−H1 (s)Y1 (s) + G2 (s)G8 (s)W (s) + G9 (s)W (s)] , or [1 + G3 (s)H1 (s)] Y1 (s) = [G3 (s)G2 (s)G8 (s)W (s) + G3 (s)G9 (s)] W (s). Considering the signal W (s) (see Figure E2.14), we determine that W (s) = G5 (s) [G4 (s)R2 (s) − H2 (s)W (s)] , © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 30 CHAPTER 2 Mathematical Models of Systems H1(s) + G1(s) R1(s) + G7(s) R2(s) G4(s)  + G2(s) G3(s) + Y1(s) G9(s) G8(s) + + G6(s) G5(s) Y2(s) W(s)  H2(s) FIGURE E2.14 Block diagram model. or [1 + G5 (s)H2 (s)] W (s) = G5 (s)G4 (s)R2 (s). Substituting the expression for W (s) into the above equation for Y1 (s) yields Y1 (s) G2 (s)G3 (s)G4 (s)G5 (s)G8 (s) + G3 (s)G4 (s)G5 (s)G9 (s) = . R2 (s) 1 + G3 (s)H1 (s) + G5 (s)H2 (s) + G3 (s)G5 (s)H1 (s)H2 (s) E2.15 For loop 1, we have di1 1 R1 i1 + L1 + dt C1 Z (i1 − i2 )dt + R2 (i1 − i2 ) = v(t) . And for loop 2, we have 1 C2 E2.16 Z di2 1 i2 dt + L2 + R2 (i2 − i1 ) + dt C1 Z (i2 − i1 )dt = 0 . The transfer function from R(s) to P (s) is P (s) 4.2 = 3 . 2 R(s) s + 2s + 4s + 4.2 The block diagram is shown in Figure E2.16a. The corresponding signal flow graph is shown in Figure E2.16b for P (s)/R(s) = s3 + 4.2 . + 4s + 4.2 2s2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 31 Exercises v1(s) R(s) v2(s) 7  q(s) 0.6 s 1 s2+2s+4 P(s) (a) R(s ) 1 V1 7 1 s2 + 2 s + 4 0.6 s V2 P (s) 1 (b) FIGURE E2.16 (a) Block diagram, (b) Signal flow graph. E2.17 A linear approximation for f is given by ∆f = ∂f ∂x ∆x = 2kxo ∆x = k∆x x=xo where xo = 1/2, ∆f = f (x) − f (xo ), and ∆x = x − xo . E2.18 The linear approximation is given by ∆y = m∆x where m= ∂y ∂x . x=xo (a) When xo = 1, we find that yo = 2.4, and yo = 13.2 when xo = 2. (b) The slope m is computed as follows: m= ∂y ∂x = 1 + 4.2x2o . x=xo Therefore, m = 5.2 at xo = 1, and m = 18.8 at xo = 2. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 32 CHAPTER 2 E2.19 Mathematical Models of Systems The output (with a step input) is Y (s) = 15(s + 1) . s(s + 7)(s + 2) The partial fraction expansion is 18 1 3 1 15 − + . 14s 7 s+7 2s+2 Y (s) = Taking the inverse Laplace transform yields y(t) = E2.20 15 18 −7t 3 −2t − e + e . 14 7 2 The inputoutput relationship is A(K − 1) Vo = V 1 + AK where K= Z1 . Z1 + Z2 Assume A ≫ 1. Then, Vo K−1 Z2 = =− V K Z1 where Z1 = R1 R1 C 1 s + 1 and Z2 = R2 . R2 C 2 s + 1 Therefore, Vo (s) R2 (R1 C1 s + 1) 2(s + 1) =− =− . V (s) R1 (R2 C2 s + 1) s+2 E2.21 The equation of motion of the mass mc is mc ẍp + (bd + bs )ẋp + kd xp = bd ẋin + kd xin . Taking the Laplace transform with zero initial conditions yields [mc s2 + (bd + bs )s + kd ]Xp (s) = [bd s + kd ]Xin (s) . So, the transfer function is bd s + kd 0.7s + 2 Xp (s) = = 2 . Xin (s) mc s2 + (bd + bs )s + kd s + 2.8s + 2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 33 Exercises E2.22 The rotational velocity is ω(s) = 2(s + 4) 1 . 2 (s + 5)(s + 1) s Expanding in a partial fraction expansion yields ω(s) = 81 1 1 3 13 1 1 + − − . 2 5 s 40 s + 5 2 (s + 1) 8 s+1 Taking the inverse Laplace transform yields ω(t) = E2.23 8 1 3 13 + e−5t − te−t − e−t . 5 40 2 8 The closedloop transfer function is Y (s) K1 K2 = T (s) = 2 . R(s) s + (K1 + K2 K3 + K1 K2 )s + K1 K2 K3 E2.24 The closedloop tranfser function is Y (s) 10 = T (s) = 2 . R(s) s + 21s + 10 E2.25 Let x = 0.6 and y = 0.8. Then, with y = ax3 , we have 0.8 = a(0.6)3 . Solving for a yields a = 3.704. A linear approximation is y − yo = 3ax2o (x − xo ) or y = 4x − 1.6, where yo = 0.8 and xo = 0.6. E2.26 The equations of motion are m1 ẍ1 + k(x1 − x2 ) = F m2 ẍ2 + k(x2 − x1 ) = 0 . Taking the Laplace transform (with zero initial conditions) and solving for X2 (s) yields X2 (s) = (m2 s2 k F (s) . + k)(m1 s2 + k) − k 2 Then, with m1 = m2 = k = 1, we have X2 (s)/F (s) = 1 . s2 (s2 + 2) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 34 CHAPTER 2 E2.27 Mathematical Models of Systems The transfer function from Td (s) to Y (s) is Y (s)/Td (s) = E2.28 G2 (s) . 1 + G1 G2 H(s) The transfer function is R2 R4 C R2 R4 Vo (s) = s+ = 24s + 144 . V (s) R3 R1 R3 E2.29 (a) If G(s) = s2 1 + 15s + 50 and H(s) = 2s + 15 , then the closedloop transfer function of Figure E2.28(a) and (b) (in Dorf & Bishop) are equivalent. (b) The closedloop transfer function is T (s) = (a) The closedloop transfer function is T (s) = G(s) 1 10 = 2 1 + G(s) s s(s + 2s + 20) where G(s) = 0.8 0.7 0.6 Amplitude E2.30 1 . s2 + 17s + 65 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 Time sec FIGURE E2.30 Step response. 4 5 6 s2 10 . + 2s + 10 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 35 Exercises (b) The output Y (s) (when R(s) = 1/s) is Y (s) = 0.5 −0.25 + 0.0573j −0.25 − 0.0573j − + . s s + 1 − 4.3589j s + 1 + 4.3589j (c) The plot of y(t) is shown in Figure E2.30. The output is given by √ √ 1 1 y(t) = 1 − e−t cos 19t − √ sin 19t 2 19 E2.31 The partial fraction expansion is V (s) = a b + s + p1 s + p2 where p1 = 4 − 19.6j and p2 = 4 + 19.6j. Then, the residues are a = −10.2j b = 10.2j . The inverse Laplace transform is v(t) = −10.2je(−4+19.6j)t + 10.2je(−4−19.6j)t = 20.4e−4t sin 19.6t . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 36 CHAPTER 2 Mathematical Models of Systems Problems P2.1 The integrodifferential equations, obtained by Kirchoff’s voltage law to each loop, are as follows: R1 i1 + 1 C1 R3 i2 + 1 C2 Z i1 dt + L1 d(i1 − i2 ) + R2 (i1 − i2 ) = v(t) dt (loop 1) and P2.2 Z i2 dt + R2 (i2 − i1 ) + L1 d(i2 − i1 ) =0 dt (loop 2) . The differential equations describing the system can be obtained by using a freebody diagram analysis of each mass. For mass 1 and 2 we have M1 ÿ1 + k12 (y1 − y2 ) + bẏ1 + k1 y1 = F (t) M2 ÿ2 + k12 (y2 − y1 ) = 0 . Using a forcecurrent analogy, the analagous electric circuit is shown in Figure P2.2, where Ci → Mi , L1 → 1/k1 , L12 → 1/k12 , and R → 1/b . FIGURE P2.2 Analagous electric circuit. P2.3 The differential equations describing the system can be obtained by using a freebody diagram analysis of each mass. For mass 1 and 2 we have M ẍ1 + kx1 + k(x1 − x2 ) = F (t) M ẍ2 + k(x2 − x1 ) + bẋ2 = 0 . Using a forcecurrent analogy, the analagous electric circuit is shown in Figure P2.3, where C→M L → 1/k R → 1/b . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 37 Problems FIGURE P2.3 Analagous electric circuit. (a) The linear approximation around vin = 0 is vo = 0vin , see Figure P2.4(a). (b) The linear approximation around vin = 1 is vo = 2vin − 1, see Figure P2.4(b). (a) (b) 0.4 4 3.5 0.3 3 0.2 2.5 0.1 2 vo vo P2.4 0 1.5 linear approximation 1 0.1 0.5 0.2 0 0.3 0.4 1 linear approximation 0.5 0.5 0 vin 0.5 FIGURE P2.4 Nonlinear functions and approximations. 1 1 1 0 1 vin 2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 38 CHAPTER 2 P2.5 Mathematical Models of Systems Given Q = K(P1 − P2 )1/2 . Let δP = P1 − P2 and δPo = operating point. Using a Taylor series expansion of Q, we have Q = Qo + ∂Q ∂δP (δP − δPo ) + · · · δP =δPo where Qo = KδPo1/2 ∂Q ∂δP and = δP =δPo K −1/2 . δP 2 o Define ∆Q = Q − Qo and ∆P = δP − δPo . Then, dropping higherorder terms in the Taylor series expansion yields ∆Q = m∆P where m= P2.6 K 1/2 2δPo . From P2.1 we have R1 i1 + 1 C1 R3 i2 + 1 C2 Z i1 dt + L1 d(i1 − i2 ) + R2 (i1 − i2 ) = v(t) dt and Z i2 dt + R2 (i2 − i1 ) + L1 d(i2 − i1 ) =0. dt Taking the Laplace transform and using the fact that the initial voltage across C2 is 10v yields [R1 + 1 + L1 s + R2 ]I1 (s) + [−R2 − L1 s]I2 (s) = 0 C1 s and [−R2 − L1 s]I1 (s) + [L1 s + R3 + 1 10 + R2 ]I2 (s) = − . C2 s s Rewriting in matrix form we have R1 + 1 C1 s + L 1 s + R2 −R2 − L1 s −R2 − L1 s L 1 s + R3 + 1 C2 s + R2 I1 (s) I2 (s) = 0 −10/s . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 39 Problems Solving for I2 yields 1 0 R2 + L 1 s 1 L1 s + R3 + C2 s + R2 = . 1 ∆ −10/s I2 (s) R2 + L 1 s R1 + C1 s + L1 s + R2 I1 (s) or I2 (s) = −10(R1 + 1/C1 s + L1 s + R2 ) s∆ where ∆ = (R1 + P2.7 1 1 + L1 s + R2 )(L1 s + R3 + + R2 ) − (R2 + L1 s)2 . C1 s C2 s Consider the differentiating opamp circuit in Figure P2.7. For an ideal opamp, the voltage gain (as a function of frequency) is V2 (s) = − Z2 (s) V1 (s), Z1 (s) where Z1 = R1 1 + R1 Cs and Z2 = R2 are the respective circuit impedances. Therefore, we obtain V2 (s) = − Z R2 (1 + R1 Cs) V1 (s). R1 Z 1 C + R1 2 R2 + + V1(s) V2(s)   FIGURE P2.7 Differentiating opamp circuit. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 40 CHAPTER 2 Let ∆= G2 + Cs −Cs −G2 −Cs G1 + 2Cs −Cs −G2 −Cs Cs + G2 . Then, Vj = ∆ij I1 ∆ or or V3 ∆13 I1 /∆ = . V1 ∆11 I1 /∆ Therefore, the transfer function is −Cs 2Cs + G1 T (s) = ∆13 V3 = = V1 ∆11 −G2 −Cs 2Cs + G1 −Cs −Cs Cs + G2 Polezero map (x:poles and o:zeros) 3 2 o 1 Imag Axis P2.8 Mathematical Models of Systems 0 x x 1 2 3 8 o 7 6 5 4 Real Axis FIGURE P2.8 Polezero map. 3 2 1 0 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 41 Problems = C 2 R1 R2 s2 + 2CR1 s + 1 . C 2 R1 R2 s2 + (2R1 + R2 )Cs + 1 Using R1 = 0.5, R2 = 1, and C = 0.5, we have T (s) = s2 + 4s + 8 (s + 2 + 2j)(s + 2 − 2j) √ √ . = 2 s + 8s + 8 (s + 4 + 8)(s + 4 − 8) The polezero map is shown in Figure P2.8. From P2.3 we have M ẍ1 + kx1 + k(x1 − x2 ) = F (t) M ẍ2 + k(x2 − x1 ) + bẋ2 = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that M s2 + 2k −k M s2 + bs + k −k X1 (s) X2 (s) = F (s) 0 , Pole zero map 0.4 0.3 0.2 0.1 Imag Axis P2.9 0  0.1 0.2 0.3 0.4 0.03 FIGURE P2.9 Polezero map. 0.025 0.02 0.015 Real Axis 0.01 0.005 0 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 42 CHAPTER 2 Mathematical Models of Systems or k F (s) 1 M s2 + bs + k = 2 ∆ X2 (s) k M s + 2k 0 X1 (s) where ∆ = (M s2 + bs + k)(M s2 + 2k) − k 2 . So, G(s) = M s2 + bs + k X1 (s) = . F (s) ∆ When b/k = 1, M = 1 , b2 /M k = 0.04, we have G(s) = s2 + 0.04s + 0.04 . s4 + 0.04s3 + 0.12s2 + 0.0032s + 0.0016 The polezero map is shown in Figure P2.9. P2.10 From P2.2 we have M1 ÿ1 + k12 (y1 − y2 ) + bẏ1 + k1 y1 = F (t) M2 ÿ2 + k12 (y2 − y1 ) = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that or M1 s2 + bs + k1 + k12 M2 s2 + k12 −k12 −k12 Y1 (s) Y2 (s) = F (s) 0 k12 F (s) 1 M2 s2 + k12 = ∆ Y2 (s) k12 M1 s2 + bs + k1 + k12 0 Y1 (s) where 2 ∆ = (M2 s2 + k12 )(M1 s2 + bs + k1 + k12 ) − k12 . So, when f (t) = a sin ωo t, we have that Y1 (s) is given by Y1 (s) = aM2 ωo (s2 + k12 /M2 ) . (s2 + ωo2 )∆(s) For motionless response (in the steadystate), set the zero of the transfer function so that (s2 + k12 ) = s2 + ωo2 M2 or ωo2 = k12 . M2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 43 Problems P2.11 The transfer functions from Vc (s) to Vd (s) and from Vd (s) to θ(s) are: K1 K2 , and (Lq s + Rq )(Lc s + Rc ) Km . θ(s)/Vd (s) = 2 (Js + f s)((Ld + La )s + Rd + Ra ) + K3 Km s Vd (s)/Vc (s) = The block diagram for θ(s)/Vc (s) is shown in Figure P2.11, where θ(s)/Vc (s) = K1 K2 Km θ(s) Vd (s) = , Vd (s) Vc (s) ∆(s) where ∆(s) = s(Lc s + Rc )(Lq s + Rq )((Js + b)((Ld + La )s + Rd + Ra ) + Km K3 ) . Vc 1 L cs+R c Ic K1 Vq 1 L qs+R q Iq K2 Vd + 1 (L d+L a)s+R d+R a Id Tm Km  1 Js+f w 1 s q Vb K3 FIGURE P2.11 Block diagram. P2.12 The openloop transfer function is Y (s) K = . R(s) s + 20 With R(s) = 1/s, we have Y (s) = K . s(s + 20) The partial fraction expansion is K Y (s) = 20 1 1 − , s s + 20 and the inverse Laplace transform is y(t) = K 1 − e−20t , 20 As t → ∞, it follows that y(t) → K/20. So we choose K = 20 so that y(t) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 44 CHAPTER 2 Mathematical Models of Systems approaches 1. Alternatively we can use the final value theorem to obtain y(t)t→∞ = lim sY (s) = s→0 K =1. 20 It follows that choosing K = 20 leads to y(t) → 1 as t → ∞. P2.13 The motor torque is given by Tm (s) = (Jm s2 + bm s)θm (s) + (JL s2 + bL s)nθL (s) = n((Jm s2 + bm s)/n2 + JL s2 + bL s)θL (s) where n = θL (s)/θm (s) = gear ratio . But Tm (s) = Km Ig (s) and Ig (s) = 1 Vg (s) , (Lg + Lf )s + Rg + Rf and Vg (s) = Kg If (s) = Kg Vf (s) . Rf + L f s Combining the above expressions yields θL (s) Kg Km = . Vf (s) n∆1 (s)∆2 (s) where ∆1 (s) = JL s2 + bL s + Jm s2 + bm s n2 and ∆2 (s) = (Lg s + Lf s + Rg + Rf )(Rf + Lf s) . P2.14 For a fieldcontrolled dc electric motor we have ω(s)/Vf (s) = Km /Rf . Js + b © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 45 Problems With a step input of Vf (s) = 80/s, the final value of ω(t) is 80Km = 2.4 Rf b ω(t)t→∞ = lim sω(s) = s→0 or Km = 0.03 . Rf b Solving for ω(t) yields 80Km −1 1 ω(t) = L Rf J s(s + b/J) = 80Km (1−e−(b/J)t ) = 2.4(1−e−(b/J)t ) . Rf b At t = 1/2, ω(t) = 1, so ω(1/2) = 2.4(1 − e−(b/J)t ) = 1 implies b/J = 1.08 sec . Therefore, ω(s)/Vf (s) = P2.15 0.0324 . s + 1.08 Summing the forces in the vertical direction and using Newton’s Second Law we obtain ẍ + k x=0. m The system has no damping and no external inputs. Taking the Laplace transform yields X(s) = s2 x0 s , + k/m where we used the fact that x(0) = x0 and ẋ(0) = 0. Then taking the inverse Laplace transform yields x(t) = x0 cos P2.16 s k t. m Using Cramer’s rule, we have 1 1.5 x1 or 2 4 x1 x2 = 6 11 1 4 −1.5 6 = ∆ −2 x2 1 11 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 46 CHAPTER 2 Mathematical Models of Systems where ∆ = 4(1) − 2(1.5) = 1 . Therefore, x1 = 4(6) − 1.5(11) = 7.5 1 and x2 = −2(6) + 1(11) = −1 . 1 The signal flow graph is shown in Figure P2.16. 11 1/4 6 1 1/2 X2 X1 1.5 FIGURE P2.16 Signal flow graph. So, x1 = P2.17 6(1) − 1.5( 11 4 ) = 7.5 3 1− 4 and x2 = 11( 41 ) + 1− −1 2 (6) 3 4 = −1 . (a) For mass 1 and 2, we have M1 ẍ1 + K1 (x1 − x2 ) + b1 (ẋ3 − ẋ1 ) = 0 M2 ẍ2 + K2 (x2 − x3 ) + b2 (ẋ3 − ẋ2 ) + K1 (x2 − x1 ) = 0 . (b) Taking the Laplace transform yields (M1 s2 + b1 s + K1 )X1 (s) − K1 X2 (s) = b1 sX3 (s) −K1 X1 (s) + (M2 s2 + b2 s + K1 + K2 )X2 (s) = (b2 s + K2 )X3 (s) . (c) Let G1 (s) = K2 + b2 s G2 (s) = 1/p(s) G3 (s) = 1/q(s) G4 (s) = sb1 , where p(s) = s2 M2 + sf2 + K1 + K2 and q(s) = s2 M1 + sf1 + K1 . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 47 Problems The signal flow graph is shown in Figure P2.17. G4 G3 X3 G1 G2 X1 K1 K1 FIGURE P2.17 Signal flow graph. (d) The transfer function from X3 (s) to X1 (s) is X1 (s) K1 G1 (s)G2 (s)G3 (s) + G4 (s)G3 (s) = . X3 (s) 1 − K12 G2 (s)G3 (s) P2.18 The signal flow graph is shown in Figure P2.18. I1 V1 Va Z2 Y3 Ia Z4 V2 Y1 Z 2 Y 1 Y 3 FIGURE P2.18 Signal flow graph. The transfer function is V2 (s) Y 1 Z2 Y 3 Z4 = . V1 (s) 1 + Y 1 Z2 + Y 3 Z2 + Y 3 Z4 + Y 1 Z2 Z4 Y 3 P2.19 For a noninerting opamp circuit, depicted in Figure P2.19a, the voltage gain (as a function of frequency) is Vo (s) = Z1 (s) + Z2 (s) Vin (s), Z1 (s) where Z1 (s) and Z2 (s) are the impedances of the respective circuits. In the case of the voltage follower circuit, shown in Figure P2.19b, we have © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 48 CHAPTER 2 Mathematical Models of Systems Z2 Z1 + vin v0 + vin (a) v0 (b) FIGURE P2.19 (a) Noninverting opamp circuit. (b) Voltage follower circuit. Z1 = ∞ (open circuit) and Z2 = 0. Therefore, the transfer function is Vo (s) Z1 = = 1. Vin (s) Z1 P2.20 (a) Assume Rg ≫ Rs and Rs ≫ R1 . Then Rs = R1 + R2 ≈ R2 , and vgs = vin − vo , where we neglect iin , since Rg ≫ Rs . At node S, we have vo = gm vgs = gm (vin − vo ) or Rs vo gm Rs = . vin 1 + gm Rs (b) With gm Rs = 20, we have vo 20 = = 0.95 . vin 21 (c) The block diagram is shown in Figure P2.20. vin(s) gmRs  FIGURE P2.20 Block diagram model. P2.21 From the geometry we find that ∆z = k l1 − l2 l2 (x − y) − y . l1 l1 vo(s) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 49 Problems The flow rate balance yields A dy = p∆z dt which implies Y (s) = p∆Z(s) . As By combining the above results it follows that l2 p l1 − l2 Y (s) = k (X(s) − Y (s)) − Y (s) . As l1 l1 Therefore, the signal flow graph is shown in Figure P2.21. Using Mason’s 1 (l 1  l 2)/l 1 k X DZ p/As Y 1 l 2 / l 1 FIGURE P2.21 Signal flow graph. gain formula we find that the transfer function is given by Y (s) = X(s) 1+ k(l1 −l2 )p l1 As k(l1 −l2 )p l2 p l1 As + l1 As = K1 , s + K2 + K1 where K1 = P2.22 k(l1 − l2 )p p l1 A and K2 = l2 p . l1 A (a) The equations of motion for the two masses are L 2 L M L θ¨1 + M gLθ1 + k (θ1 − θ2 ) = f (t) 2 2 2 L M L2 θ¨2 + M gLθ2 + k (θ2 − θ1 ) = 0 . 2 2 With θ˙1 = ω1 and θ˙2 = ω2 , we have g k k f (t) ω˙1 = − + θ1 + θ2 + L 4M 4M 2M L k g k ω˙2 = θ1 − + θ2 . 4M L 4M © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 50 CHAPTER 2 Mathematical Models of Systems a  F (t) w1 1/s 1/s 1/2ML (a) q1 b w2 1/s 1/s q2 a Imag(s) + j + j (b) g k L + 4M g k L + 2M X O X + j g L Re(s) FIGURE P2.22 (a) Block diagram. (b) Polezero map. (b) Define a = g/L + k/4M and b = k/4M . Then θ1 (s) 1 s2 + a = . F (s) 2M L (s2 + a)2 − b2 (c) The block diagram and polezero map are shown in Figure P2.22. P2.23 The inputoutput ratio, Vce /Vin , is found to be β(R − 1) + hie Rf Vce = . Vin −βhre + hie (−hoe + Rf ) P2.24 (a) The voltage gain is given by vo RL β1 β2 (R1 + R2 ) . = vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + β1 ) + R1 RL β1 β2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 51 Problems (b) The current gain is found to be ic2 = β1 β2 . ib1 (c) The input impedance is vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + β1 ) + R1 RL β1 β2 = , ib1 R1 + R2 and when β1 β2 is very large, we have the approximation vin RL R1 β1 β2 ≈ . ib1 R1 + R2 P2.25 The transfer function from R(s) and Td (s) to Y (s) is given by Y (s) = G(s) R(s) − 1 (G(s)R(s) + Td (s)) + Td (s) + G(s)R(s) G(s) = G(s)R(s) . Thus, Y (s)/R(s) = G(s) . Also, we have that Y (s) = 0 . when R(s) = 0. Therefore, the effect of the disturbance, Td (s), is eliminated. P2.26 The equations of motion for the two mass model of the robot are M ẍ + b(ẋ − ẏ) + k(x − y) = F (t) mÿ + b(ẏ − ẋ) + k(y − x) = 0 . Taking the Laplace transform and writing the result in matrix form yields M s2 + bs + k −(bs + k) −(bs + k) ms2 + bs + k X(s) Y (s) k m . = Solving for Y (s) we find that 1 Y (s) mM (bs +k) = m b F (s) s2 [s2 + 1 + M ms + ] F (s) 0 . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 52 CHAPTER 2 P2.27 Mathematical Models of Systems The describing equation of motion is mz̈ = mg − k i2 . z2 Defining f (z, i) = g − ki2 mz 2 leads to z̈ = f (z, i) . The equilibrium condition for io and zo , found by solving the equation of motion when ż = z̈ = 0 , is ki2o = zo2 . mg We linearize the equation of motion using a Taylor series approximation. With the definitions ∆z = z − zo and ∆i = i − io , ˙ = ż and ∆z ¨ = z̈. Therefore, we have ∆z ¨ = f (z, i) = f (zo , io ) + ∂f ∆z ∂z z=zo i=io ∆z + ∂f ∂i z=zo i=io ∆i + · · · But f (zo , io ) = 0, and neglecting higherorder terms in the expansion yields 2 ¨ = 2kio ∆z − 2kio ∆i . ∆z mzo3 mzo2 Using the equilibrium condition which relates zo to io , we determine that ¨ = 2g ∆z − g ∆i . ∆z zo io Taking the Laplace transform yields the transfer function (valid around the equilibrium point) ∆Z(s) −g/io = 2 . ∆I(s) s − 2g/zo © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 57 Problems R(s ) + K1 s (s+1) Y (s) 1 +K 2s 1 0.9 0.8 0.7 < time to 90% = 0.39 sec y(t) 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 time(sec) FIGURE P2.35 The equivalent block diagram and the system step response. P2.36 (a) Given R(s) = 1/s2 , the partial fraction expansion is Y (s) = s2 (s 24 3 8/3 3/4 1 13/12 = − + + 2− . + 2)(s + 3)(s + 4) s+2 s+3 s+4 s s Therefore, using the Laplace transform table, we determine that the ramp response is 8 3 13 y(t) = 3e−2t − e−3t + e−4t + t − , 3 4 12 t≥0. (b) For the ramp input, y(t) ≈ 0.21 at t = 1. second (see Figure P2.36a). (c) Given R(s) = 1, the partial fraction expansion is Y (s) = 24 12 24 12 = − + . (s + 2)(s + 3)(s + 4) s+2 s+3 s+4 Therefore, using the Laplace transform table, we determine that the impulse response is y(t) = 12e−2t − 24e−3t + 412e−4t , t≥0. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 53 Problems P2.28 The signal flow graph is shown in Figure P2.28. d G B +b P +c D +a m +e k M +g +f +h S C FIGURE P2.28 Signal flow graph. (a) The PGBDP loop gain is equal to abcd. This is a negative transmission since the population produces garbage which increases bacteria and leads to diseases, thus reducing the population. (b) The PMCP loop gain is equal to +efg. This is a positive transmission since the population leads to modernization which encourages immigration, thus increasing the population. (c) The PMSDP loop gain is equal to +ehkd. This is a positive transmission since the population leads to modernization and an increase in sanitation facilities which reduces diseases, thus reducing the rate of decreasing population. (d) The PMSBDP loop gain is equal to +ehmcd. This is a positive transmission by similar argument as in (3). P2.29 Assume the motor torque is proportional to the input current Tm = ki . Then, the equation of motion of the beam is J φ̈ = ki , where J is the moment of inertia of the beam and shaft (neglecting the inertia of the ball). We assume that forces acting on the ball are due to gravity and friction. Hence, the motion of the ball is described by mẍ = mgφ − bẋ © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 54 CHAPTER 2 Mathematical Models of Systems where m is the mass of the ball, b is the coefficient of friction, and we have assumed small angles, so that sin φ ≈ φ. Taking the Laplace transfor of both equations of motion and solving for X(s) yields X(s)/I(s) = P2.30 gk/J . + b/m) s2 (s2 Given H(s) = k τs + 1 where τ = 4µs = 4 × 10−6 seconds and 0.999 ≤ k < 1.001. The step response is Y (s) = k 1 k k · = − . τs + 1 s s s + 1/τ Taking the inverse Laplace transform yields y(t) = k − ke−t/τ = k(1 − e−t/τ ) . The final value is k. The time it takes to reach 98% of the final value is t = 15.6µs independent of k. P2.31 From the block diagram we have Y1 (s) = G2 (s)[G1 (s)E1 (s) + G3 (s)E2 (s)] = G2 (s)G1 (s)[R1 (s) − H1 (s)Y1 (s)] + G2 (s)G3 (s)E2 (s) . Therefore, Y1 (s) = G1 (s)G2 (s) G2 (s)G3 (s) R1 (s) + E2 (s) . 1 + G1 (s)G2 (s)H1 (s) 1 + G1 (s)G2 (s)H1 (s) And, computing E2 (s) (with R2 (s) = 0) we find G4 (s) E2 (s) = H2 (s)Y2 (s) = H2 (s)G6 (s) Y1 (s) + G5 (s)E2 (s) G2 (s) or E2 (s) = G4 (s)G6 (s)H2 (s) Y1 (s) . G2 (s)(1 − G5 (s)G6 (s)H2 (s)) Substituting E2 (s) into equation for Y1 (s) yields Y1 (s) = G1 (s)G2 (s) R1 (s) 1 + G1 (s)G2 (s)H1 (s) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 55 Problems + G3 (s)G4 (s)G6 (s)H2 (s) Y1 (s) . (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) Finally, solving for Y1 (s) yields Y1 (s) = T1 (s)R1 (s) where T1 (s) = G1 (s)G2 (s)(1 − G5 (s)G6 (s)H2 (s)) (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) − G3 (s)G4 (s)G6 (s)H2 (s) . . Similarly, for Y2 (s) we obtain Y2 (s) = T2 (s)R1 (s) . where T2 (s) = P2.32 G1 (s)G4 (s)G6 (s) (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) − G3 (s)G4 (s)G6 (s)H2 (s) The signal flow graph shows three loops: L1 = −G1 G3 G4 H2 L2 = −G2 G5 G6 H1 L3 = −H1 G8 G6 G2 G7 G4 H2 G1 . The transfer function Y2 /R1 is found to be Y2 (s) G1 G8 G6 ∆1 − G2 G5 G6 ∆2 = , R1 (s) 1 − (L1 + L2 + L3 ) + (L1 L2 ) where for path 1 ∆1 = 1 and for path 2 ∆ 2 = 1 − L1 . Since we want Y2 to be independent of R1 , we need Y2 /R1 = 0. Therefore, we require G1 G8 G6 − G2 G5 G6 (1 + G1 G3 G4 H2 ) = 0 . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 56 CHAPTER 2 P2.33 Mathematical Models of Systems The closedloop transfer function is G3 (s)G1 (s)(G2 (s) + K5 K6 ) Y (s) = . R(s) 1 − G3 (s)(H1 (s) + K6 ) + G3 (s)G1 (s)(G2 (s) + K5 K6 )(H2 (s) + K4 ) P2.34 The equations of motion are m1 ÿ1 + b(ẏ1 − ẏ2 ) + k1 (y1 − y2 ) = 0 m2 ÿ2 + b(ẏ2 − ẏ1 ) + k1 (y2 − y1 ) + k2 y2 = k2 x Taking the Laplace transform yields (m1 s2 + bs + k1 )Y1 (s) − (bs + k1 )Y2 (s) = 0 (m2 s2 + bs + k1 + k2 )Y2 (s) − (bs + k1 )Y1 (s) = k2 X(s) Therefore, after solving for Y1 (s)/X(s), we have Y2 (s) k2 (bs + k1 ) = . 2 X(s) (m1 s + bs + k1 )(m2 s2 + bs + k1 + k2 ) − (bs + k1 )2 P2.35 (a) We can redraw the block diagram as shown in Figure P2.35. Then, T (s) = K1 /s(s + 1) K1 = 2 . 1 + K1 (1 + K2 s)/s(s + 1) s + (1 + K2 K1 )s + K2 (b) The signal flow graph reveals two loops (both touching): L1 = −K1 s(s + 1) and L2 = −K1 K2 . s+1 Therefore, T (s) = K1 /s(s + 1) K1 = 2 . 1 + K1 /s(s + 1) + K1 K2 /(s + 1) s + (1 + K2 K1 )s + K1 (c) We want to choose K1 and K2 such that s2 + (1 + K2 K1 )s + K1 = s2 + 20s + 100 = (s + 10)2 . Therefore, K1 = 100 and 1 + K2 K1 = 20 or K2 = 0.19. (d) The step response is shown in Figure P2.35. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 58 CHAPTER 2 Mathematical Models of Systems (d) For the impulse input, y(t) ≈ 0.65 at t = 1 seconds (see Figure P2.36b). (a) Ramp input (b) Impulse input 0.8 2 1.8 0.7 1.6 0.6 1.4 0.5 y(t) y(t) 1.2 1 0.8 0.4 0.3 0.6 0.2 0.4 0.1 0.2 0 0 1 2 0 3 0 1 Time (sec) 2 3 Time (sec) FIGURE P2.36 (a) Ramp input response. (b) Impulse input response. P2.37 The equations of motion are m1 d2 x = −(k1 + k2 )x + k2 y dt2 and m2 d2 y = k2 (x − y) + u . dt2 When m1 = m2 = 1 and k1 = k2 = 1, we have d2 x = −2x + y dt2 P2.38 and d2 y =x−y+u . dt2 The equation of motion for the system is J d2 θ dθ + b + kθ = 0 , dt2 dt where k is the rotational spring constant and b is the viscous friction coefficient. The initial conditions are θ(0) = θo and θ̇(0) = 0. Taking the © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 59 Problems Laplace transform yields J(s2 θ(s) − sθo ) + b(sθ(s) − θo ) + kθ(s) = 0 . Therefore, θ(s) = (s + Jb θo ) (s2 + b Js + K J) = (s + 2ζωn )θo . + 2ζωn s + ωn2 s2 Neglecting the mass of the rod, the moment of inertia is detemined to be J = 2M r 2 = 0.5 kg · m2 . Also, s ωn = k = 0.02 rad/s J and ζ = b = 0.01 . 2Jωn Solving for θ(t), we find that q θo θ(t) = p e−ζωn t sin(ωn 1 − ζ 2 t + φ) , 1 − ζ2 where tan φ = p 1 − ζ 2 /ζ). Therefore, the envelope decay is θo θe = p e−ζωn t . 1 − ζ2 So, with ζωn = 2 × 10−4 , θo = 4000o and θf = 10o , the elapsed time is computed as t= P2.39 1 θo ln p = 8.32 hours . ζωn 1 − ζ 2 θf When t < 0, we have the steadystate conditions i1 (0) = 1A , va (0) = 2V and vc (0) = 5V , where vc (0) is associated with the 1F capacitor. After t ≥ 0, we have 2 di1 + 2i1 + 4(i1 − i2 ) = 10e−2t dt and Z i2 dt + 10i2 + 4(i2 − i1 ) − i1 = 0 . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 60 CHAPTER 2 Mathematical Models of Systems Taking the Laplace transform (using the initial conditions) yields 2(sI1 − i1 (0)) + 2I1 + 4I1 − 4I2 = 10 s+2 or (s + 3)I1 (s) − 2I2 (s) = s+7 s+2 and 1 [ I2 −vc (0)]+10I2 +4(I2 −I1 ) = I1 (s) or s −5sI1 (s)+(14s+1)I2 (s) = 5s . Solving for I2 (s) yields I2 = 5s(s2 + 6s + 13) , 14(s + 2)∆(s) where ∆(s) = s+3 −2 −5s 14s + 1 = 14s2 + 33s + 3 . Then, Vo (s) = 10I2 (s) . P2.40 The equations of motion are J1 θ̈1 = K(θ2 − θ1 ) − b(θ̇1 − θ̇2 ) + T and J2 θ̈2 = b(θ̇1 − θ̇2 ) . Taking the Laplace transform yields (J1 s2 + bs + K)θ1 (s) − bsθ2 (s) = Kθ2 (s) + T (s) and (J2 s2 + bs)θ2 (s) − bsθ1 (s) = 0 . Solving for θ1 (s) and θ2 (s), we find that θ1 (s) = (Kθ2 (s) + T (s))(J2 s + b) ∆(s) and θ2 (s) = b(Kθ2 (s) + T (s)) , ∆(s) where ∆(s) = J1 J2 s3 + b(J1 + J2 )s2 + J2 Ks + bK . P2.41 Assume that the only external torques acting on the rocket are control torques, Tc and disturbance torques, Td , and assume small angles, θ(t). Using the small angle approximation, we have ḣ = V θ © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 61 Problems J θ̈ = Tc + Td , where J is the moment of inertia of the rocket and V is the rocket velocity (assumed constant). Now, suppose that the control torque is proportional to the lateral displacement, as Tc (s) = −KH(s) , where the negative sign denotes a negative feedback system. The corresponding block diagram is shown in Figure P2.41. Td H desired=0 K + Tc + +  1 Js 2 V s H( s) FIGURE P2.41 Block diagram. P2.42 (a) The equation of motion of the motor is J dω = Tm − bω , dt where J = 0.1, b = 0.06, and Tm is the motor input torque. (b) Given Tm (s) = 1/s, and ω(0) = 0.7, we take the Laplace transform of the equation of motion yielding sω(s) − ω(0) + 0.6ω(s) = 10Tm or ω(s) = 0.7s + 10 . s(s + 0.6) Then, computing the partial fraction expansion, we find that ω(s) = A B 16.67 15.97 + = − . s s + 0.6 s s + 0.6 The step response, determined by taking the inverse Laplace transform, is ω(t) = 16.67 − 15.97e−0.6t , t≥0. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 62 CHAPTER 2 P2.43 Mathematical Models of Systems The work done by each gear is equal to that of the other, therefore Tm θm = TL θL . Also, the travel distance is the same for each gear, so r1 θ m = r2 θ L . The number of teeth on each gear is proportional to the radius, or r1 N 2 = r2 N 1 . So, θm r2 N2 = = , θL r1 N1 and N1 θ m = N2 θ L N1 θL = θm = nθm , N2 where n = N1 /N2 . Finally, Tm θL N1 = = =n. TL θm N2 P2.44 The inertia of the load is JL = πρLr 4 . 2 Also, from the dynamics we have T2 = JL ω̇2 + bL ω2 and T1 = nT2 = n(JL ω̇2 + bL ω2 ) . So, T1 = n2 (JL ω̇1 + bL ω1 ) , © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 63 Problems since ω2 = nω1 . Therefore, the torque at the motor shaft is T = T1 + Tm = n2 (JL ω̇1 + bL ω1 ) + Jm ω̇1 + bm ω1 . P2.45 Let U (s) denote the human input and F (s) the load input. The transfer function is P (s) = G(s) + KG1 (s) Gc (s) + KG1 (s) U (s) + F (s) , ∆(s) ∆(s) where ∆ = 1 + GH(s) + G1 KBH(s) + Gc E(s) + G1 KE(s) . P2.46 Consider the application of Newton’s law ( mv we obtain P F = mẍ). From the mass mv ẍ1 = F − k1 (x1 − x2 ) − b1 (ẋ1 − ẋ2 ). Taking the Laplace transform, and solving for X1 (s) yields X1 (s) = b1 s + k1 1 F (s) + X2 (s), ∆1 (s) ∆1 (s) where ∆1 := mv s2 + b1 s + k1 . From the mass mt we obtain mt ẍ2 = −k2 x2 − b2 ẋ2 + k1 (x1 − x2 ) + b1 (ẋ1 − ẋ2 ). Taking the Laplace transform, and solving for X2 (s) yields X2 (s) = b1 s + k1 X1 (s), ∆2 (s) where ∆2 := mt s2 + (b1 + b2 )s + k1 + k2 . Substituting X2 (s) above into the relationship fpr X1 (s) yields the transfer function ∆2 (s) X1 (s) = . F (s) ∆1 (s)∆2 (s) − (b1 s + k1 )2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 64 CHAPTER 2 P2.47 Mathematical Models of Systems Using the following relationships h(t) = Z (1.6θ(t) − h(t))dt ω(t) = θ̇(t) J ω̇(t) = Km ia (t) va (t) = 50vi (t) = 10ia (t) + vb (t) θ̇ = Kvb we find the differential equation is d3 h Km + 1+ 3 dt 10JK P2.48 Km dh 8Km d2 h + = vi . 2 dt 10JK dt J (a) The transfer function is V2 (s) (1 + sR1 C1 )(1 + sR2 C2 ) = . V1 (s) R1 C 2 s (b) When R1 = 100 kΩ, R2 = 200 kΩ, C1 = 1 µF and C2 = 0.1 µF , we have V2 (s) 0.2(s + 10)(s + 50) = . V1 (s) s P2.49 (a) The closedloop transfer function is T (s) = G(s) 6205 = 3 . 1 + G(s) s + 13s2 + 1281s + 6205 (b) The poles of T (s) are s1 = −5 and s2,3 = −4 ± j35. (c) The partial fraction expansion (with a step input) is Y (s) = 1 − 1.0122 0.0061 + 0.0716j 0.0061 − 0.0716j + + . s+5 s + 4 + j35 s + 4 − j35 (d) The step response is shown in Figure P2.49. The real and complex roots are close together and by looking at the poles in the splane we have difficulty deciding which is dominant. However, the residue at the real pole is much larger and thus dominates the response. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 65 Problems 1 0.9 0.8 Amplitude 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time (secs) FIGURE P2.49 Step response. P2.50 (a) The closedloop transfer function is T (s) = s3 + 45s2 14000 . + 3100s + 14500 (b) The poles of T (s) are s1 = −5 and s2,3 = −20 ± j50. (c) The partial fraction expansion (with a step input) is Y (s) = 0.9655 1.0275 0.0310 − 0.0390j 0.0310 + 0.0390j − + + . s s+5 s + 20 + j50 s + 20 − j50 (d) The step response is shown in Figure P2.50. The real root dominates the response. (e) The final value of y(t) is yss = lim sY (s) = 0.9655 . s→0 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. CHAPTER 2 Mathematical Models of Systems 1 0.9 0.8 0.7 Amplitude 66 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time (secs) FIGURE P2.50 Step response. P2.51 Consider the free body diagram in Figure P2.51. Using Newton’s Law and summing the forces on the two masses yields M1 ẍ(t) + b1 ẋ(t) + k1 x(t) = b1 ẏ(t) M2 ÿ(t) + b1 ẏ(t) + k2 y(t) = b1 ẋ(t) + u(t) k1x M1 k1 x . . b1(x  y) k2 M1 x . . b1(y  x) k2 y b1 M2 M2 y y u(t) FIGURE P2.51 Free body diagram. u(t) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 67 Advanced Problems Advanced Problems AP2.1 The transfer function from V (s) to ω(s) has the form ω(s) Km = . V (s) τm s + 1 In the steadystate, ωss = lim s s→0 Km 5 = 5Km . τm s + 1 s So, Km = 70/5 = 14 . Also, ω(t) = Vm Km (1 − e−t/τm ) where V (s) = Vm /s. Solving for τm yields τm = −t . ln(1 − ω(t)/ωss ) When t = 2, we have τm = −2 = 3.57 . ln(1 − 30/70) Therefore, the transfer function is ω(s) 14 = . V (s) 3.57s + 1 AP2.2 The closedloop transfer function form R1 (s) to Y2 (s) is Y2 (s) G1 G4 G5 (s) + G1 G2 G3 G4 G6 (s) = R1 (s) ∆ where ∆ = [1 + G3 G4 H2 (s)][1 + G1 G2 H3 (s)] . If we select G5 (s) = −G2 G3 G6 (s) then the numerator is zero, and Y2 (s)/R1 (s) = 0. The system is now decoupled. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 68 CHAPTER 2 AP2.3 Mathematical Models of Systems (a) Computing the closedloop transfer function: G(s)Gc (s) R(s) . Y (s) = 1 + Gc (s)G(s)H(s) Then, with E(s) = R(s) − Y (s) we obtain E(s) = 1 + Gc (s)G(s)(H(s) − 1) R(s) . 1 + Gc (s)G(s)H(s) If we require that E(s) ≡ 0 for any input, we need 1 + Gc (s)G(s)(H(s) − 1) = 0 or H(s) = Gc (s)G(s) − 1 n(s) = . Gc (s)G(s) d(s) Since we require H(s) to be a causal system, the order of the numerator polynomial, n(s), must be less than or equal to the order of the denominator polynomial, d(s). This will be true, in general, only if both Gc (s) and G(s) are proper rational functions (that is, the numerator and denominator polynomials have the same order). Therefore, making E ≡ 0 for any input R(s) is possible only in certain circumstances. (b) The transfer function from Td (s) to Y (s) is Gd (s)G(s) Y (s) = Td (s) . 1 + Gc (s)G(s)H(s) With H(s) as in part (a) we have Gd (s) Y (s) = Td (s) . Gc (s) (c) No. Since Y (s) = Gd (s)G(s) Td (s) = T (s)Td (s) , 1 + Gc (s)G(s)H(s) the only way to have Y (s) ≡ 0 for any Td (s) is for the transfer function T (s) ≡ 0 which is not possible in general (since G(s) 6= 0). AP2.4 (a) With q(s) = 1/s we obtain τ (s) = 1/Ct s+ QS+1/R Ct · 1 . s Define α := QS + 1/R Ct and β := 1/Ct . © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 69 Advanced Problems Then, it follows that τ (s) = β 1 −β/α β/α · = + . s+α s s+α s Taking the inverse Laplace transform yields τ (t) = β −β −αt β e + = [1 − e−αt ] . α α α 1 (b) As t → ∞, τ (t) → αβ = Qs+1/R . (c) To increase the speed of response, you want to choose Ct , Q, S and R such that α := Qs + 1/R Ct is ”large.” AP2.5 Considering the motion of each mass, we have M3 ẍ3 + b3 ẋ3 + k3 x3 = u3 + b3 ẋ2 + k3 x2 M2 ẍ2 + (b2 + b3 )ẋ2 + (k2 + k3 )x2 = u2 + b3 ẋ3 + k3 x3 + b2 ẋ1 + k2 x1 M1 ẍ1 + (b1 + b2 )ẋ1 + (k1 + k2 )x1 = u1 + b2 ẋ2 + k2 x2 In matrix form the three equations can be written as 0 M1 0 M2 AP2.6 0 0 0 0 0 ẍ1 b1 + b2 −b2 0 0 b2 + b3 −b3 ẍ2 + −b2 M3 ẍ3 −b3 b3 −k2 0 k1 + k2 + k2 + k3 −k3 −k2 −k3 k3 ẋ1 ẋ 2 ẋ3 x 1 u1 x = u . 2 2 x3 u3 Considering the cart mass and using Newton’s Law we obtain M ẍ = u − bẋ − F sin ϕ where F is the reaction force between the cart and the pendulum. Considering the pendulum we obtain m d2 (x + L sin ϕ) = F sin ϕ dt2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 70 CHAPTER 2 Mathematical Models of Systems m d2 (L cos ϕ) = F cos ϕ + mg dt2 Eliminating the reaction force F yields the two equations (m + M )ẍ + bẋ