ECE382/ME482 Spring 2005 Homework 3 Solution March 7,


 Lorena Perkins
 11 months ago
 Views:
Transcription
1 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, 2005 Solution to HW3 AP4.5 We are given a block diagram in Figure AP4.5 on page 237 of the text and asked to find steady state errors due to (a) a step disturbance and (b) a ramp input. We are then asked in part (c) to find a suitable value of the gain K, not more than 0, and plot the response to the unit step disturbance. Solution: (a) First we note that when the input R(s) is zero, the error E(s) is equal to the negative of the output. The transfer function from the disturbance input to the output is determined by noting that in this case, the block containing the transfer function /(s+2) 2 is in the feedforward path and the remaining two blocks are in the feedback path. Noting that the feedback into the first summer is negative, we thus have after simplification a block with gain K/s in the feedback path. Setting the input R(s) to zero and noting that the feedback into the second summer is positive we have transfer function from disturbance input D(s) to output of D(s) = (s+2) 2 K s(s+2) 2 = s s 3 + 4s 2 + 4s + K. () The steady state error due to a step disturbance R(s) = /s is thus, by the final value theorem, ( ) s e ss = lim s s 0 s s 3 + 4s 2 = 0. (2) + 4s + K (b) The transfer function from the input R(s) to the error E(s) is E(s) R(s) = + K = s(s+2) 2 s(s + 2)2 s(s + 2) 2 + K (3) where we have noted that all of the blocks are in the feedback path. Then the steady state error corresponding to a ramp input is ( ) s(s + 2) 2 e ss = lim s s 0 s 2 s(s + 2) 2 + K = 4 K. (4) This indicates that increasing the gain K decreases the steady state error resulting from a ramp input. (c) We don t have much information here from which to determine what a suitable value of K should be. Let s assume that since we were asked about steady state error due to a ramp input, that is important, so we should probably choose a large gain (close to our maximum of 0) so long as the system remains stable with this gain value. (We can check that in Matlab by executing the command roots([ 4 4 K]), after assigning our chosen value to the variable K, and making sure that all of the roots have negative real part. In fact they do for any value of K between 0 and 0. We will learn a method for determining this conclusively in Chapter 6.
2 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, To see whether there is a tradeoff involved, let s have Matlab plot responses to step disturbances for several values of K. Using the following Matlab code we obtain the plot shown in Figure. clf;for k=[:0];[y,t]=step(tf([ 0],[ 4 4 k]));plot(t,y),hold on; end; grid xlabel( Time (s) ) ylabel( e(t) ) title( Error due to step disturbance for K=,2,...,0 ) print deps AP4.5c Figure : Plot of Output Error Corresponding to Step Disturbance for a Range of Gain Values 0.25 Error due to step disturbance for K=,2,..., e(t) Time (s) Now by plotting the results for individual gains K we can see that the smallest error is obtained for the largest K so given the information we have, K = 0 appears to be the best choice. (That doesn t mean a smaller positive gain is unsuitable. We are not given any information to judge that.) If we had tested gains less than one, we would have seen that the maximum error increases with increasing gain. If we did not want any oscillation at all, we could see from the plot that we could achieve this by choosing a gain of, but
3 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, there would be no reason to choose a gain less than one since that would increase the maximum error. AP4.7 In this problem we are given a block diagram of a system having both disturbance and sensor noise inputs. We are asked to determine the effect of (a) the disturbance and (b) the sensor noise on the output. We are then asked to select a value of the gain K in the range to 00, inclusive that minimizes the effects of step disturbances and sensor noise on the steady state error. In all parts of the problem we will assume that the error E(s) is defined to be the output of the leftmost summer. Solution: We ll apply the final value theorem to compute the steady state errors due to step disturbance and noise inputs from the transfer functions obtained in parts (a) and (b). Then we ll use this information to obtain an answer to part (c). (a) The transfer function from the disturbance to the output is D(s) = 2s s(s + 2) + 2K (5) so a step input of D(s) = A/s leads to a steady state error of ( ) A 2s e ss = lim s = 0, (6) s 0 s s(s + 2) + 2K i.e. the value of the gain K does not affect the value of the steady state error due to a step disturbance. (b) The transfer function from the noise to the output is N(s) = 2K s(s+2) 2K s(s+2) = 2K s(s + 2) + 2K (7) so a step input of N(s) = B/s leads to a steady state error of ( ) B 2K e ss = lim s = B, (8) s 0 s s(s + 2) + 2K and again the gain K has no effect on the steady state error. (c) We ve just shown that this was a trick question. There is no best value for K in terms of minimizing the effect of step noise and disturbance inputs on steady state error. As in the previous problem, we could plot the error as a function of time for different values of K and choose one that seems a reasonable compromise between the reducing the maximum error and reducing the settling time (the time for the error to become small a more precise definition will be given in Chapter 5). AP4.8 Given the block diagram in Figure AP4.8, we are asked to determine (a) the transfer function T(s) of the system, (b) the sensitivity Sb T of the transfer function T(s) to variation in the parameter b, and finally we are asked to choose a gain K [,50] so that the disturbance effects and the sensitivity are both minimized. Solution:
4 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, (a) The transfer function is (b) The sensitivity is thus T(s) = R(s) = Kb s+ + Kb s+ S T b = SN b S D b = N b = Kb s + + Kb. (9) b N D b where N and D are the numerator and denominator of the transfer function T. We obtain Sb T = Kb Kb Kb s + + Kb = s + s + + Kb. () (c) The transfer function from the disturbance to the output is D(s) = b s+ + Kb s+ = b D b s + + Kb so if we define the error to be the output of the leftmost summer, thus e(t) = y(t) we have the following response to a unit step input: ( ) b e ss = lim s s 0 s s + + Kb = b + Kb. (3) We now have shown that both S T b and e ss resulting from a step disturbance decrease with increasing K so we should choose the maximum allowable K, namely K = 50. DP4.6 We are given the system whose block diagram is shown in Figure DP4.6 and asked to consider the effect of various choices of controllers on steady state error for unit step disturbances. In particular, we are asked to determine appropriate values of the controller gains for each case. It is noted that the parameter J represents the pitching moment of inertia, hence is always positive. Solution: (a) When G c (s) = K, the closed loop transfer function is T(s) = R(s) = (0) (2) K Js 2 + K. (4) Accordingly, if K is positive, the roots of the denominator are s = ±j K/J so both poles are on the jω axis and the system is marginally stable. If K is negative, the roots of the denominator are s = ± K/J so one pole is in the righthalf plane and the system is unstable. (b) The steady state error to a unit step disturbance when G c (s) = K can be determined by first obtaining the transfer function from D(s) to then applying the final value theorem. The transfer function is D(s) = Js 2 + K Js 2 = Js 2 + K. (5)
5 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, We can not use the final value theorem to calculate the steady state error because we have poles on the imaginary axis. (Refer to the explanation on page 50 of the conditions under which the final value theorem can be applied.) In fact, since the system will oscillate indefinitely, there is no final value. (c) Now supposing that the controller is instead G(s) = K P + K D s we obtain the closed loop system transfer function T(s) = R(s) = The poles are now K D ± KD 2 4JK P s = 2J so for stability we need to consider three cases. K p + K D s Js 2 + K D s + K P. (6) (i) The expression under the square root is negative. Then the R(s) = K D /(2J) so the system is stable if and only if K D > 0 (since J > 0). The expression under the square root is negative if 4JK P > KD 2, i.e. if K P > KD 2 /(4J), so the system is stable if both K D > 0 and K P > KD 2 /(4J) are true. (ii) The expression under the square root is negative. This occurs when KD 2 > 4JK P. First, suppose K P > 0. Then the square root will always be less than K D, so, as long as both K P and K D are positive, the system will be stable. (This is a weaker condition than the one we found above.) Next, suppose K P is negative. In that case, the square root would be greater than K D so regardless of whether K D were positive or negative, one of the roots would end up positive. Finally, suppose K P = 0. If K P = 0 then one of the roots will be zero and the system will be marginally stable. (iii) The expression under the square root is zero. Then the system is stable if and only if K D > 0. In summary, we ve found that for stability we need both gains strictly positive. (d) We assume here that K D and K P are positive. Now the transfer function from the disturbance D(s) to the output Θ(s) is Θ(s) D(s) = Js 2 + K P +K D s Js 2 = (7) Js 2 + K D s + K P (8) so the steady state error resulting from a step disturbance is ( ) e ss = lim s s 0 s Js 2 = (9) + K D s + K P K P MP4.3 We are given a closed loop transfer function T(s) = 0K s s + 0K and asked to obtain and plot step responses for K = 0, 00, and 500. We are also asked to develop a table comparing percent overshoot, settling time, and steady state error for (20)
6 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, these values of K. This is a little complicated since we haven t yet defined settling time and percent overshoot, so you were allowed some leeway in your answers. Here s the answer to the problem using the definitions given in Chapter 5. Solution: First, let s look at how these values of K affect the poles (roots of the denominator) of the system transfer function. Using Matlab we find that for K = 0 we obtain unequal poles in the left half plan; for K = 00 we obtain a repeated pole of multiplicity 2 in the left half plane; and for K = 500 we obtain a complex pair of poles in the left half plane as shown in the following Matlab transcript. >> roots([ 20 0]) ans = >> roots([ 20 00]) ans = >> roots([ ]) ans = i i Accordingly, we expect different behavior in the three cases. By calculating the damping ratio and natural frequency we can tell more. The natural frequency will be ω n = K and the damping ratio will be ζ = 20/(2ω n ) = 0/ K so the system is overdamped (ζ > ) if K = 0, critically damped (ζ = ) if K = 00, and underdamped (ζ < ) if K = 500. Accordingly we expect to see overshoot only in the case K = 500. Note that our findings agree with the discussion in Section 2.4, on p. 50 of the text. There it was noted that if ζ < there are (distinct) real roots; if ζ = there are real repeated roots; and if ζ < there is a complex conjugate pair of roots. Before we plot the step responses, let s find out more about the sort of responses we expect. The steady state output for a step input will be ( ) y ss = lim s T(s) = T(0) = 0. (2) s 0 s (This means that the steady state error e ss = y ss = 9 does not depend on the value of K.)
7 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, The 2% settling time given by (5.3) in the text is T s,2% = 4/(ζω n ). Since ω n = K and ζ = 20/(2ω n ) here, we would seem to have T s,2% = 4 0 independent of K. However, the plot of the step responses clearly shows that this is not the case. >> t=[0:.0:2] ; >> clf; >> for k=[0,00,500];[y,t]=step(tf([0*k],[ 20 k]),t);plot(t,y),hold on;end; >> xlabel( Time (s) ); >> ylabel( y(t) ) >> title( MP4.3 Step responses for K=0, 00, and 000 ) >> print deps MP4.3.eps What went wrong here? The problem is that equation (5.3) for the settling time was derived assuming that the system response was a sinusoid multiplied by e ζωnt. This is only true if the second order system is underdamped. From the analysis above and the computed step responses we determine the values shown in Table for the percent overshoot, settling time and steady state error. (22) Table : Table of Response Parameters for MP4.3 Gain Percent Overshoot 2% Settling Time (s) SteadyState Error MP4.7 We are given a block diagram in Figure MP4.7 of a closed loop system with a controller G c (s). We are asked to compare the response of the closedloop system to a unit step input for each of the two possible controllers. Solution: We will compare the steady state errors first by hand and then using the Matlab code shown in the transcript below. (a) Using the proportional controller G c (s) = K = 2 we obtain ( ) e ss = lim s 0 s + 0K = + K = 3. (23) s+0 (b) Using the proportionalintegral controller G c (s) = K 0 + K /s = /2 we obtain e ss = lim s 0 ( ) s + 0K 0+K /s s+0 s 2 + 0s = lim s 0 s 2 = 0. (24) + 0( + K 0 )s + K
8 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, Figure 2: Plot of Output Corresponding to Step Disturbance for a Range of Gain Values in MP4.3 4 MP4.3 Step responses for K=0, 00, and y(t) Time (s) (c) Obviously by using a more complex controller we have managed to reduce the steady state error for a step input to zero. Here are the Matlab calculations, which were, actually, superfluous except that the problem requested them. The plot of the step responses for both controllers is shown in Figure 3 >> %MP4.7 >> sysa = feedback(tf([20],[ 0]),) Transfer function: s + 30 >> sysb = feedback(tf(0*[2 20],[ 0 0]),)
9 ECE382/ME482 Spring 2005 Homework 3 Solution March 7, Transfer function: 20 s s^ s >> step(sysa, r,sysb, b. ) >> print deps MP4.7.eps Figure 3: Plots of Step Responses for P and PI Controllers in MP4.7 Step Response Amplitude Time (sec)
First Order System. Transfer function: Response to a unit step input is: Partial Fraction Expansion leads to: Inverse Laplace transform leads to:
First Order System Transfer function: Response to a unit step input is: Partial Fraction Expansion leads to: Inverse Laplace transform leads to: First Order System At t = T, the output is: T represents
More informationUnderstanding Poles and Zeros
MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.14 Analysis and Design of Feedback Control Systems Understanding Poles and Zeros 1 System Poles and Zeros The transfer function
More informationPositive Feedback and Oscillators
Physics 3330 Experiment #6 Fall 1999 Positive Feedback and Oscillators Purpose In this experiment we will study how spontaneous oscillations may be caused by positive feedback. You will construct an active
More informationROUTH S STABILITY CRITERION
ECE 680 Modern Automatic Control Routh s Stability Criterion June 13, 2007 1 ROUTH S STABILITY CRITERION Consider a closedloop transfer function H(s) = b 0s m + b 1 s m 1 + + b m 1 s + b m a 0 s n + s
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationLecture 5 Rational functions and partial fraction expansion
S. Boyd EE102 Lecture 5 Rational functions and partial fraction expansion (review of) polynomials rational functions polezero plots partial fraction expansion repeated poles nonproper rational functions
More informationG(s) = Y (s)/u(s) In this representation, the output is always the Transfer function times the input. Y (s) = G(s)U(s).
Transfer Functions The transfer function of a linear system is the ratio of the Laplace Transform of the output to the Laplace Transform of the input, i.e., Y (s)/u(s). Denoting this ratio by G(s), i.e.,
More informationΣ _. Feedback Amplifiers: One and Two Pole cases. Negative Feedback:
Feedback Amplifiers: One and Two Pole cases Negative Feedback: Σ _ a f There must be 180 o phase shift somewhere in the loop. This is often provided by an inverting amplifier or by use of a differential
More informationOverdamped system response
Second order system response. Im(s) Underdamped Unstable Overdamped or Critically damped Undamped Re(s) Underdamped Overdamped system response System transfer function : Impulse response : Step response
More informationROOT LOCUS TECHNIQUES
ROOT LOCUS TECHNIQUES In this lecture you will learn the following : The definition of a root locus How to sketch root locus How to use the root locus to find the poles of a closed loop system How to use
More informationEE 402 RECITATION #13 REPORT
MIDDLE EAST TECHNICAL UNIVERSITY EE 402 RECITATION #13 REPORT LEADLAG COMPENSATOR DESIGN F. Kağan İPEK Utku KIRAN Ç. Berkan Şahin 5/16/2013 Contents INTRODUCTION... 3 MODELLING... 3 OBTAINING PTF of OPEN
More informationChapter 9: Controller design
Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback
More informationEECE 460 : Control System Design
EECE 460 : Control System Design PID Controller Design and Tuning Guy A. Dumont UBC EECE January 2012 Guy A. Dumont (UBC EECE) EECE 460 PID Tuning January 2012 1 / 37 Contents 1 Introduction 2 Control
More informationECE 3510 Final given: Spring 11
ECE 50 Final given: Spring This part of the exam is Closed book, Closed notes, No Calculator.. ( pts) For each of the timedomain signals shown, draw the poles of the signal's Laplace transform on the
More information3.2 Sources, Sinks, Saddles, and Spirals
3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients
More informationRoot Locus. E(s) K. R(s) C(s) 1 s(s+a) Consider the closed loop transfer function:
Consider the closed loop transfer function: Root Locus R(s) +  E(s) K 1 s(s+a) C(s) How do the poles of the closedloop system change as a function of the gain K? The closedloop transfer function is:
More informationFrequency response of a general purpose singlesided OpAmp amplifier
Frequency response of a general purpose singlesided OpAmp amplifier One configuration for a general purpose amplifier using an operational amplifier is the following. The circuit is characterized by:
More informationPart IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Dr Glenn Vinnicombe HANDOUT 3. Stability and pole locations.
Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Dr Glenn Vinnicombe HANDOUT 3 Stability and pole locations asymptotically stable marginally stable unstable Imag(s) repeated poles +
More informationMATLAB Control System Toolbox Root Locus Design GUI
MATLAB Control System Toolbox Root Locus Design GUI MATLAB Control System Toolbox contains two Root Locus design GUI, sisotool and rltool. These are two interactive design tools for the analysis and design
More informationPID Control. Chapter 10
Chapter PID Control Based on a survey of over eleven thousand controllers in the refining, chemicals and pulp and paper industries, 97% of regulatory controllers utilize PID feedback. Desborough Honeywell,
More informationController Design in Frequency Domain
ECSE 4440 Control System Engineering Fall 2001 Project 3 Controller Design in Frequency Domain TA 1. Abstract 2. Introduction 3. Controller design in Frequency domain 4. Experiment 5. Colclusion 1. Abstract
More informationMotor Modeling and Position Control Lab Week 3: Closed Loop Control
Motor Modeling and Position Control Lab Week 3: Closed Loop Control 1. Review In the first week of motor modeling lab, a mathematical model of a DC motor from first principles was derived to obtain a first
More informationSecond Order Systems
Second Order Systems Second Order Equations Standard Form G () s = τ s K + ζτs + 1 K = Gain τ = Natural Period of Oscillation ζ = Damping Factor (zeta) Note: this has to be 1.0!!! Corresponding Differential
More informationDrivetech, Inc. Innovations in Motor Control, Drives, and Power Electronics
Drivetech, Inc. Innovations in Motor Control, Drives, and Power Electronics Dal Y. Ohm, Ph.D.  President 25492 Carrington Drive, South Riding, Virginia 20152 Ph: (703) 3272797 Fax: (703) 3272747 ohm@drivetechinc.com
More informationLoop Analysis. Chapter 7. 7.1 Introduction
Chapter 7 Loop Analysis Quotation Authors, citation. This chapter describes how stability and robustness can be determined by investigating how sinusoidal signals propagate around the feedback loop. The
More informationQNET Experiment #06: HVAC Proportional Integral (PI) Temperature Control Heating, Ventilation, and Air Conditioning Trainer (HVACT)
Quanser NIELVIS Trainer (QNET) Series: QNET Experiment #06: HVAC Proportional Integral (PI) Temperature Control Heating, Ventilation, and Air Conditioning Trainer (HVACT) Student Manual Table of Contents
More informationAnalog Filter Design Demystified
FILTER CIRCUITS (ANALOG) VIDEO CIRCUITS Dec 03, 2002 Analog Filter Design Demystified This article shows the reader how to design analog filters. It starts by covering the fundamentals of filters, it then
More information3 0 + 4 + 3 1 + 1 + 3 9 + 6 + 3 0 + 1 + 3 0 + 1 + 3 2 mod 10 = 4 + 3 + 1 + 27 + 6 + 1 + 1 + 6 mod 10 = 49 mod 10 = 9.
SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) (1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid
More informationParameter Definition with the reader. Since the scope of this article is practical in nature all theoretical derivations have been omitted, hoping to
Freescale Semiconductor Application Note Document Number: AN535 Rev. 1.0, 02/2006 PhaseLocked Loop Design Fundamentals by: Garth Nash Applications Engineering Abstract The fundamental design concepts
More informationResponse to Harmonic Excitation Part 2: Damped Systems
Response to Harmonic Excitation Part 2: Damped Systems Part 1 covered the response of a single degree of freedom system to harmonic excitation without considering the effects of damping. However, almost
More informationChapter 3: Analysis of closedloop systems
Chapter 3: Analysis of closedloop systems Control Automático 3º Curso. Ing. Industrial Escuela Técnica Superior de Ingenieros Universidad de Sevilla Control of SISO systems Control around an operation
More information23 Identification of a Response Amplitude Operator from Data
23 IDENTIFICATION OF A RESPONSE AMPLITUDE OPERATOR FROM DATA 75 23 Identification of a Response Amplitude Operator from Data Load the data file homework5.dat from the course website. The first column is
More informationManufacturing Equipment Modeling
QUESTION 1 For a linear axis actuated by an electric motor complete the following: a. Derive a differential equation for the linear axis velocity assuming viscous friction acts on the DC motor shaft, leadscrew,
More informationHITACHI INVERTER SJ/L100/300 SERIES PID CONTROL USERS GUIDE
HITACHI INVERTER SJ/L1/3 SERIES PID CONTROL USERS GUIDE After reading this manual, keep it for future reference Hitachi America, Ltd. HAL1PID CONTENTS 1. OVERVIEW 3 2. PID CONTROL ON SJ1/L1 INVERTERS 3
More informationLecture 220 Compensation of Op Amps (3/27/10) Page 2201
Lecture 220 Compensation of Op Amps (3/27/0) Page 220 LECTURE 220 INTRODUCTION TO OP AMPS LECTURE OUTLINE Outline Op Amps Categorization of Op Amps Compensation of Op Amps Miller Compensation Other Forms
More informationdspace DSP DS1104 based State Observer Design for Position Control of DC Servo Motor
dspace DSP DS1104 based State Observer Design for Position Control of DC Servo Motor Jaswandi Sawant, Divyesh Ginoya Department of Instrumentation and control, College of Engineering, Pune. ABSTRACT This
More informationTEACHING AUTOMATIC CONTROL IN NONSPECIALIST ENGINEERING SCHOOLS
TEACHING AUTOMATIC CONTROL IN NONSPECIALIST ENGINEERING SCHOOLS J.A.Somolinos 1, R. Morales 2, T.Leo 1, D.Díaz 1 and M.C. Rodríguez 1 1 E.T.S. Ingenieros Navales. Universidad Politécnica de Madrid. Arco
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationChapter 11 Current Programmed Control
Chapter 11 Current Programmed Control Buck converter v g i s Q 1 D 1 L i L C v R The peak transistor current replaces the duty cycle as the converter control input. Measure switch current R f i s Clock
More informationECE302 Spring 2006 HW5 Solutions February 21, 2006 1
ECE3 Spring 6 HW5 Solutions February 1, 6 1 Solutions to HW5 Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics
More informationMAE143A Signals & Systems, Final Exam  Wednesday March 16, 2005
MAE43A Signals & Systems, Final Exam  Wednesday March 6, 25 Instructions This quiz is open book. You may use whatever written materials you choose including your class notes and the textbook. You may
More informationChapter Eight Root Locus Control Design. 8.3 Common Dynamic Controllers
Chapter Eight Root Locus Control Design 8.3 Common Dynamic Controllers Several common dynamic controllers appear very often in practice. They are known as PD, PI, PID, phaselag, phaselead, and phaselaglead
More informationSpeed Control of DC Motor using Pid Controller Based on Matlab
Speed Control of DC Motor using Pid Controller Based on Matlab Aditya Pratap Singh Asst Prof, EX Dept. IEI BHOPAL, MP Udit Narayan Student of BE, IEI, BHOPAL, MP Akash Verma Student of BE, IEI, BHOPAL,
More informationRealTime Systems Versus CyberPhysical Systems: Where is the Difference?
RealTime Systems Versus CyberPhysical Systems: Where is the Difference? Samarjit Chakraborty www.rcs.ei.tum.de TU Munich, Germany Joint work with Dip Goswami*, Reinhard Schneider #, Alejandro Masrur
More informationLaboratory 4: Feedback and Compensation
Laboratory 4: Feedback and Compensation To be performed during Week 9 (Oct. 2024) and Week 10 (Oct. 2731) Due Week 11 (Nov. 37) 1 PreLab This PreLab should be completed before attending your regular
More informationEDUMECH Mechatronic Instructional Systems. Ball on Beam System
EDUMECH Mechatronic Instructional Systems Ball on Beam System Product of Shandor Motion Systems Written by Robert Hirsch Ph.D. 9989 All Rights Reserved. 999 Shandor Motion Systems, Ball on Beam Instructional
More informationTesting a power supply for line and load transients
Testing a power supply for line and load transients Powersupply specifications for line and load transients describe the response of a power supply to abrupt changes in line voltage and load current.
More information2.2 Magic with complex exponentials
2.2. MAGIC WITH COMPLEX EXPONENTIALS 97 2.2 Magic with complex exponentials We don t really know what aspects of complex variables you learned about in high school, so the goal here is to start more or
More informationSystem Modeling and Control for Mechanical Engineers
Session 1655 System Modeling and Control for Mechanical Engineers Hugh Jack, Associate Professor Padnos School of Engineering Grand Valley State University Grand Rapids, MI email: jackh@gvsu.edu Abstract
More informationLecture 8 : Dynamic Stability
Lecture 8 : Dynamic Stability Or what happens to small disturbances about a trim condition 1.0 : Dynamic Stability Static stability refers to the tendency of the aircraft to counter a disturbance. Dynamic
More informationMatlab and Simulink. Matlab and Simulink for Control
Matlab and Simulink for Control Automatica I (Laboratorio) 1/78 Matlab and Simulink CACSD 2/78 Matlab and Simulink for Control Matlab introduction Simulink introduction Control Issues Recall Matlab design
More informationFractions and Decimals
Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first
More informationTime Response Analysis of DC Motor using Armature Control Method and Its Performance Improvement using PID Controller
Available online www.ejaet.com European Journal of Advances in Engineering and Technology, 5, (6): 566 Research Article ISSN: 394658X Time Response Analysis of DC Motor using Armature Control Method
More informationPID Controller Tuning: A Short Tutorial
PID Controller Tuning: A Short Tutorial Jinghua Zhong Mechanical Engineering, Purdue University Spring, 2006 Outline This tutorial is in PDF format with navigational control. You may press SPACE or, or
More information2. THE xy PLANE 7 C7
2. THE xy PLANE 2.1. The Real Line When we plot quantities on a graph we can plot not only integer values like 1, 2 and 3 but also fractions, like 3½ or 4¾. In fact we can, in principle, plot any real
More informationFirst, we show how to use known design specifications to determine filter order and 3dB cutoff
Butterworth LowPass Filters In this article, we describe the commonlyused, n th order Butterworth lowpass filter. First, we show how to use known design specifications to determine filter order and
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More informationChapter 7  Roots, Radicals, and Complex Numbers
Math 233  Spring 2009 Chapter 7  Roots, Radicals, and Complex Numbers 7.1 Roots and Radicals 7.1.1 Notation and Terminology In the expression x the is called the radical sign. The expression under the
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More informationIntroduction to Control Systems
CHAPTER 1 Introduction to Control Systems 1.1 INTRODUCTION In this Chapter, we describe very briefly an introduction to control systems. 1.2 CONTROL SYSTEMS Control systems in an interdisciplinary field
More informationStability of Linear Control System
Stabilit of Linear Control Sstem Concept of Stabilit Closedloop feedback sstem is either stable or nstable. This tpe of characterization is referred to as absolte stabilit. Given that the sstem is stable,
More informationVCO K 0 /S K 0 is tho slope of the oscillator frequency to voltage characteristic in rads per sec. per volt.
Phase locked loop fundamentals The basic form of a phase locked loop (PLL) consists of a voltage controlled oscillator (VCO), a phase detector (PD), and a filter. In its more general form (Figure 1), the
More information2.6 The driven oscillator
2.6. THE DRIVEN OSCILLATOR 131 2.6 The driven oscillator We would like to understand what happens when we apply forces to the harmonic oscillator. That is, we want to solve the equation M d2 x(t) 2 + γ
More informationPhysics 9 Fall 2009 Homework 2  Solutions
Physics 9 Fall 009 Homework  s 1. Chapter 7  Exercise 5. An electric dipole is formed from ±1.0 nc charges spread.0 mm apart. The dipole is at the origin, oriented along the y axis. What is the electric
More informationEngineering Sciences 22 Systems Summer 2004
Engineering Sciences 22 Systems Summer 24 BODE PLOTS A Bode plot is a standard format for plotting frequency response of LTI systems. Becoming familiar with this format is useful because: 1. It is a standard
More informationSimple Pendulum 10/10
Physical Science 101 Simple Pendulum 10/10 Name Partner s Name Purpose In this lab you will study the motion of a simple pendulum. A simple pendulum is a pendulum that has a small amplitude of swing, i.e.,
More informationDomain of a Composition
Domain of a Composition Definition Given the function f and g, the composition of f with g is a function defined as (f g)() f(g()). The domain of f g is the set of all real numbers in the domain of g such
More informationTechnical Guide No. 100. High Performance Drives  speed and torque regulation
Technical Guide No. 100 High Performance Drives  speed and torque regulation Process Regulator Speed Regulator Torque Regulator Process Technical Guide: The illustrations, charts and examples given in
More informationPHYS 210 Electronic Circuits and Feedback
PHYS 210 Electronic Circuits and Feedback These notes give a short introduction to analog circuits. If you like to learn more about electronics (a good idea if you are thinking of becoming an experimental
More informationlaboratory guide 2 DOF Inverted Pendulum Experiment for MATLAB /Simulink Users
laboratory guide 2 DOF Inverted Pendulum Experiment for MATLAB /Simulink Users Developed by: Jacob Apkarian, Ph.D., Quanser Hervé Lacheray, M.A.SC., Quanser Michel Lévis, M.A.SC., Quanser Quanser educational
More informationLecture 230 Design of TwoStage Op Amps (3/27/10) Page 2301
Lecture 230 Design of TwoStage Op Amps (3/27/0) Page 230 LECTURE 230 DESIGN OF TWOSTAGE OP AMPS LECTURE OUTLINE Outline Steps in Designing an Op Amp Design Procedure for a TwoStage Op Amp Design Example
More information19 LINEAR QUADRATIC REGULATOR
19 LINEAR QUADRATIC REGULATOR 19.1 Introduction The simple form of loopshaping in scalar systems does not extend directly to multivariable (MIMO) plants, which are characterized by transfer matrices instead
More information7. Beats. sin( + λ) + sin( λ) = 2 cos(λ) sin( )
34 7. Beats 7.1. What beats are. Musicians tune their instruments using beats. Beats occur when two very nearby pitches are sounded simultaneously. We ll make a mathematical study of this effect, using
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MT OpenCourseWare http://ocw.mit.edu.6 Signal Processing: Continuous and Discrete Fall 00 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
More informationOpAmp Simulation EE/CS 5720/6720. Read Chapter 5 in Johns & Martin before you begin this assignment.
OpAmp Simulation EE/CS 5720/6720 Read Chapter 5 in Johns & Martin before you begin this assignment. This assignment will take you through the simulation and basic characterization of a simple operational
More informationA C O U S T I C S of W O O D Lecture 3
Jan Tippner, Dep. of Wood Science, FFWT MU Brno jan. tippner@mendelu. cz Content of lecture 3: 1. Damping 2. Internal friction in the wood Content of lecture 3: 1. Damping 2. Internal friction in the wood
More informationStabilizing a Gimbal Platform using SelfTuning Fuzzy PID Controller
Stabilizing a Gimbal Platform using SelfTuning Fuzzy PID Controller Nourallah Ghaeminezhad Collage Of Automation Engineering Nuaa Nanjing China Wang Daobo Collage Of Automation Engineering Nuaa Nanjing
More informationCurrent Loop Tuning Procedure. Servo Drive Current Loop Tuning Procedure (intended for Analog input PWM output servo drives) General Procedure AN015
Servo Drive Current Loop Tuning Procedure (intended for Analog input PWM output servo drives) The standard tuning values used in ADVANCED Motion Controls drives are conservative and work well in over 90%
More informationChapter 24 Physical Pendulum
Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...
More informationMATH 52: MATLAB HOMEWORK 2
MATH 52: MATLAB HOMEWORK 2. omplex Numbers The prevalence of the complex numbers throughout the scientific world today belies their long and rocky history. Much like the negative numbers, complex numbers
More informationSystem Identification and State Feedback Controller Design of Magnetic Levitation System
International Journal of Engineering and Technical Research (IJETR) ISSN: 23210869, Volume2, Issue6, June 2014 System Identification and State Feedback Controller Design of Magnetic Levitation System
More informationMotor Control. Suppose we wish to use a microprocessor to control a motor  (or to control the load attached to the motor!) Power supply.
Motor Control Suppose we wish to use a microprocessor to control a motor  (or to control the load attached to the motor!) Operator Input CPU digital? D/A, PWM analog voltage Power supply Amplifier linear,
More informationNotch Filter Design. August 29, 2005
Notch Filter Design William East Brian Lantz August 29, 2005 1 Introduction This report summarizes an investigation made into designing notch filters with the control system for seismic isolation of Advanced
More informationSECTION 19 CONTROL SYSTEMS
ChristiansenSec.9.qxd 6:8:4 6:43 PM Page 9. The Electronics Engineers' Handbook, 5th Edition McGrawHill, Section 9, pp. 9.9.3, 5. SECTION 9 CONTROL SYSTEMS Control is used to modify the behavior of
More informationPartial Fraction Decomposition for Inverse Laplace Transform
Partial Fraction Decomposition for Inverse Laplace Transform Usually partial fractions method starts with polynomial long division in order to represent a fraction as a sum of a polynomial and an another
More informationSession 7 Bivariate Data and Analysis
Session 7 Bivariate Data and Analysis Key Terms for This Session Previously Introduced mean standard deviation New in This Session association bivariate analysis contingency table covariation least squares
More informationPID Control. 6.1 Introduction
6 PID Control 6. Introduction The PID controller is the most common form of feedback. It was an essential element of early governors and it became the standard tool when process control emerged in the
More informationSAMPLE CHAPTERS UNESCO EOLSS PID CONTROL. Araki M. Kyoto University, Japan
PID CONTROL Araki M. Kyoto University, Japan Keywords: feedback control, proportional, integral, derivative, reaction curve, process with selfregulation, integrating process, process model, steadystate
More informationProcess Instrumentation Terminology
Process Instrumentation Terminology Accuracy  Degree of conformity of an indicated value to a recognized accepted standard value, or ideal value. Accuracy, measured  The maximum positive and negative
More informationMAS.836 HOW TO BIAS AN OPAMP
MAS.836 HOW TO BIAS AN OPAMP OpAmp Circuits: Bias, in an electronic circuit, describes the steady state operating characteristics with no signal being applied. In an opamp circuit, the operating characteristic
More informationTime series Forecasting using HoltWinters Exponential Smoothing
Time series Forecasting using HoltWinters Exponential Smoothing Prajakta S. Kalekar(04329008) Kanwal Rekhi School of Information Technology Under the guidance of Prof. Bernard December 6, 2004 Abstract
More informationActive Vibration Isolation of an Unbalanced Machine Spindle
UCRLCONF206108 Active Vibration Isolation of an Unbalanced Machine Spindle D. J. Hopkins, P. Geraghty August 18, 2004 American Society of Precision Engineering Annual Conference Orlando, FL, United States
More informationInverted Pendulum Experiment
Introduction Inverted Pendulum Experiment This lab experiment consists of two experimental procedures, each with sub parts. Experiment 1 is used to determine the system parameters needed to implement a
More informationDCMS DC MOTOR SYSTEM User Manual
DCMS DC MOTOR SYSTEM User Manual release 1.3 March 3, 2011 Disclaimer The developers of the DC Motor System (hardware and software) have used their best efforts in the development. The developers make
More informationYear 9 set 1 Mathematics notes, to accompany the 9H book.
Part 1: Year 9 set 1 Mathematics notes, to accompany the 9H book. equations 1. (p.1), 1.6 (p. 44), 4.6 (p.196) sequences 3. (p.115) Pupils use the Elmwood Press Essential Maths book by David Raymer (9H
More informationIntroduction to Diophantine Equations
Introduction to Diophantine Equations Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles September, 2006 Abstract In this article we will only touch on a few tiny parts of the field
More informationNoise Canceling Headphones Shizhang Wu Supervisor: Ed Richter, Arye Nehorai, Walter Chen
Noise Canceling Headphones Shizhang Wu Supervisor: Ed Richter, Arye Nehorai, Walter Chen Department of Electrical and Systems Engineering Washington University in St. Louis Fall 2008 Abstract In this undergraduate
More informationPID Control. Proportional Integral Derivative (PID) Control. Matrix Multimedia 2011 MX009  PID Control. by Ben Rowland, April 2011
PID Control by Ben Rowland, April 2011 Abstract PID control is used extensively in industry to control machinery and maintain working environments etc. The fundamentals of PID control are fairly straightforward
More information5.1 Radical Notation and Rational Exponents
Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots
More informationOperational Amplifiers: Part 2. Nonideal Behavior of Feedback Amplifiers DC Errors and LargeSignal Operation
Operational Amplifiers: Part 2 Nonideal Behavior of Feedback Amplifiers DC Errors and LargeSignal Operation by Tim J. Sobering Analog Design Engineer & Op Amp Addict Summary of Ideal Op Amp Assumptions
More information