ECE382/ME482 Spring 2005 Homework 8 Solution December 11,


 Gregory Rose
 3 years ago
 Views:
Transcription
1 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 1 Solution to HW8 P1.21 We are given a system with open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1) and unity negative feedback. We are asked to design a pair of identical cascaded phaselead networks to compensate the system such that the closedloop system has velocity error constant K v = 2, phase margin near 45 degrees, and closed loop bandwidth at least ω = 4 rad/s. Solution: We might ask why we would want to design a pair of identical phaselead networks rather than a single network or two not necessarily identical ones. Hardware constraints would be most likely the reason we could not use a single network to obtain the entire phase margin adjustment. As for the request that the networks be identical, that minimizes the opportunity for installation errors (from the controls standpoint, they re in series so it doesn t matter which is first, but from a hardware standpoint, they may contain active components which require power sources or other connections which should not be confused). Also, if we wanted to keep a spare component on the shelf in case of failure, making the two networks the same means we only need to keep one spare rather than two. Now, to solve the stated problem, we first note that the definition of the velocity error constant is K v = lim sg(s), (2) s (see Section 5.7, page 262, of the textbook). Thus, to meet the design specification K v = 2 we must choose K = 2. Our next step (following the outline on page 69 of the text) is to determine the phase margin of the uncompensated system. Using the Matlab command margin(tf([2],[1/12 1/2+1/6 1 ])) we obtain the Bode plot shown in Figure 1, which indicates that the phase margin of the uncompensated system is 2.7 degrees at 5.29 rad/s. Next, we determine α. If we were designing a single lead compensator to add the entire needed phase margin degrees (1) α 1 = sin 66 deg (3) α + 1 but since we want to use a pair of identical compensators, we let each of them provide half the required phase margin, so solve instead using the Matlab command α 1 = sin33 deg, (4) α + 1
2 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 1: Open Loop for P1.21 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) alfa = (1+sin(33*pi/18))/(1sin(33*pi/18)) to obtain α = , so we ll let α = 3.4. Note that we must not call our variable alpha because Matlab has a function called alpha. Next, we find the frequency at which the magnitude for the uncompensated network is 1log 1 α or 5.3 db. This gives us ω m = 6.8 rad/s, as shown in Figure 2. We must also multiply by α to compensate for the attenuation, therefore each of our twin compensators is α(s + ωm / α) 3.4(s + 6.8/ 3.4) G c (s) = = s + ω m α s (5) 3.4 Our combined open loop transfer function is then G 2 c(s)g(s) = 2(3.4)(s + 6.8) 2 s(s ) 2 (s/2 + 1)(s/6 + 1) for which the phase margin is degrees, as shown in Figure 3. (6)
3 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 2: Finding ω m in the Open Loop for P1.21 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) System: untitled1 : 6.81 : Now we have to make a decision based on our application. Is degrees close enough to 45 degrees? If not, we would adjust the estimated phase margin needed, recompute α, and repeat the procedure above. Not having a specific application in mind, we ll assume that degrees will do. Finally, we must check to see that we have achieved the required closed loop bandwidth. As shown in Figure 4, we have achieved a bandwidth of ω BW = 7.75 rad/s which exceeds 4 rad/s so we have met our design criteria. Here s the Matlab script that I used, so that you can see how I automated the steps of checking the Bode plot to obtain the needed phase margin and the new crossover frequency. P1_21.m solves part of problem P1_21 from Dorf and Bishop 1th ed. 3 April 5 sk clear all
4 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 4 Figure 3: Open Loop for Compensated System of P1.21 Gm = 14.5 db (at 12.9 rad/sec), Pm = deg (at 4.26 rad/sec) G = tf([2],[1/12 1/2+1/6 1 ]); figure(1) margin(tf([2],[1/12 1/2+1/6 1 ])) print deps p1_21a having determined that the phase margin of the open loop system is 2.7 degrees, we solve for alpha to obtain half the needed additional phase margin of 66 degrees. (To automate things, I ve reiterated the margin command, this time with arguments to obtain the phase margin in the variable Pm.) [Gm,Pm,Wcg,Wcp] = margin(g); ph_needed = (45Pm)/2; alfa = (1+sin(ph_needed*pi/18))/(1sin(ph_needed*pi/18)) mmag = 1*log1(alfa)
5 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 5 Figure 4: Closed Loop for Compensated System of P1.21 Gm = 12.7 db (at 12.9 rad/sec), Pm = 76.3 deg (at 6.9 rad/sec) System: untitled1 : 7.75 : having obtained alpha I automate the process of finding the frequency omega_m at which the magnitude is mmag and then determine Gc wrange = logspace(2,3,1); [mag,ph] = bode(g,wrange); wm = interp1(2*log1(squeeze(mag)),wrange,mmag) Gc = tf(sqrt(alfa)*[1 wm/sqrt(alfa)],[1 wm*sqrt(alfa)]) GcGcG = series(series(gc,gc),g) figure(2) margin(gcgcg) print deps p1_21ol figure(3) margin(feedback(gcgcg,1)) print deps p1_21cl
6 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 6 P1.22 We are asked to design a phaselag network for the same system, having open loop transfer function K G(s) = (7) s(s/2 + 1)(s/6 + 1) and unity negative feedback, such that the closedloop system has velocity error constant K v = 2, phase margin near 45 degrees, and closed loop bandwidth at least ω = 2 rad/s. What s changed from P1.21 is that the bandwidth requirement has been relaxed. Solution: As in the previous problem, we first note that the definition of the velocity error constant is K v = lim sg(s), (8) s (see Section 5.7, page 262, of the textbook). Thus, to meet the design specification K v = 2 we must choose K = 2. As determined in the previous problem from the Bode plot shown in Figure 1, the phase margin of the uncompensated system is 2.7 degrees at 5.29 rad/s. From the Bode plot in Figure 1, we find the frequency ω c at which the phase margin (plus 5 degrees for the phase lag network) would be satisfied if the magnitude curve crossed db there. As shown in Figure 2, we choose ω c to be that frequency at which the phase is = 13 degrees, namely ω c = 1.12 rad/s. We see that we will need to adjust the attenuation by a gain of 23.7 db in order to have the magnitude curve cross db at ω c. We place the zero of the compensator one decade below ω c, i.e. at ω z =.11. We determine the value of α from 2log 1 α = 24 (9) to be α = The pole of the lag network is then placed at ω z /α.74, so lag compensator has transfer function G c = (s +.11) 15.22(s +.74) (1) and the closedloop transfer function will be T(s) = G c(s)g(s) 1 + G c (s)g(s) = 2(s +.11) 15.21(s +.74)(s/2 + 1)(s/6 + 1) + 2(s +.11). (11) As shown in Figure 5, the phase margin requirement is met. Figure 6 shows that the bandwidth requirement is met to within one percent, which we will declare to be adequate. Otherwise, we d have to fiddle with gains and pole locations to improve it. The Matlab script used to generate these results is shown below.
7 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 5: Open Loop for Compensated System of P1.22 Gm = 15.1 db (at 3.34 rad/sec), Pm = 44.6 deg (at 1.13 rad/sec) P1_22.m solves part of problem P1_22 from Dorf and Bishop 1th ed. 3 April 5 sk clear all G = tf([2],[1/12 1/2+1/6 1 ]) figure(1) margin(tf([2],[1/12 1/2+1/6 1 ])) print deps p1_22a wrange = logspace(2,3,1); [mag,ph] = bode(g,wrange); wc = interp1(squeeze(ph),wrange,13) mmag = 2*log1(interp1(wrange,squeeze(mag),wc)) alfa = 1^(mmag/2)
8 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 8 Figure 6: Closed Loop for Compensated System of P1.22 Gm = 13.4 db (at 3.34 rad/sec), Pm = 62 deg (at 1.67 rad/sec) System: untitled1 : 1.99 : Gc = tf([1 wc/1],alfa*[1 (wc/1)/alfa]) figure(2) margin(series(gc,g)) print deps p1_22ol figure(3) margin(feedback(series(gc,g),1)) print deps p1_22cl P1.23 We are again given the system with unity negative feedback and forward path transfer function open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1). (12) We are asked to design a leadlag compensator such that the closedloop system has
9 ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 9 velocity error constant K v = 2, phase margin near 45 degrees, and closed loop bandwidth at least ω = 2 rad/s but not more than 1 rad/s. We are given the general form of the leadlag compensator transfer function, with α = 1 for both the lead and lag portions, that is, G c (s) = (1 + s 1a )(1 + s b ) (1 + s a )(1 + s (13) ), and told to select a to characterize the lag portion of the compensator and b to characterize the lead portion of the compensator. Actually, if we check the text, we find that a should correspond to the lead portion and b should correspond to the lag portion. Solution: As in the previous problem, we first note that the definition of the velocity error constant is K v = lim sg(s), (14) s (see Section 5.7, page 262, of the textbook). Thus, to meet the design specification K v = 2 we must choose K = 2. We are told on page 612 of the textbook that the idea of using a leadlag network is to [provide] the attenuation of a phaselag network and the leadphase angle of a phaselead network. Let s see what this means. The network structure that s been specified in (13) fixes α = 1 for both the lead network and the lag network. Looking first at the lead network, we can solve 1b sinφ m = α 1 α + 1 (15) for the phase angle that we will obtain from the lead filter (irrespective of the actual values of the zero and pole that we eventually choose). The resulting phase contribution is φ m = sin 1 (9/11) = 55 deg. (16) On the other hand, the attenuation introduced by the phase lag network is 2log 1 α at the chosen crossover frequency, (see Step 6 on page 69 of the text). This is 2 db. The uncompensated system has magnitude 2 db at 1.54 rad/s as shown in Figure 7. Then from Step 6, we choose crossover frequency ω c = 1.54 and (as in Step 4), place the zero of the lag filter one decade below the crossover frequency at.154 rad/s. The pole is then placed at.154/α =.154 rad/s. Let s see what we have so far. The phase margin of the combined system is 22.2 degrees as shown in Figure 8. Now we have to correct the phase margin. Our lead filter will give us 3 degrees of phase margin at its crossover frequency, so we must choose the zero and pole accordingly. The crossover frequency should be ω m = zp with p = ω m α. From the Bode plot of this open loop system, we choose ω m to be that frequency at which the phase is = 165 rad/s. This gives us ω m = 2.35 rad/s. Accordingly, we choose p = and
10 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 7: Open Loop for System of P1.23 Gm = 7.96 db (at 3.46 rad/sec), Pm = 2.7 deg (at 5.29 rad/sec) System: G : 1.54 : z = 2.35/ 1 and obtain phase margin of 47 degrees, as shown in Figure 9, and closedloop bandwidth of around 6.35 rad/s as show in Figure 1. The leadlag compensator obtained has transfer function G c = The Matlab script used is given below. (s +.154)(s +.154). (17) s +.743)(s ) P1_23.m solves part of problem P1_23 from Dorf and Bishop 1th ed. 3 April 5 sk clear all G = tf([2],[1/12 1/2+1/6 1 ]) figure(1)
11 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 8: Open Loop for Lag Compensated System of P1.23 Gm = 11.2 db (at 3.3 rad/sec), Pm = 32.8 deg (at 1.54 rad/sec) System: untitled1 : 2.35 : margin(g) print deps p1_23a alfa=1; mmag = 2*log1(alfa) wrange = logspace(2,3,1); [mag,ph] = bode(g,wrange); wcp = interp1(2*log1(squeeze(mag)),wrange,mmag) Gc_lag = tf(1/1*[1 wcp/1],[1 (wcp/1)/alfa]) figure(2) margin(series(gc_lag,g)) print deps p1_23ol1 wm=2.35 where phase is =165 Gc_lead = tf(alfa*[1 wm/sqrt(alfa)],[1 wm*sqrt(alfa)])
12 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 9: Open Loop for Compensated System of P1.23 Gm = 1.7 db (at 7.69 rad/sec), Pm = 47 deg (at 3.69 rad/sec) figure(3) margin(series(gc_lead,series(gc_lag,g))) print deps p1_23ol2 figure(4) margin(feedback(series(gc_lead,series(gc_lag,g)),1)) print deps p1_23cl
13 ECE382/ME482 Spring 25 Homework 8 Solution December 11, Figure 1: Closed Loop for Compensated System of P1.23 Gm = 7.74 db (at 7.69 rad/sec), Pm = 41.5 deg (at 5.59 rad/sec) System: untitled1 : 6.37 :
ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =
ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.
More informationModule 5: Design of Sampled Data Control Systems Lecture Note 8
Module 5: Design of Sampled Data Control Systems Lecture Note 8 Laglead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationLINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad
LINEAR CONTROL SYSTEMS Ali Karimpour Associate Professor Ferdowsi University of Mashhad Controller design in the frequency domain Topics to be covered include: Lag controller design 2 Dr. Ali Karimpour
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationControl Systems. Control Systems Design LeadLag Compensator.
Design LeadLag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response
More informationBoise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain
Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism:
More informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationDesign via Frequency Response
4th Edition E L E V E N Design via Frequency Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Gain Design a. The required phase margin for 25% overshoot (ζ = 0.404), found from Eq. (10.73),
More informationECE382/ME482 Spring 2005 Homework 1 Solution February 10,
ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Solution to HW1 P2.33 For the system shown in Figure P2.33 on p. 119 of the text, find T(s) = Y 2 (s)/r 1 (s). Determine a relationship that
More information= rad/sec. We can find the last parameter, T, from ωcg new
EE572 Solution to HW#22. Keep working on your project!! 1. Consider the following system: W(s) + T s =1 msec G lead (z) G zoh (z) 8 ( s+ 4)  a) Design a lead compensator, G lead (z), which meets the following
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More informationFrequency (rad/s)
. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer
More informationMAE 143B  Homework 9
MAE 43B  Homework 9 7.2 2 2 3.8.6.4.2.2 9 8 2 2 3 a) G(s) = (s+)(s+).4.6.8.2.2.4.6.8. Polar plot; red for negative ; no encirclements of, a.s. under unit feedback... 2 2 3. 4 9 2 2 3 h) G(s) = s+ s(s+)..2.4.6.8.2.4
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
More informationPM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
More information6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson
Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closedloop behavior what we want it to be. To review:  G c (s) G(s) H(s) you are here! plant For
More informationProblem Weight Score Total 100
EE 350 EXAM IV 15 December 2010 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationLoop shaping exercise
Loop shaping exercise Excerpt 1 from Controlli Automatici  Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA  La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More information2.010 Fall 2000 Solution of Homework Assignment 8
2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)
More informationDesired Bode plot shape
Desired Bode plot shape 0dB Want high gain Use PI or lag control Low freq ess, type High low freq gain for steady state tracking Low high freq gain for noise attenuation Sufficient PM near ω gc for stability
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually
More information6.302 Feedback Systems Recitation 17: Black s Formula Revisited, and Lead Compensation Prof. Joel L. Dawson
Recitation 7: Black s Formula Revisited, and Lead Compensation By now, applying Black s Formula to a feedback system is almost a reflex: x(s) G(s) Y(s) H(s) This formula actually lends itself rather naturally
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationAnalysis and Design of Analog Integrated Circuits Lecture 12. Feedback
Analysis and Design of Analog Integrated Circuits Lecture 12 Feedback Michael H. Perrott March 11, 2012 Copyright 2012 by Michael H. Perrott All rights reserved. Open Loop Versus Closed Loop Amplifier
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More informationMAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ)
MAE 42 Homework #2 (Design Project) SOLUTIONS. Top Dynamics (a) The main body s kinematic relationship is: ω b/a ω b/a + ω a /a + ω a /a ψâ 3 + θâ + â 3 ψˆb 3 + θâ + â 3 ψˆb 3 + C 3 (ψ) θâ ψ + C 3 (ψ)c
More informationECE 388 Automatic Control
Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More information6.302 Feedback Systems
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Fall Term 2005 Issued : November 18, 2005 Lab 2 Series Compensation in Practice Due
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
More informationEngraving Machine Example
Engraving Machine Example MCE44  Fall 8 Dr. Richter November 24, 28 Basic Design The Xaxis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer
More informationTopic # Feedback Control
Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More information] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command prefilter [ 0.
EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965372 Problem (Analysis of a Feedback System) Consider the feedback system
More informationProblem 1 (Analysis of a Feedback System  Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1.
1 EEE480 Final Exam, Spring 2016 A.A. Rodriguez Rules: Calculators permitted, One 8.5 11 sheet, closed notes/books, open minds GWC 352, 9653712 Problem 1 (Analysis of a Feedback System  Bode, Root Locus,
More informationLecture 11. Frequency Response in Discrete Time Control Systems
EE42  Discrete Time Systems Spring 28 Lecturer: Asst. Prof. M. Mert Ankarali Lecture.. Frequency Response in Discrete Time Control Systems Let s assume u[k], y[k], and G(z) represents the input, output,
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More information1 Chapter 9: Design via Root Locus
1 Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B); b. responses from poles at A and B 2
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequencydomain analysis and control design (15 pt) Given is a
More informationCDS 101/110a: Lecture 81 Frequency Domain Design
CDS 11/11a: Lecture 81 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More information(Refer Slide Time: 2:11)
Control Engineering Prof. Madan Gopal Department of Electrical Engineering Indian institute of Technology, Delhi Lecture  40 Feedback System Performance based on the Frequency Response (Contd.) The summary
More informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationFEEDBACK AND STABILITY
FEEDBCK ND STBILITY THE NEGTIVEFEEDBCK LOOP x IN X OUT x S + x IN x OUT Σ Signal source _ β Open loop Closed loop x F Feedback network Output x S input signal x OUT x IN x F feedback signal x IN x S x
More informationActive Control? Contact : Website : Teaching
Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationUnit 11  Week 7: Quantitative feedback theory (Part 1/2)
X reviewer3@nptel.iitm.ac.in Courses» Control System Design Announcements Course Ask a Question Progress Mentor FAQ Unit 11  Week 7: Quantitative feedback theory (Part 1/2) Course outline How to access
More informationIMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION. K. M. Yanev A. Obok Opok
IMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multistage compensation is suggested. A
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More information1(b) Compensation Example S 0 L U T I 0 N S
S 0 L U T I 0 N S Compensation Example I 1U Note: All references to Figures and Equations whose numbers are not preceded by an "S"refer to the textbook. (a) The solution of this problem is outlined in
More informationx(t) = x(t h), x(t) 2 R ), where is the time delay, the transfer function for such a e s Figure 1: Simple Time Delay Block Diagram e i! =1 \e i!t =!
1 TimeDelay Systems 1.1 Introduction Recitation Notes: Time Delays and Nyquist Plots Review In control systems a challenging area is operating in the presence of delays. Delays can be attributed to acquiring
More informationDepartment of Aerospace Engineering and Mechanics University of Minnesota Written Preliminary Examination: Control Systems Friday, April 9, 2010
Department of Aerospace Engineering and Mechanics University of Minnesota Written Preliminary Examination: Control Systems Friday, April 9, 2010 Problem 1: Control of Short Period Dynamics Consider the
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationFeedback design for the Buck Converter
Feedback design for the Buck Converter Portland State University Department of Electrical and Computer Engineering Portland, Oregon, USA December 30, 2009 Abstract In this paper we explore two compensation
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability
More informationEE451/551: Digital Control. Final Exam Review Fall 2013
EE45/55: Digital Control Final Exam Review Fall 03 Exam Overview The Final Exam will consist of four/five questions for EE45/55 students based on Chapters 7 and a bonus based on Chapters 8 9 (students
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationElectronics II. Final Examination
f3fs_elct7.fm  The University of Toledo EECS:3400 Electronics I Section Student Name Electronics II Final Examination Problems Points.. 3 3. 5 Total 40 Was the exam fair? yes no Analog Electronics f3fs_elct7.fm
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationThe Nyquist Stability Test
Handout X: EE24 Fall 2002 The Nyquist Stability Test.0 Introduction With negative feedback, the closedloop transfer function A(s) approaches the reciprocal of the feedback gain, f, as the magnitude of
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationAutomatic Control (TSRT15): Lecture 7
Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Outline 2 Feedforward
More information