Exercises for lectures 13 Design using frequency methods

Save this PDF as:
Size: px
Start display at page:

Download "Exercises for lectures 13 Design using frequency methods"

Transcription

1 Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control

2 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition) Value is L( jc ) T( jc ) 1 L( j ) c but it also depends on the phase L( j c ), thus on PM (Phase Margin) For PM = 90 is L( j ) j and small phase - 90, so c j 1 T( jc ) j 2 L( j ) 1 In this case the closed loop bandwidth equals to the transition frequency of open loop exactly! BW c For small PM, the value T( j c ) rises and appears a resonance peak. This bandwidth BW moves to the right, but usually does not exceed 2 c. It is usually c BW 2c. Therefore, we set (OL!!!) in order to ensure the required (CL!!!) c BW Michael Šebek ARI c

3 Relationship of ω c and ω BW Bode plot T( j) with marked and values c BW for various PM It is usually 2 c BW c For a 2 nd order system without zeros, the dependency on is show in the figure c BW C , BW (1 2 ) Michael Šebek ARI

4 Repetition: steady behavior of the Bode plot >> L=(1+s)/(2+s)/(3+s), M 15dB v=value(l,0),l=l/v*10^(15/20),k=value(l,0),bode(l) L = s / 6 + 5s + s^2 K = >> KpdB=20*log10(abs(value(L,j*.01))), Kp=10^(15/20) KpdB = , Kp = >> einfty = 1/(1+Kp) einfty = Initial slope is 0 so the system is of type 0 initial value (without a pole at 0) asymptote is 15 db and thus K 15d B p steady-state error to a step is estep,ss 1 1 K p L=(1+s)/(2+s)/(3+s)/s,v=value(coprime(s*L),0);L=L/v*10, L = s / 6s + 5s^2 + s^3 Kv=value(coprime(s*L),0),bode(L) Kv = Initial slope is 20 db/dek and so the system is of type 1 (with one pole at 0) stretched "initial asymptote" intersects the zero line for frequency 10 and thus K 10 Steady-state error to a ramp v e ( ) ramp K v Michael Šebek Pr-ARI

5 Comparison of time and frequency responses Michael Šebek Pr-ARI

6 Example: Setting K p by a P regulator System is 5 Gs () s 2 32dB K e p ss 2.5, K 20 log 2.5 8dB 1 1 K p p,db 0.29 We want K e ss,2 ss, Kp,2 99 ess,2 p,2,db We use K 1 e 20 log 99 40dB K p,2 K p db p,2, p,db 39.6 K K K dB (Beware - the result is very fast, with a large action peak) 6

7 58390 System Gs () has s s 36 s 100 Example: Setting K v by a P regulator eramp ( ) K If we want to reduce the steady-state error to ramp 10x, we must set Kv We increase the gain 10x, what leads to Ls () We obtain s s 36 s 100 K v v but beware, the result is unstable! Here, P controller will not solve the task! 7

8 Example: Setting of gain for required PM For the position control system in the figure set the preamp gain so that the resulting system reaches 9.5% overshoot by a step of reference. From the required overshoot we calculate the damping (of dominant poles) ln(%os 100) ln(0, 095) , ln (%OS 100) ln (0,095) and from that we obtain PM PM arctan arctan (0.6) 1 4 (0.6) The open loop transfer function has an indefinite K To draw the Bode plot and perform a design on the plot we have to choose some K. Let choose K = 3.6 and obtain Ls () s s L K K 36s 100 () s s s s 100 8

9 Example: Setting of gain for required PM We draw a Bode plot LK 3.6 ( s) 360 s s 36s 100 and find a frequency, for which L( j) From the graph we subtract 14.8rad s For this frequency the amplitude is L( ) M( ) dB and therefore we must increase gain by 44.2 db, so cca 162.2x. Then we obtain Ls () s s s 100 Simulation verifying the correct design is necessary. We continue later with this example and for this purpose we measure Kv e ( ) ramp 44.2dB rad s 9

10 Example: Setting PD System transfer function (aircraft attitude) Requirements e K 1 e 2257 ramp, ss v ramp, ss PM 80 First set K p = , to increase K v,1 = 12.5 to K v =2258 and to ensure the required regulation error. Then search for the part 1 KDs Obtained PD regulator for the system is K G() s P ss Gs () K p 45dB Kv, ss Kv,

11 Example: Setting PD We draw the Bode plot for system L( s) KP 1 KDs G( s) s s for K d = 0. We find ω D, at which PM = required (regulator phase at ω D ) = = 35 where the phase is = = It is D 516. We calculate 1 1 KD Resulting L has a Bode plot. The requirement is fulfilled: PM = K Ds D 145 Phase of PD regulator D K P K D D KP KD 10K P K D 11

12 One more example: Setting PD For a transfer function Gs () s s Consider, we already designed K P = 1 and now we set K D in PD regulator for good PM We draw Bode plot for following values KD 0, 0.002, 0.005, 0.02 Uncompensated system (K d = 0) has PM = 7.78 To reach PM 58.5 PM = 80, regulator should PM add 72,22 to the new ω c From figure it follows, that it is impossible. High regulator gain PM 7.78 shifts ω c to higher frequencies, where phase of the uncompensated system declining faster than it is increased by the compensator s D K s PM 25.9

13 For a transfer function Gs () Find a PI regulator, that increases PM = 22.6 to PM new = 65 Draw a Bode graph Ls () K s K K 2 s s First for K p = 1 and K I = 0 From requirement PM new =65 find ω c,new = 170 rad/s and calculate K P G j K I choose so that the corner freq. is less than a decade ω c,new K K P I P ( c, new ) db K I P c, new I P c, new 10 K ss Example: Setting PI PM new 65 c, new c PM 22.6

14 Example: Setting PI For this K 1.42 calculate the transfer function and draw the Bode plot Ls () We obtain PM new =59, what does not satisfy the requirement. Lets try to use a smaller K I (= move the corner frequency to left). For example K I = 0.07 leads to the transfer function L () s 2 2 with PM new = 64.3 I KP s KI KP 68489s s s s s s s s

15 PID See the attached document. 15

16 Example: Lag regulator design Task: For a plat give by a transfer function Fs () 1 2s 30 s s design a Lag regulator satisfying these requirements: e, 0.05, PM 45 ss ramp Solution: 1. Find the value of the gain providing the desired deviation: ( ) ( ) K L1 s KF s s s 2 30 e ss, ramp s K 1200 Kv lim sl1 ( s) K K 0.05 s0 230 This OL transfer function gives incorrect PM and GM >> K=1200;F=1/s/(s+2)/(s+30);L1=K*F L1 = 1200 / 60s + 32s^2 + s^3 >> [GM,PM,om_cp,om_cg]=margin(tf(L1)) GM = PM = om_cp = om_cg = >> GM_dB = 20*log10(GM) GM_dB = Michael Šebek ARI

17 Phase (deg) Magnitude (db) Example: Lag regulator design 2. Draw a Bode plot L L1 () s s s s 30 From the required PM we calculate necessary phase and we find new ω c,new = 1.28 rad/s At this frequency, we find the necessary attenuation c, new Bode Diagram C( j ) 22.1dB db System: untitled1 Frequency (rad/s): 1.31 Magnitude (db): 22.1 System: untitled1 Frequency (rad/s): 1.28 Phase (deg): Frequency (rad/s) 3. We calculate the parameter a from the measurements or from a transfer fcn. a C( j ) 22.1dB c, new db C( j ) c, new = C( j ) 10 db c, new >> aa=1/abs(value(l1,j*1.28)) aa = Michael Šebek ARI

18 4. We calculate zero and pole 5. The final regulator is p c z c 10 c, new Example: Lag regulator design az c C lag () s as pc s s p s c 6. Finally, we verify if the regulator satisfies the requirements. Michael Šebek ARI

19 Phase (deg) Magnitude (db) Example: Lag regulator design Bode Diagram rad/s rad/s System: untitled3 Phase Margin (deg): 49 Delay Margin (sec): 0.65 At frequency (rad/s): 1.32 Closed loop stable? Yes Frequency (rad/s) >> Kv=value(coprime(s*L2),0), e_ss_ramp=1/kv Kv = , e_ss_ramp = Michael Šebek ARI

20 Other Example: Lag compensation In the positioning control system, Ls () the gain was by previous method adjusted so ss 36s 100 that the system has overshoot 9.5% and 1 Kv eramp ( ) Kv Add Lag compensation so that the steady state value to the ramp is 10x smaller and the overshoot does not increase The steady state leads to Kv 162.2, so we have to increase the gain 10 and then we obtain Ls () The overshot requirement 9.5% leads to ss 36s PM 59.2 Because Lag decreases PM only little, but still (we expect a deteriorateon PM 5 12 ), we consider rather PM Lets find a frequency, for which the phase is L( j)

21 Lag compensation From required phase we determine frequency 9.8rad s 24dB Ls () 36s 100 s s and then the value 20log M( ) 24dB rad s From the definition, PM for should be 20log M( ) 0dB 9.8rad s Lag should have at the frequency attenuation 24dB 21

22 Lag compensation Draw the asymptote for higher frequencies in 20log M( ) 24dB 1 T 0.062rad s 20dB dek 24dB 1 T 0.98rad s The upper corner frequency is chosen by a decade left from 9.8rad s, it is 1 T 0.98rad s From there we continue with the slope 20dB dek to 0dB, what we reach for 1 T 0.062rad s After substitution we obtain s 1 T s 0.98 Cs () s 1 T s It has correct shape, but not the gain, so we set up the DC gain of the compensator s 0.98 K DC( s) KCC( s) C 1 p z DC(0) 1 0dB s

23 Lag compensation The result is s 100 s s 0.063( s 0.98) s ( s 0.98) s s 36s100 s Compensated system Lag compensator Amplified uncompensated system Step response Ramp response 23

24 Lets get back to the positioning control system and design a regulator according to specifications: OS 20%, K v = 40, T p = 0,1s First set up gain so, that K 40 K lim sl( s) K 40 K 1440 v s0 Lets substitute it and continue From the given specification we calculate PM a ω BW : v Example: Lead compensation Ls () s s Ls () s s 100K 36s s 100 ln(%os 100) PM arctan ln (%OS 100) BW 2 1 T p rad s 24

25 Lead compensation Draw a Bode plot for This uncompensated system has PM = 34,1 By Lead compensation we increase PM to required value Since Lead also increases ω C, we add also some compensation factor. To compensate the smaller phase for larger frequencies ω C we choose the factor as 10º. Ls () s 100 s s We require the regulator phase increase of

26 Lead compensation We require the regulator phase increase of 48, = 24,1 Generally the compensated system should have PM 48.1 a BW 46.6rad s It should not produce satisfactory results, we have to repeat the design with other correction factor. From the phase growth requirement we have max 24.1 and from it It follows that 1 sin 1 sin max max D( max ) 3.76dB If we choose C, new max, then at this frequency the amplitude of the uncompensated system should be -3,76 db According to that we find ω max Michael Šebek Pr-ARI

27 Lead compensation On the Bode plot Ls () s s s 100 We measure max 39rad s. Then from and 0.42 we calculate max 3.76dB max 39rad s max T , 60.2 T T and in the end, we obtain the search factor 1 s ( ) T s Ds s s 60.2 T 27

28 Lead compensation The result is: Compensated system Uncompensated system Lead compensator Simulation: OS% 22.6, PM 45.5, 39 rad s 68.8rad s, T 0.075s, K 40 BW p v C 28

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

EE 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 information

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such

More information

Homework 7 - Solutions

Homework 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 information

Transient 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

Transient 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 information

Systems Analysis and Control

Systems 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 information

Exercise 1 (A Non-minimum Phase System)

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Non-minimum Phase System) To increase the rise time of the system, we

More information

Exercise 1 (A Non-minimum Phase System)

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Non-minimum Phase System) To decrease the rise time of the system,

More information

Systems Analysis and Control

Systems 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 information

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

The 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 information

Module 5: Design of Sampled Data Control Systems Lecture Note 8

Module 5: Design of Sampled Data Control Systems Lecture Note 8 Module 5: Design of Sampled Data Control Systems Lecture Note 8 Lag-lead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.

More information

Outline. Classical Control. Lecture 5

Outline. 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 information

LINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad

LINEAR 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 information

Automatic 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 Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback

More information

1 (20 pts) Nyquist Exercise

1 (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 information

Lecture 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 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 information

CDS 101/110a: Lecture 8-1 Frequency Domain Design

CDS 101/110a: Lecture 8-1 Frequency Domain Design CDS 11/11a: Lecture 8-1 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 information

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

EE 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 information

PID controllers. Laith Batarseh. PID controllers

PID controllers. Laith Batarseh. PID controllers Next Previous 24-Jan-15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time

More information

MAS107 Control Theory Exam Solutions 2008

MAS107 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 information

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain

Boise 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 information

Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types

Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This

More information

ECE 388 Automatic Control

ECE 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 information

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

(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

Analysis and Design of Analog Integrated Circuits Lecture 12. Feedback

Analysis 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 information

Active Control? Contact : Website : Teaching

Active 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 information

CDS 101/110 Homework #7 Solution

CDS 101/110 Homework #7 Solution Amplitude Amplitude CDS / Homework #7 Solution Problem (CDS, CDS ): (5 points) From (.), k i = a = a( a)2 P (a) Note that the above equation is unbounded, so it does not make sense to talk about maximum

More information

FREQUENCY-RESPONSE DESIGN

FREQUENCY-RESPONSE DESIGN ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins

More information

EE3CL4: Introduction to Linear Control Systems

EE3CL4: 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 information

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10

Today (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 information

Stability of CL System

Stability 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 information

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

VALLIAMMAI 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 information

Desired Bode plot shape

Desired 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 information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open

More information

PM 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 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 information

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability

Frequency 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 information

Frequency Response Techniques

Frequency 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 information

(a) Find the transfer function of the amplifier. Ans.: G(s) =

(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 closed-loop system

More information

EE 4343/ Control System Design Project LECTURE 10

EE 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 Phase-lead and Phase-lag compensators using

More information

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

CHAPTER 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 information

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD 206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)

More information

Frequency Response Analysis

Frequency Response Analysis Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions

More information

Engraving Machine Example

Engraving Machine Example Engraving Machine Example MCE44 - Fall 8 Dr. Richter November 24, 28 Basic Design The X-axis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional

More information

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage 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, Force-current and Force-Voltage analogies..

More information

ECE317 : Feedback and Control

ECE317 : Feedback and Control ECE317 : Feedback and Control Lecture : Steady-state error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace

More information

Asymptotic Bode Plot & Lead-Lag Compensator

Asymptotic Bode Plot & Lead-Lag Compensator Asymptotic Bode Plot & Lead-Lag Compensator. Introduction Consider a general transfer function Ang Man Shun 202-2-5 G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer

More information

AMME3500: System Dynamics & Control

AMME3500: 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 information

Root Locus Design Example #3

Root Locus Design Example #3 Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll

More information

Performance of Feedback Control Systems

Performance of Feedback Control Systems Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type

More information

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018

More information

Reglerteknik: Exercises

Reglerteknik: Exercises Reglerteknik: Exercises Exercises, Hints, Answers Liten reglerteknisk ordlista Introduktion till Control System Toolbox ver. 5 This version: January 3, 25 AUTOMATIC CONTROL REGLERTEKNIK LINKÖPINGS UNIVERSITET

More information

Chapter 9: Controller design

Chapter 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 information

Root Locus Methods. The root locus procedure

Root Locus Methods. The root locus procedure Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain

More information

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral

More information

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method .. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to

More information

Step Response Analysis. Frequency Response, Relation Between Model Descriptions

Step Response Analysis. Frequency Response, Relation Between Model Descriptions Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step

More information

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

SAMPLE 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: Frequency-domain analysis and control design (15 pt) Given is a

More information

ECE382/ME482 Spring 2005 Homework 8 Solution December 11,

ECE382/ME482 Spring 2005 Homework 8 Solution December 11, 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

More information

Prüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Prüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid

More information

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES CHAPTER 7 STEADY-STATE 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 information

Control Systems. Control Systems Design Lead-Lag Compensator.

Control Systems. Control Systems Design Lead-Lag Compensator. Design Lead-Lag 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 information

16.30/31, Fall 2010 Recitation # 2

16.30/31, 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 information

Digital Control Systems

Digital 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 information

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design. Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and

More information

Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...

Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature... Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................

More information

Electronics II. Final Examination

Electronics 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 information

Automatic Control (TSRT15): Lecture 7

Automatic 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: 13-282226 Office: B-house extrance 25-27 Outline 2 Feedforward

More information

Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus - 1

Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus - 1 Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position

More information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III

More information

ECE 486 Control Systems

ECE 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 information

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) = 1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot

More information

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got

More information

Positioning Servo Design Example

Positioning 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 pick-and-place robot to move the link of a robot between two positions. Usually

More information

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1 Code No: R06 R0 SET - II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics

More information

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2

More information

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators. Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8- am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total

More information

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor 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 information

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0. 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)

More information

DIGITAL CONTROLLER DESIGN

DIGITAL CONTROLLER DESIGN ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steady-state accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some

More information

LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693

LABORATORY 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 information

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and

More information

Step input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system?

Step input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system? IC6501 CONTROL SYSTEM UNIT-II TIME RESPONSE PART-A 1. What are the standard test signals employed for time domain studies?(or) List the standard test signals used in analysis of control systems? (April

More information

Controller Design using Root Locus

Controller Design using Root Locus Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers

More information

Table of Laplacetransform

Table of Laplacetransform Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e- at, an exponential function s + a sin wt, a sine fun

More information

STABILITY OF CLOSED-LOOP CONTOL SYSTEMS

STABILITY OF CLOSED-LOOP CONTOL SYSTEMS CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 10-1 Road Map of the Lecture X Stability of closed-loop control

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007

MASSACHUSETTS 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

Due Wednesday, February 6th EE/MFS 599 HW #5

Due Wednesday, February 6th EE/MFS 599 HW #5 Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]

More information

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 5. 2. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -

More information

MAE 143B - Homework 9

MAE 143B - Homework 9 MAE 143B - Homework 9 7.1 a) We have stable first-order 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 straight-line

More information

Control Systems I Lecture 10: System Specifications

Control Systems I Lecture 10: System Specifications Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture

More information

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Radar 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 information

MAE 143B - Homework 9

MAE 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 information

INTRODUCTION TO DIGITAL CONTROL

INTRODUCTION TO DIGITAL CONTROL ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant

More information

Solutions to Skill-Assessment Exercises

Solutions to Skill-Assessment Exercises Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part

More information

If you need more room, use the backs of the pages and indicate that you have done so.

If you need more room, use the backs of the pages and indicate that you have done so. EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty

More information

Robust Performance Example #1

Robust 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 information

D(s) G(s) A control system design definition

D(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 information

Compensator Design to Improve Transient Performance Using Root Locus

Compensator Design to Improve Transient Performance Using Root Locus 1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning

More information

Dynamic Compensation using root locus method

Dynamic 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 information

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2) Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)

More information