# Root Locus. Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering

Size: px
Start display at page:

Download "Root Locus. Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering"

Transcription

1 Root Locus Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering Recall, the example of the PI controller car cruise control system. A consequence of choosing the parameter for the integral control, K i, is that it affects the location of the poles and zeros of the cruise control system on the s-plane. This different pole/zero locations for K i = 50 and K i = 100 are shown below Figure 1: Pole/Zeros Diagrams in the car cruise control example for K i = 50 and K i = 100 Knowing the locations of the poles and zeros is important because 1. It affects the transient response. 2. It indicates whether the system is stable or not. 1

2 Root Locus An important property of the control system is the variation of these locations as the controller gain changes. Therefore we need to quantify these variations. In this handout we introduce the idea the root locus which as a graphical means of quantifying the variations in pole locations (but not the zeros). 3C1 Signals and Systems 2 Control Systems

3 Root Locus 1 CLOSED LOOP SYSTEM STABILITY 1 Closed Loop System Stability Recall that any system is stable if all the poles lie on the LHS of the s-plane. Ensuring stability for an open loop control system, where H(s) = C(s)G(s), is straightforward as it is sufficient merely to use a controller such that the cascade C(s)G(s) only has poles on the LHS of the complex plane. The poles of the open loop system will not be affected by the gain of the controller. i.e. If C(s)G(s) is stable then KC(s)G(s) will be stable for all values of K. However for a closed loop system this will not be the case. Considering the following block diagram we get a transfer function H(s) = = = C(s)G(s) 1 + C(s)G(s) K (s+1)(s+2)(s+3) K 1 + (s+1)(s+2)(s+3) K s 3 + 6s s K (1) Looking at the open loop transfer(h(s) = C(s)G(s)) for this system, all the poles (s = 1, s = 2, s = 3) lie in the LHS of the s-plane and hence the system is stable. For the closed-loop system if the gain K = 10 the poles lie in the LHS of the plane and so it is a stable system. However, if the gain is increased to K = 200 then there are 2 poles on the RHS of the s-plane and so the system is unstable. 3C1 Signals and Systems 3 Control Systems

4 Root Locus 1 CLOSED LOOP SYSTEM STABILITY Figure 2: Pole Zero Plots for the system transfer function in Eq. (1) for K = 10 (left) and K = 200 (right). It is also possible to have an unstable open loop system and a stable closed loop system. For example, consider the previous example where G(s) is now G(s) = s + 1 (s 1)(s + 2)(s + 3). (2) As G(s) has a pole at s = 1, the open loop system C(s)G(s) is clearly unstable. However in the closed loop system, if we choose K = 10 then the poles are located as shown in Fig. 3 and the system is stable. Figure 3: Pole Zero Plots for the system transfer function in Eq. (2) for K = 10. 3C1 Signals and Systems 4 Control Systems

5 Root Locus 2 ROOT LOCUS 2 Root Locus Revisiting the example in the last page, the system above has a transfer function H(s) = K s 3 + 6s s K. Starting at a value of K = 0 calculate the locations of the poles of H(s) and repeat for increasingly large values of K. If we trace the locations of these poles as they move across the s-plane the following plot is obtained. This plot is called the root locus. 3C1 Signals and Systems 5 Control Systems

6 Root Locus 2 ROOT LOCUS Observations Because we have a 3 rd Order System, there are 3 separate plots on the root locus, one for each root. The plot is symmetric about the Real Axis. This is because complex roots occur in conjugate pairs. Each plot starts at a location equal to the location of a root of the plant transfer function. Each plot ends at infinity. The asymptotes of each plot as they tend infinity, share a common centre. Their angles are evenly spread out over 360. Two of the plots cross the imaginary axis. This corresponds to the system becoming unstable with increasing K. All parts of the real axis to the left of an odd number of poles are part of one of the plots. 2 of the plots break off from the real axis when they meet. Before they break off the system has 3 real poles, after they break off 2 roots are complex and one is real. A root locus plot also exists for negative values of K. The root locus plot for 0 K > shown below. In 3c1 we focus almost exclusively on the root locus for positive K. 3C1 Signals and Systems 6 Control Systems

7 Root Locus 2 ROOT LOCUS 2.1 The Root Locus Concept We will now look at the root locus in general and look in more detail at some of the observations we made from our example. Consider a closed loop system above with a simple feedback loop that uses simple proportional controller. It has a transfer function H(s) = KG(s) 1 + KG(s) = n(s) d(s) The poles occur at the roots of d(s). For system transfer function above these roots occur where 1 + KG(s) = 0. (3) This is referred to as the characteristic equation of the system. Therefore, to determine the poles it is necessary that KG(s) = 1 (4) G(s) = 180 ± k360 for k Z. (5) 3C1 Signals and Systems 7 Control Systems

8 Root Locus 2 ROOT LOCUS These two equations are referred to as the Magnitude and Angle criteria respectively. Therefore, the root locus is the path of the roots of the characteristic equation with respect to K as K is varied between 0 and infinity. 3C1 Signals and Systems 8 Control Systems

9 Root Locus 2 ROOT LOCUS Example We will show that by manipulating the denominator polynomial it is possible to generate a root locus plot for the variation of other transfer function parameters. For the system above the characteristic equation of the root locus due to variations in K can be written directly from Eq. 3 as 1 + K 1 s(s + τ) = 0 for a fixed τ. However we can also rewrite the characteristic equation if we wished to see the effect of variations of τ on the roots of the control system K s(s + τs) = 0 s 2 + τs + K = 0 s 1 + τ = 0 for a fixed K. s 2 + K Comment In general, it is often possible to generate a characteristic equation of the form in Eq. 3 for another choice of parameter. It can be also be used for more complicated closed loop system architectures. Although it is not necessary to generate the equation in this form to plot a root locus, it allows the keypoints of the root locus to be easily identified. 3C1 Signals and Systems 9 Control Systems

10 3 Root Locus Procedure We will now show the procedure for rapid sketching of the root locus of a transfer function H(s) as values of the tuning parameter K varies from 0 K <. Before we start we note that as the complex poles of H(s) will always occur in conjugate pairs the root locus will be symmetric about horizontal real axis. Step 1: Prepare the sketch Firstly, identify the parameter to be varied (K) and generate the appropriate characteristic equation of the form 1 + KP (s) = 0 (6) Secondly, we consider the positions of the roots of the characteristic equation for the start (K = 0) and end (K = ) points for the root locus. To do this, we factorise P (s) and write the polynomial in terms of its poles and zeros as follows M i=1 1 + K (s z i) N j=1 (s p j) = 0 where z i and p j are the zeros and poles of P (s) respectively. When K = 0, we can find the roots of the characteristic equation by rewriting it as N M (s p j ) + K (s z i ) = 0, j=1 N (s p j ) = 0. j=1 Hence when K = 0, the roots of the transfer function H(s) are given by the the poles of P (s). 3C1 Signals and Systems 10 Control Systems i=1

11 When K =, we can find the roots of the characteristic equation by re writing it as 1 K N (s p j ) + j=1 M (s z i ) = 0. i=1 M (s z i ) = 0, Hence when K =, the roots of the transfer function H(s) are given by the the zeros of P (s). i=1 Observations The number of separate plots on the root locus sketch will be the same as the number of poles of P (s). Each plot will start at a pole of P (s) for K = 0 and end at a zero of P (s) for K =. In general, system could have more poles than zeros. In this case, there will be a number of plots that tend to infinity, with the number being determined by the difference between the number of poles and zeros (i.e.. N M). Such a system is said to contain N M zeros at infinity. 3C1 Signals and Systems 11 Control Systems

12 Case Study Consider the closed loop system with a simple proportional controller as follows The transfer function for this system is H(s) = Ks s 2 + (2 + K)s + 2. To get the root locus as K varies we obtain the characteristic equation from the denominator of H(s) as follows s 2 + (2 + K)s + 2 = 0 s 2 + 2s Ks = 0 s 1 + K s 2 + 2s + 2 = 0. We then find the zeros and poles of P (s) which mark the start and end points of each branch. For s P (s) = s 2 + 2s + 2 we have one zero at s = 0 and two poles at 1 ± j. These points are marked on the root locus plot 3C1 Signals and Systems 12 Control Systems

13 3C1 Signals and Systems 13 Control Systems

14 Step 2: Determine the Parts of the Real Axis that are on the Root Locus The root locus lies at all points on the real axis to the left of an odd number of poles and zeros that lie on the real axis. This arises because of the angle criterion (Eq. 5) and the symmetry of the root locus. For example if we had a characteristic equation of 1 + K s + 2 s(s + 4) = 0 in which P (s) has a zero at s = 2 and poles at s = 0 and s = 4, then the parts of the root locus marked on the real axis are marked by the solid line in the plot below 3C1 Signals and Systems 14 Control Systems

15 Case Study Revisiting our example from Step 1, we see that we have only one zero on the real axis (at s = 0) and so all points on the left of s = 0 must lie on the root locus. We add that to the root locus plot. 3C1 Signals and Systems 15 Control Systems

16 Step 3: Determine the Asymptotes for the loci tending to zeros at infinity We know that if we have a characteristic equation P (s) that has more poles N than zeros M then N M of the root locus bracnches tend to zeros at infinity. These asymptotes intercept the real axis at a point, σ A, given by N j=1 σ A = p j M i=1 z i. N M In other words σ A = (finite poles) (finite zeros) N M The angles of the asymptotes φ k are given by φ A = 2k 1 π for k = 1, 2,..., N M N M In other words the orientations of the assymptotes are evenly spread out over 360. For example, a 4 th Order System with a characteristic equation s K s(s + 2)(s + 3)(s + 5) = 0, with 1 zero and 4 poles, 3 branches of the root locus will tend to zeros at infinity. The centre of the 3 asymptotes is and have an angles at σ A = ( ) ( 1) 4 1 φ A = 2k 1 π for k = 1, 2, 3 3 = 9 3 = 3 Therefore the angles of the asymptotes are π/3, π and 5π/3 = π/3. The asymptotes are marked for this system are shown in Fig. 4 in the left plot and are superimposed on the full root locus for this system shown on the right. 3C1 Signals and Systems 16 Control Systems

17 Figure 4: Left: Asymptotes shown for the above example. Right: Asymptotes superimposed on the completed root locus. Case Study In the case study example we have N M = 2 1 = 1. Therefore, there is only one asymptote that has an angle and intercepts the real axis at φ A = π = π = 180 σ A = (( 1 + j) + ( 1 j) (0) 1 = 2 As the asymptote is parallel to the real axis it does not intersect with it, the value of σ A is meaningless. However, it does lie on the real axis and so is consistent with the rule. A casual glance at the partially completed root locus shows that step 3 merely confirms the result of step 2. All points along the real axis to the left of s = 0 are on the root locus. 3C1 Signals and Systems 17 Control Systems

18 Step 4: Determining the Breakaway Points From the examples we have seen so far, we can see that often there is a point on the real axis at which the root locus either breaks off or onto the real axis. Considering the root locus of a system with P (s) = 1 (s + 2)(s + 4) shown below, we can see that there is a breakaway point on the real axis at a value of s = σ = 3 and that the angle of the branches as they leave the imaginary axis is perpendicular to the real axis. It turns out that K = 1 at the breakaway point. This corresponds from the system transitioning from being overdamped K < 1 to being underdamped K > 1 and so marks a transition in the transient response of the system from non-oscillatory to oscillatory behaviour. 3C1 Signals and Systems 18 Control Systems

19 It is possible to determine the breakaway point as follows. Firstly, identify the portions of the real axis where a breakaway point must exist. Assuming we have already marked the segments of the real axis that are on the root locus, we need to find the segments that are bookended by either two poles or two zeros (either finite zeros or zeros at infinity). For example, lets look again at the root locus shown in Figure 4 There are 3 portions of the root locus on the real axis yet only one portion (the middle one) contains a breakaway point. This is because that segment of the root locus is terminated on either side by poles. The other two have a pole at one end and a zero at the other and hence do not contain breakaway points. 3C1 Signals and Systems 19 Control Systems

20 To estimate the values of s at the breakaway points, the characteristic equation is rewritten in terms of K as 1 + KP (s) = 0 (7) K = 1 P (s) = K(s). To find the breakaway points we find the values of s corresponding to the maxima in K(s). i.e. where dk ds = K (s) = 0. As the last step we check the roots of K (s) that lie on the real axis segments of the locus. The roots that lie in these intervals are the breakaway points. 3C1 Signals and Systems 20 Control Systems

21 Case Study Revisiting our example, after stage 3 we have a partially completed root locus shown below. There is one segment of the real axis on the root locus. As it is bookended by a zero at s = 0 and a zero at infinity (s = ) it must contain a breakaway point. Since s P (s) = s 2 + 2s + 2 we have breakaway points where K (s) = d ( ) s2 + 2s + 2 = 0. ds s = s(2s + 2) (s2 + 2s + 2) s 2 = s2 2 = 0 s 2 The roots K (s) are ± 2. As the root at 2 does not fall on the root locus segment, the breakaway point occurs at s = 2. 3C1 Signals and Systems 21 Control Systems

22 The value of K at the breakaway point is K(s) = 1 P ( 2) = 2 = Therefore, we can update the root locus by marking the breakaway point. As the root locus will leave the breakaway points at an angle perpendicular to the real axis. We add this tangent to the root locus sketch. 3C1 Signals and Systems 22 Control Systems

23 Step 5: Find the angles of departure/arrival for complex poles/zeros In our case study, we can see that there are two complex poles (at 1 ± j) from which branches of the root locus must start. This step wishes to determine the angle at which the root locus leaves such poles. When there are complex poles or zeros of P (s), the root locus branches will either depart or arrive at an angle θ where, for a complex pole at s = p or zero at s = z, θ is given by θ = P (p) π for s = p or θ = π P (z) for s = z where P (p) = P (z) = M (p z i ) i=1 N (p p j ) j=1 p p j M (z z i ) i=1 z z i N (z p j ) j=1 In other words, P (p) is (s p) P (s) evaluated at s = p and P (z) is (s z) 1 P (s) evaluated at s = z 3C1 Signals and Systems 23 Control Systems

24 Therefore, exploiting the rules of complex numbers, we can rewrite P (p) and P (z) as P (p) = P (z) = M (p z i ) i=1 M (z z i ) i=1 z z i N (p p j ) j=1 p p j N (z p j ) j=1 3C1 Signals and Systems 24 Control Systems

25 Case Study Here we have two complex poles at s = 1 ± j. Therefore, for the pole at s = 1 + j, the root locus departs the pole at an angle θ where P ( 1 + j) θ = π (s + 1 j)p (s) θ = π s= 1+j s s ( 1 j) θ = π s= 1+j 1 + j 1 + j ( 1 j) θ = π 3π 4 0 π 2 θ = π θ = 3π 4 We could repeat this calculation to estimate θ for s = 1 j. However, exploiting the symmetry of the root locus, it is quicker to notice that the value of θ 1 j = θ 1+j = 3π 4. Therefore we can update the root locus plot by adding tangents corresponding to the angles of departure for the complex poles. 3C1 Signals and Systems 25 Control Systems

26 Step 6: Complete the Sketch This can be done by hand from all the information gained from the previous steps. You could make it simpler by finding the poles of the characteristic equation for more values of K and placing them on the locus. The root locus for our case study is shown below. 3C1 Signals and Systems 26 Control Systems

27 3.1 Summary The process of generating the root locus for 0 K < can be summarised as follows Step Estimate the transfer function and write the characteristic equation of that the tuning parameter, K, appears as a multiplier. Mark the start and end points of the loci. Determine the number of separate loci Locate the segments of the real axis that are root loci Find the asymptotes of the loci tending to zeros at infinity Rule 1 + KP (s) = 0 Find the poles (start points) and zeros (end points) of P (s). It is the number of poles of P (s). Locus lies to the left of an odd number of zeros or poles. The intersection point of the asymptotes is σ A = pj z i N M. The angles are φ A = 2k 1 N M π Determine the breakaway points. Set K(s) = 1 P (s) and find K (s) = 0 Determine the angles of arrival/departure at complex zeros/poles of P (s) For poles the angle of departure is P (p k ) π. For zeros the angle of arrival is π P (z k ) Complete the Sketch 3C1 Signals and Systems 27 Control Systems

28 Root Locus 4 ROOT LOCUS FOR NEGATIVE K 4 Root Locus for Negative K A similar set of rules exists for plotting the root locus for 0 K >. The summary for plotting the negative K root is shown below with differences from the positive K root locus highlighted in bold. Step Estimate the transfer function and write the characteristic equation of that the tuning parameter, K, appears as a multiplier. Mark the start and end points of the loci. Determine the number of separate loci Locate the segments of the real axis that are root loci Find the asymptotes of the loci tending to zeros at infinity Rule 1 + KP (s) = 0 Find the poles (start points) and zeros (end points) of P (s). It is the number of poles of P (s). Locus lies to the left of an even number of zeros or poles. The intersection point of the asymptotes is σ A = pj z i N M. The angles are φ A = 2 2k 1 N M π Determine the breakaway points Set K(s) = 1 P (s) and find K (s) = 0 Determine the angles of arrival/departure at complex zeros/poles of P (s) For poles the angle of departure is P (p k ). For zeros the angle of arrival is P (z k ) Complete the Sketch 3C1 Signals and Systems 28 Control Systems

29 Root Locus 5 CONCLUSION Case Study We will now plot the negative K root locus for a system with Ks H(s) = s 2 + (2 + K)s + 2. Hence, P (s) = The root locus is plotted below. s s 2 + 2s + 2. Notice that for a value of K < 2 the roots move into the RHS of the s-plane and so the system is unstable. 5 Conclusion In this handout we showed how to determine the location of the poles of a control system through a root locus plot. These plots are important because they indicate whether a system is stable or not and can also be used to estimate the transient response. 3C1 Signals and Systems 29 Control Systems

### ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8

Learning Objectives ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Dr. Oishi oishi@unm.edu November 2, 203 State the phase and gain properties of a root locus Sketch a root locus, by

### Root locus Analysis. P.S. Gandhi Mechanical Engineering IIT Bombay. Acknowledgements: Mr Chaitanya, SYSCON 07

Root locus Analysis P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: Mr Chaitanya, SYSCON 07 Recap R(t) + _ k p + k s d 1 s( s+ a) C(t) For the above system the closed loop transfer function

### Lecture 1 Root Locus

Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the

### Software Engineering 3DX3. Slides 8: Root Locus Techniques

Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007

### 7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)

### School of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:

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: θ D 0.09 See

### Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 02-760-4253 Fax:02-760-4435 Introduction In this lesson, you will learn the following : The

### SECTION 5: ROOT LOCUS ANALYSIS

SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path

### Chemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University

Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this

### a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a

Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation

### Course roadmap. ME451: Control Systems. What is Root Locus? (Review) Characteristic equation & root locus. Lecture 18 Root locus: Sketch of proofs

ME451: Control Systems Modeling Course roadmap Analysis Design Lecture 18 Root locus: Sketch of proofs Dr. Jongeun Choi Department of Mechanical Engineering Michigan State University Laplace transform

### CHAPTER # 9 ROOT LOCUS ANALYSES

F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closed-loop system is closely related to the location of the closed-loop poles. If the system

### Alireza Mousavi Brunel University

Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched

### Chapter 7 : Root Locus Technique

Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci

### I What is root locus. I System analysis via root locus. I How to plot root locus. Root locus (RL) I Uses the poles and zeros of the OL TF

EE C28 / ME C34 Feedback Control Systems Lecture Chapter 8 Root Locus Techniques Lecture abstract Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley

### Module 07 Control Systems Design & Analysis via Root-Locus Method

Module 07 Control Systems Design & Analysis via Root-Locus Method Ahmad F. Taha EE 3413: Analysis and Desgin of Control Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha March

### Unit 7: Part 1: Sketching the Root Locus

Root Locus Unit 7: Part 1: Sketching the Root Locus Engineering 5821: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland March 14, 2010 ENGI 5821 Unit 7: Root

### Module 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions

Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus

### Automatic Control Systems, 9th Edition

Chapter 7: Root Locus Analysis Appendix E: Properties and Construction of the Root Loci Automatic Control Systems, 9th Edition Farid Golnaraghi, Simon Fraser University Benjamin C. Kuo, University of Illinois

### Lecture 3: The Root Locus Method

Lecture 3: The Root Locus Method Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 56001 This draft: March 1, 008 1 The Root Locus method The Root Locus method,

### Example on Root Locus Sketching and Control Design

Example on Root Locus Sketching and Control Design MCE44 - Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We

### An Introduction to Control Systems

An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a

### EE302 - Feedback Systems Spring Lecture KG(s)H(s) = KG(s)

EE3 - Feedback Systems Spring 19 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 1.. 1.1 Root Locus In control theory, root locus analysis is a graphical analysis method for investigating the change of

### Methods for analysis and control of dynamical systems Lecture 4: The root locus design method

Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.inpg.fr www.gipsa-lab.fr/ o.sename 5th February 2015 Outline

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 13: Root Locus Continued Overview In this Lecture, you will learn: Review Definition of Root Locus Points on the Real Axis

### Root Locus U R K. Root Locus: Find the roots of the closed-loop system for 0 < k < infinity

Background: Root Locus Routh Criteria tells you the range of gains that result in a stable system. It doesn't tell you how the system will behave, however. That's a problem. For example, for the following

### Methods for analysis and control of. Lecture 4: The root locus design method

Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.inpg.fr www.lag.ensieg.inpg.fr/sename Lead Lag 17th March

### Unit 7: Part 1: Sketching the Root Locus. Root Locus. Vector Representation of Complex Numbers

Root Locus Root Locus Unit 7: Part 1: Sketching the Root Locus Engineering 5821: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland 1 Root Locus Vector Representation

### MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN

MAK 391 System Dynamics & Control Presentation Topic The Root Locus Method Student Number: 9901.06047 Group: I-B Name & Surname: Göksel CANSEVEN Date: December 2001 The Root-Locus Method Göksel CANSEVEN

### (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.

### 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

### 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

### Root Locus Techniques

Root Locus Techniques 8 Chapter Learning Outcomes After completing this chapter the student will be able to: Define a root locus (Sections 8.1 8.2) State the properties of a root locus (Section 8.3) Sketch

### 2.004 Dynamics and Control II Spring 2008

MT OpenCourseWare http://ocw.mit.edu.004 Dynamics and Control Spring 008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts nstitute of Technology

### 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

### "APPENDIX. Properties and Construction of the Root Loci " E-1 K ¼ 0ANDK ¼1POINTS

Appendix-E_1 5/14/29 1 "APPENDIX E Properties and Construction of the Root Loci The following properties of the root loci are useful for constructing the root loci manually and for understanding the root

### 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

### SECTION 8: ROOT-LOCUS ANALYSIS. ESE 499 Feedback Control Systems

SECTION 8: ROOT-LOCUS ANALYSIS ESE 499 Feedback Control Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed-loop transfer function is KKKK ss TT ss = 1 + KKKK ss HH ss GG ss

### Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus

Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign

### ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s)

C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s) - H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closed-loop system when the gain K changes from 0 to 1+ K G ( s)

### 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

### 5 Root Locus Analysis

5 Root Locus Analysis 5.1 Introduction A control system is designed in tenns of the perfonnance measures discussed in chapter 3. Therefore, transient response of a system plays an important role in the

### CHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System

CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages

### Root Locus Techniques

4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since

### EXAMPLE PROBLEMS AND SOLUTIONS

Similarly, the program for the fourth-order transfer function approximation with T = 0.1 sec is [num,denl = pade(0.1, 4); printsys(num, den, 'st) numlden = sa4-2o0sa3 + 1 80O0sA2-840000~ + 16800000 sa4

### Class 11 Root Locus part I

Class 11 Root Locus part I Closed loop system G(s) G(s) G(s) Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K We shall assume here K >,

### MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

### Lecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore A Fundamental Problem in Control Systems Poles of open

### Some special cases

Lecture Notes on Control Systems/D. Ghose/2012 87 11.3.1 Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first

### Control Systems. Root Locus & Pole Assignment. L. Lanari

Control Systems Root Locus & Pole Assignment L. Lanari Outline root-locus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS - Root

### ESE319 Introduction to Microelectronics. Feedback Basics

Feedback Basics Stability Feedback concept Feedback in emitter follower One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability

### MODERN CONTROL SYSTEMS

MODERN CONTROL SYSTEMS Lecture 1 Root Locu Emam Fathy Department of Electrical and Control Engineering email: emfmz@aat.edu http://www.aat.edu/cv.php?dip_unit=346&er=68525 1 Introduction What i root locu?

### Time Response Analysis (Part II)

Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary

### Block Diagram Reduction

Block Diagram Reduction Figure 1: Single block diagram representation Figure 2: Components of Linear Time Invariant Systems (LTIS) Figure 3: Block diagram components Figure 4: Block diagram of a closed-loop

### Chapter 7. Root Locus Analysis

Chapter 7 Root Locu Analyi jw + KGH ( ) GH ( ) - K 0 z O 4 p 2 p 3 p Root Locu Analyi The root of the cloed-loop characteritic equation define the ytem characteritic repone. Their location in the complex

### EE402 - Discrete Time Systems Spring Lecture 10

EE402 - Discrete Time Systems Spring 208 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 0.. Root Locus For continuous time systems the root locus diagram illustrates the location of roots/poles of a closed

### 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

### Root locus 5. tw4 = 450. Root Locus S5-1 S O L U T I O N S

Root Locus S5-1 S O L U T I O N S Root locus 5 Note: All references to Figures and Equations whose numbers are not preceded by an "S" refer to the textbook. (a) Rule 2 is all that is required to find the

### Nyquist Criterion For Stability of Closed Loop System

Nyquist Criterion For Stability of Closed Loop System Prof. N. Puri ECE Department, Rutgers University Nyquist Theorem Given a closed loop system: r(t) + KG(s) = K N(s) c(t) H(s) = KG(s) +KG(s) = KN(s)

### Problems -X-O («) s-plane. s-plane *~8 -X -5. id) X s-plane. s-plane. -* Xtg) FIGURE P8.1. j-plane. JO) k JO)

Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 -X-O

### Essence of the Root Locus Technique

Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general set-up, namely for the case when the closed-loop

### Class 12 Root Locus part II

Class 12 Root Locus part II Revising (from part I): Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K Comple plane jω ais 0 real ais Thus,

### Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Guzzella 9.1-3, Emilio Frazzoli

Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Guzzella 9.1-3, 13.3 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 3, 2017 E. Frazzoli (ETH)

### Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:

### Control of Electromechanical Systems

Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance

### Root Locus. 1 Review of related mathematics. Ang Man Shun. October 30, Complex Algebra in Polar Form. 1.2 Roots of a equation

Root Locus Ang Man Shun October 3, 212 1 Review of relate mathematics 1.1 Complex Algebra in Polar Form For a complex number z, it can be expresse in polar form as z = re jθ 1 Im z Where r = z, θ = tan.

### EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO

EE 380 EXAM II 3 November 2011 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

### Control Systems. Frequency Method Nyquist Analysis.

Frequency Method Nyquist Analysis chibum@seoultech.ac.kr Outline Polar plots Nyquist plots Factors of polar plots PolarNyquist Plots Polar plot: he locus of the magnitude of ω vs. the phase of ω on polar

### Lecture Sketching the root locus

Lecture 05.02 Sketching the root locus It is easy to get lost in the detailed rules of manual root locus construction. In the old days accurate root locus construction was critical, but now it is useful

### ME2142/ME2142E Feedback Control Systems

Root Locu Analyi Root Locu Analyi Conider the cloed-loop ytem R + E - G C B H The tranient repone, and tability, of the cloed-loop ytem i determined by the value of the root of the characteritic equation

### Laplace Transform Analysis of Signals and Systems

Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.

### 1 Closed Loop Systems

Harvard University Division of Engineering and Applied Sciences ES 45 - Physiological Systems Analysis Fall 2009 Closed Loop Systems and Stability Closed Loop Systems Many natural and man-made systems

### 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

### EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation

EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginally-stable

EC6405 - CONTROL SYSTEM ENGINEERING Questions and Answers Unit - I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform

### 1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant

### Andrea Zanchettin Automatic Control AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Linear systems (frequency domain)

1 AUTOMATIC CONTROL Andrea M. Zanchettin, PhD Spring Semester, 2018 Linear systems (frequency domain) 2 Motivations Consider an LTI system Thanks to the Lagrange s formula we can compute the motion of

### CISE302: Linear Control Systems

Term 8 CISE: Linear Control Sytem Dr. Samir Al-Amer Chapter 7: Root locu CISE_ch 7 Al-Amer8 ١ Learning Objective Undertand the concept of root locu and it role in control ytem deign Be able to ketch root

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain

### ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system

### Introduction to Root Locus. What is root locus?

Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response

### Numerical Methods. Equations and Partial Fractions. Jaesung Lee

Numerical Methods Equations and Partial Fractions Jaesung Lee Solving linear equations Solving linear equations Introduction Many problems in engineering reduce to the solution of an equation or a set

### Control Systems I. Lecture 9: The Nyquist condition

Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control

### Course Summary. The course cannot be summarized in one lecture.

Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: Steady-State Error Unit 7: Root Locus Techniques

### Remember that : Definition :

Stability This lecture we will concentrate on How to determine the stability of a system represented as a transfer function How to determine the stability of a system represented in state-space How to

### CONTROL SYSTEMS. Chapter 5 : Root Locus Diagram. GATE Objective & Numerical Type Solutions. The transfer function of a closed loop system is

CONTROL SYSTEMS Chapter 5 : Root Locu Diagram GATE Objective & Numerical Type Solution Quetion 1 [Work Book] [GATE EC 199 IISc-Bangalore : Mark] The tranfer function of a cloed loop ytem i T () where i

### 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

### Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0

Objective Root Locu Diagram Upon completion of thi chapter you will be able to: Plot the Root Locu for a given Tranfer Function by varying gain of the ytem, Analye the tability of the ytem from the root

### A marks are for accuracy and are not given unless the relevant M mark has been given (M0 A1 is impossible!).

NOTES 1) In the marking scheme there are three types of marks: M marks are for method A marks are for accuracy and are not given unless the relevant M mark has been given (M0 is impossible!). B marks are

### Control Systems. University Questions

University Questions UNIT-1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write

### 2. If the values for f(x) can be made as close as we like to L by choosing arbitrarily large. lim

Limits at Infinity and Horizontal Asymptotes As we prepare to practice graphing functions, we should consider one last piece of information about a function that will be helpful in drawing its graph the

### 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

### ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled

### Root Locus (2A) Young Won Lim 10/15/14

Root Locus (2A Copyright (c 2014 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version

### Root Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples

Root Locu Content Root locu, ketching algorithm Root locu, example Root locu, proof Root locu, control example Root locu, influence of zero and pole Root locu, lead lag controller deign 9 Spring ME45 -

### Transient Response of a Second-Order System

Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop

### Test 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010

Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the

### Proportional plus Integral (PI) Controller

Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on

### Systems 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