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## Program for North American Mobility In Higher Education

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**NAMP**Program for North American Mobility In Higher Education Module 5 Controllability Analysis PIECE Introducing Process integration for Environmental Control in Engineering Curricula Module 5 – Controllability Analysis**PIECE**University of Ottawa École Polytechnique de Montréal Instituto Mexicano del Petróleo North Carolina State University Paprican Universidad Autónoma de San Luis Potosí Texas A&M University Universidad de Guanajuato Process integration for Environmental Control in Engineering Curricula NAMP Program for North American Mobility in Higher Education Module 5 – Controllability Analysis**University of Ottawa**University of Ottawa Universidad de Guanajuato Universidad de Guanajuato Module 5 This module was created by: Stacey Woodruff From Host University Carlos Carreón Module 5 – Controllability Analysis**Project Summary**Objectives • Create web-based modules to assist universities to address the introduction to Process Integration into Engineering curricula • Make these modules widely available in each of the participating countries Participating institutions • Six universities in three countries (Canada, Mexico and the USA) • Two research institutes in different industry sectors: petroleum (Mexico) and pulp and paper (Canada) • Each of the six universities has sponsored 7 exchange students during the period of the grant subsidised in part by each of the three countries’ governments Module 5 – Controllability Analysis**What is the structure of this module?**All modules are divided into 3 tiers, each with a specific goal: Tier I: Background Information Tier II: Case Study Applications Tier III: Open-Ended Design Problem These tiers are intended to be completed in that particular order. In the first tier, students are quizzed at various points to measure their degree of understanding, before proceeding to the next two tiers. Structure of Module 5 Module 5 – Controllability Analysis**What is the purpose of this module?**It is the objective of this module to cover the basic aspects of Controllability Analysis. It is targeted to be an integral part of a fundamental/and or advanced Control course. This module is intended for students with some basic understanding of the fundamental concepts of control. Purpose of Module 5 Module 5 – Controllability Analysis**Tier IBackground Information**Module 5 – Controllability Analysis**Statement of Intent**Define Stability Demonstrate simple methods for stability analysis, mostly for Single-Input Single-Output (SISO) systems Understand interaction between control loops in Multiple-Input Multiple-Output (MIMO) systems Demonstrate the Relative Gain Array Investigate controllability analysis for continuous and discrete systems Comprehend singular value decomposition (SVD) Module 5 – Controllability Analysis**Stability**A dynamic system is stable if the system output response is bounded for all bounded inputs. A stable system will tend to return to its equilibrium point following a disturbance. Conversely, an unstable system will have the tendency to move away from its equilibrium point following a disturbance. Module 5 – Controllability Analysis**Why is the stability of a system important??**When a system becomes unstable it can be A DISASTER!!!!! Module 5 – Controllability Analysis**(a)**(b) (c) • Example The concept of stability is illustrated in the following figure. The sphere in (a) is stable as it will return to its original equilibrium after a small disturbance whereas the sphere in (b) is unstable as it moves away from its equilibrium point and never comes back. The sphere in (c) is said to be marginally stable. Module 5 – Controllability Analysis**Quiz #1**• Why is it important that a system is stable? • List two examples of systems that have become unstable. Module 5 – Controllability Analysis**There are many ways of determining if a system is stable**such as : • Roots of Characteristic Equation • Bode Diagrams • Nyquist Plots • Simulation Module 5 – Controllability Analysis**Roots of Characteristic Equation**One can determine if a system is stable based on the nature of the roots of its characteristic equations. Consider the following system: D(s) U(s) + Y(s) Y*(s) M(s) (s) + - + Ym(s) Module 5 – Controllability Analysis**From the previous diagram, we can see that the output Y is**influenced in the following manner. GOLis the open loop transfer function. Module 5 – Controllability Analysis**For the moment, let’s consider that there is only a change**in set point, therefore, the previous equation reduces to the closed loop transfer function, The roots r1, r2, r3… rn are those of the characteristic equation 1+GcG1G2G4 =0 and (s) is a function that arises from the rearrangement. The roots of the characteristic equation (denominator) are the poles of the transfer function whereas the roots of the numerator are the zeros. Module 5 – Controllability Analysis**The nature of the roots of the characteristic equation can**dictate if a system is stable or not due to the fact that if there is one (or more) root on the right half of the complex plane, the response will contain a term that grows exponentially, leading to an unstable system. Imaginary Part Imaginary Part Imaginary Part Real Part Real Part φ φ time time Negative real root Positive real root Stable Region Unstable Region Real Part Imaginary Part Imaginary Part Stable Region Real Part Real Part φ φ time time Complex Roots (Negative real parts) Complex Roots (Positive real parts) Module 5 – Controllability Analysis**Routh Test**The Routh test (Routh stability criterion) is a very useful tool in determining whether or not a closed-loop system is stable provided the characteristic equation is available. The Routh stability criterion is based on a characteristic equation that is in the form A necessary (but not sufficient) condition of stability is that all of the coefficients (a0, a1, a2, …etc.) must be positive. Module 5 – Controllability Analysis**Routh Array**When all coefficients are positive, a Routh Array must be constructed as follows: The system is stable if ALL the elements in the first column are positive! The first two rows are filled in using the coefficients of the characteristic equation. Subsequent rows are calculated as shown in the next page. } Module 5 – Controllability Analysis**After the coefficients of the characteristic equation are**input in the array, the coefficients, b1, b2 … bn and subsequently c1…cn should be calculated as follows and input into the array. Routh Array Pivot to calculate all bi Module 5 – Controllability Analysis**Routh Test Theorems**Theorem 1-The necessary and sufficient condition for stability (i.e. All roots with negative real parts) is that all elements of the first column of the Routh Array must be positive and non zero. Routh Test Example 1- Consider the following characteristic equation: All of the elements in the first column of this Routh Array are positive, therefore the system is stable. Row 1(s3) 1 6.38 2(s2) 4.583 15.625 3(s1) 2.97 0 4(s0) 15.6250 Module 5 – Controllability Analysis**Routh Test Example 2- It is possible to determine for which**values of Kc the system remains stable Row 1(s3) 1 6.38 2(s2) 4.583 (1+Kc)/0.384 3(s1) 0 4(s0) (1+Kc)/0.384 0 29.24-(1-Kc)/0.384>0 → Kc <10.23 1+Kc >0 → Kc>-1 (Kc is positive) Module 5 – Controllability Analysis**Theorem 2- If some of the elements of the first column are**negative, the number of roots on the right hand side of the imaginary axis is equal to the number of sign changes in the first column. Routh Test Example 3 – If the characteristic equation of a system is given by the following equation, is the system stable? Row 1(s4) 1 11 120 2(s3) 6 36 0 3(s2) 5 120 4(s1) -108 0 5(s0) 120 There are 2 sign changes. Therefore, the system has two roots in the right-hand plane, and the system is unstable. Module 5 – Controllability Analysis**Theorem 3- If one pair of roots is on the imaginary axis,**equidistant from the origin, and all the other roots are in the left-hand plane, all the elements of the nth row will vanish. The location of the pair of imaginary roots can be found by solving the auxiliary equation: where the coefficients C and D are the elements of the array in the (n-1)th row. These roots are also the roots of the characteristic equation. Cs2+D=0 Module 5 – Controllability Analysis**Routh Test Example 4 – Determine the stability of the**system having the following characteristic equation: Row 1(s4) 1 6 8 2(s3) 3 12 3(s2) 2 8 4(s1) 0 4(s1) 4 5(s0) 8 The derivative taken indicates that a 4 should be placed in the s row (Row 4). The procedure is carried out. There are no sign changes in the first column, indicating that there are no roots located on the right-hand side of the plane. Module 5 – Controllability Analysis**Quiz #2**In what cases can the Routh test be used to determine stability? Is the system having the following characteristic equation stable? If a system has two negative real roots, is the system stable? If a system has one negative real root and one positive real root is the system stable? Module 5 – Controllability Analysis**Frequency Response**One very useful method of determining system stability, even when transportation lags exist, is Frequency Response. Frequency response is a method concerning the response of a process or system to a sustained sinusoidal plot. Frequency Response Stability Criteria Two principal criteria: 1. Bode Stability Criterion 2. Nyquist Stability Criterion Module 5 – Controllability Analysis**Bode stability criterion**A closed-loop system is unstable if the Frequency Response of the open-loop Transfer Function, GOL=GCG1G2G4, has an amplitude ratio greater than one at the critical frequency, ωc. Otherwise the closed-loop system is stable. Note: ωc is the value of ω where the open-loop phase angle is -1800. Thus, The Bode Stability criterion provides information on the closed-loop stability from open-loop frequency response information. Module 5 – Controllability Analysis**Bode Stability Criterion- Example 1**A process has the following transfer function: With a value of G1=0.1 and G4=10. If proportional control is used, determine closed-loop stability for 3 values of Kc: 1, 4, and 20. GOL=GCG1G2G4 Solution: You will find the Bode plots on the next slide Module 5 – Controllability Analysis**Bode plots for GOL = 2Kc/(0.5s + 1)3**Module 5 – Controllability Analysis**Nyquist Stability Criterion**The Nyquist stability criterion is the most powerful stability test that is available for linear systems described by transfer function models. Consider an open-loop transfer function, GOL(s) that is proper and has no unstable pole-zero cancellations. Let N be the number of times that the Nyquist plot of GOL(s) encircles the (-1, 0) point in a clockwise direction. Also, let P denote the number of poles of GOL(s) that lie to the right of the imaginary axis. Then, Z=N+P, where Z is the number of roots (or zeros) of the characteristic equation that lie to the right of the imaginary axis. The closed-loop system is stable, if and only if Z=0. Module 5 – Controllability Analysis**U(s)**X(s) Z(s) Y(s) • Example 9.2 – Find the amplitude ratio and the phase lag of the following process for = 0.1 and 0.4. Module 5 – Controllability Analysis**If (0.4 – 2.33) < 0 then – or –180o**• Example 9.2 – Find AR and (from known equations) Module 5 – Controllability Analysis**=0.4**=0.1 • Example 9.2 – Find AR and … Nyquist plot Im Re Module 5 – Controllability Analysis**Quiz #3**Name two methods of determining stability using frequency response. What does an amplitude ratio (AR) of 1 signify? An amplitude ratio of less than 1? What does a value of Z=0 signify? Module 5 – Controllability Analysis**Multiple Input Multiple Output (MIMO) Systems**Module 5 – Controllability Analysis**When dealing with Multiple Input Multiple Output systems, we**have to ask ourselves two main questions. 1. How to pair the input and output variables 2. How to design the individual single-loop controllers Module 5 – Controllability Analysis**Let’s consider the following system:**Loop 1 - y1 m1 G11 Gc1 + + + G12 G21 + + y2 + m2 G22 Gc2 - Loop 2 y1(s) = G11(s)m1(s) + G12(s)m2(s) y2(s) = G21(s)m1(s) + G22(s)m2(s) Module 5 – Controllability Analysis**We will perform 2 small “experiments” to demonstrate**MIMO system interactions. Let´s consider m1 as a candidate to pair with y1. Experiment #1 When a unit step change is made to the input variable m1, with all loops open, the output y1 will change, and so will y2, but for now, we are primarily concerned with the effect on y1. After steady-state is reached, let’s consider the change in y1 as a result of the change in m1, y1m ; this will represent the main effect of m1 on y1. Δy1m = K11 Keep in mind that no other input variables have been changed, and that all loops are open, so no feedback control is required. Module 5 – Controllability Analysis**Experiment #2-Unit step change in m1 with Loop 2 closed.**These things will happen as a result of the unit step change in m1. 1- y1 changes because of G11, but because of interactions via the element G21, y2 changes as well. 2- Under feedback control, Loop 2 wards off this interaction effect on y2 by manipulating m2 until y2 is returned to its initial state before the disturbance. 3-The changes in m2 will now affect y1 via the G12 transfer element. The changes in y1 are from two different sources. (1) the DIRECT INFLUENCE of m1 on y1 (Δy1m) (2) the Indirect Influence, from the retaliatory action from Loop 2 in warding off the interaction effect of m1 on y2 (Δy1r) Module 5 – Controllability Analysis**After dynamic transients die away and steady-state is**reached, the net change observed in y1 is given by: Δy1*= Δy1m+Δy1r This net change is the sum of the main effect of m1 on y1 and the interactive effect provoked by m1 interacting with the other loop. A good measure of how well a system can be controlled (λ) if m1 is used to control y1 is: Module 5 – Controllability Analysis**Loop Pairing on the Basis of Interaction Analysis**Case 1 : λ11=1 This case is only possible if y1r is equal to zero. In physical terms, this means that the main effect of m1 on y1, when all the loops are opened, and the total effect, measured when the other loop is closed, are identical. This will be the case if: • m1 does not affect y2, and thus, there is no retaliatory control action from m2, or • m1 does affect y2, but the retaliatory control action from m2 does not cause any change in y1 because m2 does not affect y1. Under these circumstances, m1 is the perfect input variable to control y1 because there will be NO interaction problems. Module 5 – Controllability Analysis**Case 2 : λ11=0**This condition indicates that m1 has no effect on y1, therefore y1m willbe zero in response to a change in m1. Note that under these circumstances, m2 is the perfect input variable for controlling y2, NOT y1. Since m1 does not affect y1, y1 can be controlled with m2 without any interaction with y1. Module 5 – Controllability Analysis**Case 3 : 0 < λ11< 1**This condition indicates that the direction of the interaction effect is in the same direction as that of the main effect. In this case the total effect is greater than the main effect. For λ11>0.5, the main effect contributes MORE to the total effect than the interaction effect, and as the contribution of the main effect increases, the closer to a value of 1 λ11 becomes. For λ11<0.5, the contribution from the interaction effect dominates, as this contribution increases, λ11 moves closer to zero. For λ11=0.5, the contributions of the main effect and the interaction effect are equal. Module 5 – Controllability Analysis**Case 4 : λ11>1**This is the condition where y1r is the opposite sign of y1m, but it is smaller in absolute value. In this case y1* (y1r +y1m) is less than the main effect y1m, and therefore a larger controller action m1 is needed to achieve a given change in y1 in the closed loop than in the open loop. For a very large and positive λ11 the interaction effect almost cancels out the main effect and closed-loop control of y1 using m1 will be very difficult to achieve. Case 5 : λ11< 0 This is the case when y1r is not only opposite in sign, but also larger in absolute value to y1m. The pairing of m1 with y1 in this case is not very desirable because the direction of the effect of m1 on y1 in the open loop is opposite to the direction in the closed loop. The consequences of using such a pairing could be catastrophic. Module 5 – Controllability Analysis**Quiz#4**• What is a MIMO system? • What does λ11=1 signify? If this is the case, is m1 a good input variable to control y1? • If λ11 is very large and positive, is m1 a good input variable to control y1? Module 5 – Controllability Analysis**Relative Gain Array (RGA)**The quantity λ11 is defined as the Relative Gain between input m1 and output y1. λij is defined as the relative gain between output yi and input mj, as the ratio of two steady-state gains: Module 5 – Controllability Analysis**When the relative gain is calculated for all of the**input/output combinations of a multivariable system, the results are placed into a matrix as follows and this array produces THE RELATIVE GAIN ARRAY Module 5 – Controllability Analysis**PROPERTIES OF THE RELATIVE GAIN ARRAY**• Properties of the Relative Gain Array 1. The elements of the RGA across any row, or down any column sum up to 1. i.e.: 2. λij is dimensionless; therefore, neither the units, nor the absolute value actually taken by the variables mj, or yi affect it. Module 5 – Controllability Analysis**PROPERTIES OF THE RELATIVE GAIN ARRAY**3. The value λij is a measure of the steady-state interaction expected in the ith loop of the multivariable system if its output (yi) is paired with input (mj); in particular, λij =1 indicates that mj affects yi without interacting with the other loops. Conversely, if λij=0 this indicates that mj has no effect on yi. Module 5 – Controllability Analysis