Lecture 2.3 Stability of Feedback Control Systems CH158P Process Dynamics and Control Contents • Stability • Concept • Definition • Criterion • Routh Test for Stability 2 Stability 3 Concept of Stability The first- and second-order systems discussed so far were inherently stable—the responses were bound. We now consider the problem of stability in a control system that is only slightly complicated than those studied before. Consider the proportional control of two stirred-tank heaters (in series) with a measurement lag: We are going to consider only the servo problem (U = 0, set-point tracking). The overall transfer function for this is: 4 Concept of Stability In terms of the individual transfer functions, It can be seen that the denominator is a third-order polynomial. 5 Concept of Stability For a unit-step change in R, the response in the Laplace domain is: Inversion of the previous equation provides the equation for the response in the time domain. This requires obtaining the roots of the third-order polynomial in the original transfer function, which entails lengthy algebraic methods or the use of math software, such as MATLAB. It is, however, apparent that the roots will depend on the values of πΎπΆ , π1 , π2 , and π3 . These roots determine the nature of the transient response (constant, oscillatory, exponentially increasing, decaying, etc.). 6 Concept of Stability Assuming the time constants π1 , π2 , and π3 are constant, and only πΎπΆ is varied, the step responses of the closed-loop system will look like these: As πΎπΆ is increased, the response becomes more oscillatory. It can be seen, however, that at high enough values of πΎπΆ , the oscillations become unbounded (e.g. for πΎπΆ = 12). When the response grows rather than decays, it is deemed unstable. In the design of control systems, we are interested in being able to determine quickly the values of πΎπΆ that give unstable responses. 7 Definition of Stability For our purposes, we define stable control systems as follows: A bounded input function is a function of time that always falls within certain bounds during the course of time (e.g., step, sinusoidal). The ramp function π π‘ = π΄π‘ is an example of an unbounded function. This definition is only true in the mathematical sense. In reality, physical bounds or constraints are always exhibited, such as the fully closed and fully open positions of a valve. We describe such limitations using the term saturation. A physical system, when unstable, may not follow the response of its linear mathematical model beyond certain physical bounds, but rather may saturate. We still need to predict system stability since operation is always within the saturating boundaries; operation at saturation implied unsatisfactory control. 8 Stability Criterion We now translate the definition of stability into a simpler criterion, one that can be used to ascertain the stability of control systems of the form shown: The response of this closed-loop system to changes in both U and R is: 9 Stability Criterion To determine under what conditions this system is stable, it is necessary to test the response to a bounded input. Suppose a unit-step change in set point is applied. The response is: where π1 , π2 , … , ππ are the n roots of the characteristic equation: and πΉ π is a function that arises in the rearrangement to the single-term form of the response. 10 Stability Criterion For example, for the control system being considered earlier, the step response is: which may be rearranged to: 11 Stability Criterion Comparing this final result to the original form of the response, For this case, the function πΉ π is: and the characteristic equation, whose roots are π1 , π2 , and π3 , is: 12 If there are any roots in the right half of the complex plane, the response will contain a term that grows exponentially in time—the system is unstable. If there is an π π in the characteristic equation, where π ≥ 2, there is more than one root at the origin and the response is again unbounded. If there is a repeated pair of conjugate roots on the imaginary axis, the contribution is a sinusoid with an amplitude that increases with time. The right-half plane, including the imaginary axis, is the unstable region of roots. 13 Stability Criterion The mathematical definition of stability is therefore: Note that the characteristic equation of a control system is same for set point or load changes. Furthermore, this definition is applicable for any input—not only for the step input. 14 Stability Criterion Exercise 2.3.1 In terms of the figure shown, a control system has the transfer functions Find the characteristic equation and its roots, and determine whether the system is stable. 15 Stability Criterion Exercise 2.3.1 16 Stability Criterion Exercise 2.3.1 17 Routh Test for Stability 18 Routh Test The Routh test is a purely algebraic method for determining how many roots of the characteristic equation have positive real parts: from this it can also be determined whether the system is stable. If there are no roots with positive real parts, the system is stable. The test is limited to systems with polynomial characteristic equations (without an exponential terms, such as those for systems with transportation lag). The procedure for examining the roots is to write the characteristic equation in the form: where π0 is positive. In this form, it is necessary that all the coefficients π0 , π1 , π2 , … , ππ−1 , ππ be positive if all the roots are to lie in the left half-plane. If any coefficient is negative, the system is definitely unstable, and the Routh test is not needed to determine stability. If all the coefficients are positive, the system may stable or unstable. 19 Routh Test If the coefficients are all positive, the procedure for the Routh test is as follows: 1. Arrange the coefficients of the characteristic equation into the first two rows of the Routh array. The array shown has been filled in for n = 7. In general, there are π + 1 rows. For n even, the first row has one more element than the second row. 2. The elements in the remaining rows are found from the formulas During the computation of the Routh array, any row can be divided by a positive constant without changing the results of the test. 20 Routh Test The following theorems are then applied to determine stability. THEOREM 1 The necessary and sufficient condition for all the roots of the characteristic equation to have negative real parts is that all elements of the first column of the Routh array (π0 , π1 , π1 , π1 , etc.) be positive and nonzero. THEOREM 2 If some of the elements in the first column are negative, the number of roots with a positive real part (in the right half-plane) is equal to the number of sign changes in the first column. THEOREM 3 If one pair of roots is on the imaginary axis, equidistant from the origin, and all other roots are in the left half-plane, then all the elements of the nth row will vanish and none of the elements of the preceding row will vanish. The location of the pair of imaginary roots can be found by solving the equation πΆπ 2 + π· = 0, where the coefficients C and D are the elements of the array in the (n-1)st row as read from left to right, respectively. 21 Routh Test Exercise 2.3.2 Given the characteristic equation π 4 + 3π 3 + 5π 2 + 4π + 2 = 0, determine the stability by the Routh criterion. 22 Routh Test Exercise 2.3.3 1 1 Using π1 = 1, π2 = , and π3 = , determine the values of πΎπΆ for which the control system is stable. 2 3 For the value of πΎπΆ for which the system is on the threshold of stability, determine the roots of the characteristic equation with the help of Theorem 3. 23 Routh Test Exercise 2.3.3 1 2 π1 = 1, π2 = , π3 = 1 3 24 Routh Test Exercise 2.3.3 1 2 π1 = 1, π2 = , π3 = 1 3 25 Routh Test Exercise 2.3.3 1 2 π1 = 1, π2 = , π3 = 1 3 26 Routh Test Exercise 2.3.4 1 2 Determine the stability of the system for which a PI controller is used. Use π1 = 1, π2 = , π3 = 1 , πΎπΆ 3 = 5, and ππΌ = 0.25. 27 Routh Test Exercise 2.3.4 1 2 Determine the stability of the system for which a PI controller is used. Use π1 = 1, π2 = , π3 = 1 , πΎπΆ 3 = 5, and π1 = 0.25. 28 Lecture 2.3 Problems Solve the following problems from the textbook (LeBlanc & Coughanowr, 3E): 1. Problem 13.1 2. Problem 13.9 3. Problem 13.10 29 Lecture 2.3 Stability of Feedback Control Systems CH158P Process Dynamics and Control -end-