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

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1 Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1

2 Chapter 4 System Stability 2

3 Chapter Objectives End of this chapter, you should know : 1. Explain the concepts of stability feedback control systems 2. Determine the stability of control systems using the Direct Method and Routh Hurwitz Criteria 3

4 Introduction Industrial processes which are stable without feedback controllers, are said to be open-loop stable or selfregulating For a control system to be satisfactory it should meet two requirements: Stability i.e it shows stable behavior Good control quality (good dynamics that deals with the transient response) 4

5 Definition A system is stable if its impulse response approaches zero as time approaches infinity i.e a system is stable if and only if the absolute value of its impulse response g(t), integrated over an infinite range is finite g t dt 0 Where M is a finite quantity = M Dynamically: (response to inputs) 5 An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise, it is said to be unstable.

6 Bounded variable = stays within upper and lower limits for all values of time. If some excitation causes the controlled variable to increase continuously or to oscillate with growing amplitude it is unstable (ξ < 0) Theoretically: 6 It can be said that the control systems up to 2 nd order are stable since it is impossible for ξ to be made ve and only in higher order systems it is possible for ξ < 0 (unstable system).

7 Mathematically: A system is stable when all the roots of its characteristic equation lie on the left hand portion of the s-plane (all the roots of the characteristic equation must be real and negative or be complex with negative parts). 7

8 Relative stability is the degree of stability of a system that provides valuable information about its behavior (if it is stable, how close is it to being unstable?). The system stability can be tested by considering its response to a finite input signal. This means the analysis of system dynamics in the actual time domain which is usually cumbersome and time consuming. The shortcut methods deduce the system stability from its characteristic equation 8

9 The Short-Cut Method Include: 1. Direct Method: Mathematical evaluation of the roots of the characteristic equation 2. Algebraic Method (Routh Test) 3. Graphical Method (Root Locus Method) 9

10 Direct Method It depends only upon the open loop transfer function OPLF (relates measured variable b to reference input r). OLTF = b r = G cg v G p H Characteristic equation: 1 + OLTF =0 The closed loop transfer function CLTF, relates the control variable c to either the reference input r or the load variable u 10 C R = Y = Y sp C U = G c G v G p 1 + G c G v G p H = G 1 + OLTF K L G p 1 + G c G v G p H = K LG p 1 + OLTF

11 If OLTF is a ratio of polynomials in s, then the closedloop transfer function in is also a rational function. Then, it can be factored into poles (p i ) and zeroes (z i ) as To have a physically realizable system, the number of poles must be greater than or equal to the number of zeroes The poles are also the roots of the Characteristic equation 11

12 Partial fractionation: Taking inverse Laplace: 12

13 A linear control system is unstable if any root of its characteristic equation lies to the right of the right of the imaginary axis on Argand diagram, otherwise the system is either stable or critically stable. The response of a control system can be predicted from the locations of the roots of its characteristic equation. Each root location, s, generates a certain response c(t) as shown c t = A e αt sin βt Where the roots are: 13

14 s 1, s 2 = α ± jβ Regarding the real part α: i. α < 0 the response contain e αt (decaying response) ii. α = 0 a constant or neutral response iii. α > 0 the response contain e αt (growing response) To ensure stability, roots should be excluded not only from the right half plane but also from the imaginary axis. Limitations: In general the C.E.s are polynomials of higher order and therefore not simple to evaluate their roots (mathematical difficulties) 14

15 15

16 Example: The roots of a characteristic equation is for a control system are: r 1 = - 8, r 2, r 3 = - 4 ± 6 j Express the general form of the equation describing the response y(t) of the system when the input f (t) is a unit step function. 16

17 Routh Test Uses an analytical technique for determining whether any roots of a polynomial have positive real parts Characteristic equation where a n >0. According to the Routh criterion, if any of the coefficients a o, a 1, a K, a n-1 are negative or zero, then at least one root of the characteristic equation lies in the RHP and thus the system is unstable 17

18 On the other hand, if all of the coefficients are positive, then one must construct the Routh Array shown below: For stability, all elements in the first column must be positive 18

19 The first two rows of the Routh Array are comprised of the coefficients in the characteristics equation. The elements in the remaining rows are calculated from coefficients by the using the formulas: Note: Recall the determinant of a 2x2 matrix. 19 n+1 rows must be constructed n = order of the characteristic eqn.

20 Routh Criterion Rules Rule (1) The necessary and sufficient condition for all the roots of C.E. to have ve real parts (stable system) is that all elements of the first column of the Routh array be +ve and non-zero. It states that that the number of change of algebraic sign in the 1 st column of the array is equal to the number of roots having +ve real parts (lie in the RHP) Example 20 Use Routh criterion to investigate the stability of the system with the following characteristic equation: s 5 + 2s 4 + 3s 3 + 7s 2 + 4s + 4 = zero

21 Rule (2) 21 If one pair of roots is purely imaginary (0 ± jβ) and all the other roots have ve real parts, all the elements of the n th row will vanish and non of the element of the preceding will vanish. In this case one pair of roots is on the imaginary axis and all the other roots are in the left hand plane and the system is critically stable. The values of the pair of imaginary roots from As 2 + B = zero. Where A and B are the first and second elements respectively in the (n-1) th row.

22 Example Examine the stability of a system with with the following characteristic equation: s 3 + 3s 2 + 4s + 12 = zero Rule (3) Any row can be multiplied or divided through out by +ve constant for simplification without changing the sign of the first column. Hence this dose not affect the results of the test 22

23 Rule (4) If the coefficient of the last row is zero, the C. E. has a root at the origin. If the coefficient of the last two rows are zeros, then the C. E. has a double roots at the origin Rule (5) 23 When the first term in a row is zero, but not all the other terms are zeros, one of these methods can be used: a) Replace the zero obtained by a very small +ve number ε, then after completing the array allow ε to approach zero.

24 a) Replace the zero obtained by a very small +ve number ε, then after completing the array allow ε to approach zero. b) In the original equation substitute s by 1/x and solve. The number of roots of x with +ve real parts is the same as the number of roots of s with +ve real parts. c) Multiply the original polynomial by the factor (s+1) and solve. (an additional ve root) 24

25 Quiz: Use the Routh criterion to investigate the stability of the system which has the following C. E. 4s s s s 2 + 3s + 1 = 0 Application of Routh-Hurwitz method as a tuning method Use Routh criterion to determine the adjustable parameters for tuning a PID controller: s 4 + 2s 3 + 3s 2 + s k c = zero 25

26 Solution: No. of roots: 4+1=5 Routh-array: 26

27 Solution: 27 For the system to be critically stable: (0.5 2kc) /2.5 = 0 kc = 0.25 This is the ultimate gain: k u = 0.25 To get the ultimate period: Set s = iω Substitute for s and kc in the C. E. ω 2 2iω 3 3ω 2 + iω = zero Imaginary part 2iω 3 + i ω = 0 or 2 ω 2 = 1 Then ω = 0.5 = 0.71 rad/sec

28 The ultimate period: P u = 2π/w = sec Use Z-N to tune the PID- controller: k c = 0.6 k u, k c = 0.6*0.25 = 0.15 τ i = P u /2, τ D = P u / 8, τ i = 8.845/ 2 = sec τ D = 8.845/ 8 = sec 28

29 The relative stability If the system satisfies the Routh criterion and absolute stable, it is desirable to determine The Relative Stability Relative Stability is represented by real part of each root. The more negative stability of each root is clearly necessary A first approach using an s plane formulation is to extend the Routh criterion to ascertain relative stability. This can by simply accomplished by shifting the s plane axis in order to utilize the Routh criterion 29

30 To determine how many roots of the characteristic equation lie in right of some vertical line, a distance α from the imaginary axis (i.e. to determine the numbers of roots that have a real part greater than α where α may be positive or negative depending upon whether the new axis of the s 1 plane is on the right or left of the original axis. Substitute (s 1 +α) for (s) in the C.E. and apply Routh s criterion 30

31 This is a useful approach, particularly for higher order systems with several pairs of closed-loop complex conjugate roots 31

32 Example : Determine the number of the roots that lie to the right of the imaginary axis of Argand diagram, and then determine the number of the roots that have a real part greater than or equal to ( 4), for the following C. E.: P s = s 4 + 3s 3 + 4s 2 + 6s = 0 Solution: Number of rows = 5 The Routh array is 32

33 Since all roots in first column are positive with only one root lies on the origin, the system is critically stable. Thus Routh criterion provides only the absolute stability i.e. stable, unstable or critically stable 33

34 If it is required to determine the relative damping of the system represent by the above eq. relative stability is to be applied. To go to relative stability put: s = s 1 + α in the C. E. and let α = - 4 P s = s s s s = 0 34

35 From Routh array after transformation to relative stability there are 4 changes of sign in the first column there are 4 roots in the RH of the s 1 plane i.e. the shifted plane. These roots are - 4 < s < 0, note that Routh array will not give any information of the values of relevant roots, but the relative stability determines the region in which the roots are located 35

36 Root Locus Method Very useful graphical techniques for identifying the roots of the C. E. 1 + K G s H s = 0 When the C. E. is of degree n, then there are n roots which are continuous functions of the open loop sensitivity K. As K varies from 0 to, each root traces a continuous curve (locus). When K = 0 the roots lies at the poles of the OLTF. It is based on the relationship that exists between the roots of the CLTF and the poles and zeros of the OLTF. 36

37 The root locus diagram can be plotted either: 1. Directly by solving the C. E. several times for various values of K, or 2. By application of Evan s rules Evan s Rules These are 11 steps to draw root locus on a complex plane (s plane). 37 Step 1: Write the OLTF in the following form: OLTF = k m c. j=1(s z j ) n j=1(s p j ) Where n = no. of poles and m = no. of zeros

38 Step 2: 1 + K(s + 6) (s + 1)(s + 3) = 0 Plot poles and zeros (x, 0) on complex s plane. Step 3: Odd number of poles and zeros to the right Draw the locus on the real axis to the left of an odd number of real poles and zeros 38 Step 4: Number of asymptotes N = n m Centered at the center of gravity C. G = n P j m j=1 j=1 Z j n m

39 39 Leaving at angles asy = ± (2k+1)π n m for k = 0, 1, 2, (n m 1) Step 5: Compute and locate the breakaway and break-in points if any exist, on the real axis using one of three approaches: 1. Differentiation method From C. E. find K = f (s) and put 2. Geometrical method Solve for σ : n j=1 1 σ P j = m j=1 1 σ Z j dk(s) ds = 0

40 3. Circular Loci When 2 poles lie on the same side of a single zero 40

41 41 Step 6: Compute the departure angles φ j from complex or compound poles together with entry or arrival angles ψ j at complex or compound zeros: From a complex pole P a : φ a = 2k + 1 π + (P a Z j ) m j=1 n (P a P j ) j=1 j a From q th order pole the departure angle is φ a = 1 q 2k o + m j=1 Where k = 0, 1, 2,.., (q 1) ψ j n j=1 j a φ j

42 Similarly from a complex zero Z b : ψ b = 2k + 1 π + (Z b P j ) n j=1 m (Z b Z j ) j=1 j b From v th order zero the arrival angle ψ b is ψ b = 1 v 2k o + n j=1 Where k = 0, 1, 2,.., (v 1) Step 7: φ j m j=1 j b Compute the crossover points where loci cross the imaginary axis. (by using Routh test) ψ j 42

43 Step 8: Apply the angle criterion to find more points on the root locus plot using: For all K > 0 Step 9: n j=1 (s P j ) m (s Z j ) j=1 = (2k + 1)π Apply the magnitude criterion to evaluate gain K at any point along the loci using: K = c s P 1 s Z 1 Or as vectors: s P 2 s P n s Z 2 s Z n 43 K = c s P 1 s Z 1 s P 2 s P n s Z 2 s Z n

44 Step 10: Grant s Rule states that the sum of the open loop poles is equal to the sum of the closed loop roots n m Where n m+2 j=1 P j = r j j=1 Step 11: Complete the locus 44

45 Example: A temperature controlled polymerization process has a transfer function: G s = k (s 40)(s + 80)(s + 100) Sketch the root locus diagram and show by the Routh test that the system is conditionally stable. Determine the upper and the lower values of Kc for proportional control with unity feedback and regulating elements. 45

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