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1. Dynamic Behavior of Closed-Loop Control Systems Chapter 11

2. Chapter 11

3. Next, we develop a transfer function for each of the five elements in the feedback control loop. For the sake of simplicity, flow rate w 1 is assumed to be constant, and the system is initially operating at the nominal steady rate. Process In section 4.1 the approximate dynamic model of a stirred-tank blending system was developed: where Chapter 11

4. Chapter 11

5. The symbol denotes the internal set-point composition expressed as an equivalent electrical current signal. is related to the actual composition set point by the composition sensor-transmitter gain K m : Chapter 11

6. Current-to-Pressure (I/P) Transducer The transducer transfer function merely consists of a steady-state gain K IP : Control Valve As discussed in Section 9.2, control valves are usually designed so that the flow rate through the valve is a nearly linear function of the signal to the valve actuator. Therefore, a first-order transfer function is an adequate model Chapter 11

7. Composition Sensor-Transmitter (Analyzer) We assume that the dynamic behavior of the composition sensor-transmitter can be approximated by a first-order transfer function, but τ m is small so it can be neglected. Controller Suppose that an electronic proportional plus integral controller is used. where and E ( s ) are the Laplace transforms of the controller output and the error signal e ( t ). K c is dimensionless. Chapter 11

8. Chapter 11

9. 1. Summer 2. Comparator 3. Block Blocks in Series are equivalent to... Chapter 11

10. Chapter 11

11. “Closed-Loop” Transfer Functions Indicate dynamic behavior of the controlled process (i.e., process plus controller, transmitter, valve etc.) Set-point Changes (“Servo Problem”) Assume Y sp  0 and D = 0 (set-point change while disturbance change is zero) (11-26) Disturbance Changes (“Regulator Problem”) Assume D  0 and Y sp = 0 (constant set-point) (11-29) *Note same denominator for Y/D, Y/Y sp . Chapter 11

12. Chapter 11

13. Chapter 11

14. Chapter 11 Figure 11.16 Block diagram for level control system.

15. Chapter 11

16. Chapter 11

17. Chapter 11

18. EXAMPLE 1: P.I. control of liquid level Block Diagram: Chapter 11

19. Assumptions 1. q 1 , varies with time; q 2 is constant. 2. Constant density and x-sectional area of tank, A. 3. (for uncontrolled process) 4. The transmitter and control valve have negligible dynamics (compared with dynamics of tank). 5. Ideal PI controller is used (direct-acting). For these assumptions, the transfer functions are: Chapter 11

20. The closed-loop transfer function is: Substitute, Simplify, Characteristic Equation: Recall the standard 2 nd Order Transfer Function: (11-68) (2) (3) (4) (5) Chapter 11

21. For 0 <  < 1 , closed-loop response is oscillatory. Thus decreased degree of oscillation by increasing K c or  I (for constant K v , K M , and A). To place Eqn. (4) in the same form as the denominator of the T.F. in Eqn. (5), divide by K c , K V , K M : Comparing coefficients (5) and (6) gives: Substitute, unusual property of PI control of integrating system better to use P only Chapter 11

22. Stability of Closed-Loop Control Systems Chapter 11

23. Proportional Control of First-Order Process Set-point change: Chapter 11

24. Set-point change = M Offset = See Section 11.3 for tank example Chapter 11

25. Closed-Loop Transfer function approach: First-order behavior closed-loop time constant (faster, depends on K c ) Chapter 11

26. Chapter 11

27. General Stability Criterion Most industrial processes are stable without feedback control. Thus, they are said to be open-loop stable or self-regulating . An open-loop stable process will return to the original steady state after a transient disturbance (one that is not sustained) occurs. By contrast there are a few processes, such as exothermic chemical reactors, that can be open-loop unstable . Definition of Stability. 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. Chapter 11

28. Effect of PID Control on a Disturbance Change For a regulator (disturbance change), we want the disturbance effects to attenuate when control is applied. Consider the closed-loop transfer function for proportional control of a third-order system (disturbance change). K c is the controller function, i.e., . is unspecified Chapter 11

29. Let If K c = 1, Since all of the factors are positive, , the step response will be the sum of negative exponentials, but will exhibit oscillation. Chapter 11 If K c = 8, Corresponds to sine wave (undamped), so this case is marginally stable.

30. If K c = 27 Since the sign of the real part of the root is negative, we obtain a positive exponential for the response. Inverse transformation shows how the controller gain affects the roots of the system. Offset with proportional control (disturbance step-response; D(s) =1/s ) Chapter 11

31. Therefore, if K c is made very large, y(t) approaches 0 , but does not equal zero. There is some offset with proportional control, and it can be rather large when large values of K c create instability. Chapter 11 Integral Control: For a unit step load-change and K c =1, no offset (note 4 th order polynomial)

32. adjust K c and  I to obtain satisfactory response (roots of equation which is 4 th order). PI Control: no offset PID Control: (pure PID) No offset, adjust K c ,  I ,  D to obtain satisfactory result (requires solving for roots of 4 th order characteristic equation). Analysis of roots of characteristic equation is one way to analyze controller behavior Chapter 11

33. Rule of Thumb: Closed-loop response becomes less oscillatory and more stable by decreasing K c or increasing t I . General Stability Criterion Consider the “characteristic equation,” Note that the left-hand side is merely the denominator of the closed-loop transfer function. The roots (poles) of the characteristic equation (s - p i ) determine the type of response that occurs: Complex roots  oscillatory response All real roots  no oscillations ***All roots in left half of complex plane = stable system Chapter 11

34. Chapter 11 Figure 11.25 Stability regions in the complex plane for roots of the characteristic equation.

35. Stability Considerations Feedback control can result in oscillatory or even unstable closed-loop responses. Typical behavior (for different values of controller gain, K c ). Chapter 11

36. Roots of 1 + G c G v G p G m (Each test is for different value of K c ) Chapter 11 (Note complex roots always occur in pairs) Figure 11.26 Contributions of characteristic equation roots to closed-loop response.

37. Chapter 11

38. Routh Stability Criterion Characteristic equation Where a n >0 . According to the Routh criterion, if any of the coefficients a , a 1 , …, 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. On the other hand, if all of the coefficients are positive, then one must construct the Routh Array shown below: (11-93) Chapter 11

39. For stability, all elements in the first column must be positive. Chapter 11

40. The first two rows of the Routh Array are comprised of the coefficients in the characteristic equation. The elements in the remaining rows are calculated from coefficients by using the formulas: (n+1 rows must be constructed; n = order of the characteristic eqn.) . . (11-94) (11-95) (11-96) (11-97) Chapter 11

41. The important constraint is K c <8. Any K c  8 will cause instability. Application of the Routh Array: Characteristic Eqn is We want to know what value of K c causes instability, I.e., at least one root of the above equation is positive. Using the Routh array, Conditions for Stability Chapter 11

42. Figure 11.29 Flowchart for performing a stability analysis. Chapter 11

43. 1. Bode Stability Criterion Ch. 14 - can handle time delays 2. Nyquist Stability Criterion Ch. 14 Additional Stability Criteria Chapter 11

44. Direct Substitution Method Imaginary axis is the dividing line between stable and unstable systems. Substitute s = j w into characteristic equation Solve for K cm and w c (a) one equation for real part (b) one equation for imaginary part Example (cf. Example 11.11) characteristic equation: 1 + 5 s + 2 K c e -s = 0 (11-101) set s = j w 1 + 5 j w + 2 K c e -j w = 0 1 + 5 j w + 2 K c ( cos( w) – j sin( w) ) = 0 Chapter 11

45. Direct Substitution Method (continued) Re: 1 + 2 K c cos w = 0 (1) Im: 5 w – 2 K c sin w = 0 (2) solve for K c in (1) and substitute into (2): Solve for w : w c = 1.69 rad/min (96.87 ° /min) from (1) K cm = 4.25 (vs. 5.5 using Pade approximation in Example 11.11) Chapter 11

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