Process Control Stability: Concepts, Cases, and Definitions

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Explore the concept of stability in process control systems, including different stability cases such as stable, critically stable, and unstable. Understand the definition of stability as the ability of a system to provide bounded output for bounded input, known as the BIBO definition.

  • Process Control
  • Stability
  • Control Systems
  • Bounded Input Bounded Output
  • System Stability
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  1. Process Control Course II Lecture 8 The Stability Part I By Prof. Alaa Kareem Mohammed 1

  2. Outline 1- Concept of stability 2- Stability Cases 3- Definition of Stability 4- Characteristic Equation 5- S-Plane 6- Criteria of Stability 2

  3. Stability 1- Concept of Stability Consider the heating tank system shown below. Assume that the temperature of the exit stream To is controlled at value 40 oC . Now we wish to change the set point from 40 oC to 45oC, so the exit temperature (To ) should varied to reach the value 45oC. Temperature measuring element m cp Ti M m cp To Process Steam in Q P E Tm Controller Final control element Control Valve Comparator Tsp 3

  4. 1 ?i (s) ?? + 1 + ?(s) ?(s) ?(s) + + ?sp (s) ?? ? ?? (s) ?? ??? + 1 _ ?? + 1 ?? ?m (s) ??? + 1 According to the numerical values of the transfer functions in the block diagram, the response will be one of the following of three cases 4

  5. 1- The response of ( To) converge to 45oC in one of the forms shown below. Note that in all cases the response (To) reaches a certain value at ? . So in this case we said the system is stable. To To 45oC 44oC 45oC Offset =1 40oC 40oC t t 45oC 45oC 40oC 40oC t t 5

  6. 2- The response of ( To) oscillate around to 45oC as shown below. Note that the response (To) neither converge nor diverge from a certain value at ? . So in this case we said the system is critically stable. To 45oC 40oC t The response neither converge nor diverge 6

  7. 3- The response of ( To) diverge from 45oC shown below. Note that the response (To) does not reach a certain value at ? . So in this case we said the system is unstable. To 45oC 40oC t 7

  8. 2- Stability Cases Stability Unstable Stable Critically stable 8

  9. 3- Definition of Stability The stability of a control system is defined as the ability of any system to provide a bounded output when a bounded input is applied to it. A system is stable if every bounded input yields a bounded output. We call this statement a bounded-input, bounded-output (BIBO) definition of stability. More specifically, we can say, that stability allows the system to reach the steady-state and remain in that state for that particular input even after variation in the parameters of the system. Stability The system reach a certain value called steady state value 9

  10. 4- Characteristic Equation Characteristic equation is the denominator of the transfer function of the closed loop equal to zero. 1 + ????????= 0 Characteristic equation 1 + ???= 0 ?L (s) ?? ? ??? ? = ?????? + + + ?sp (s) ?? (s) ?? ?? ?? _ ?? 10

  11. Example 1 Find the characteristic equation for the closed loop shown below. 0.5 ? + 1 ?2(s) Solution 1 + ???= 0 ???(s) + ?2(s) + + 0.5 0 0.5 2 0 ? + 12 1 + ????????= 0 2 0.5 2 (2? + 1) 1 + (2)(0.5)( (? + 1)2)( 2? + 1) = 0 2?3+ 5?2+ 4? + 2 = 0 Characteristic equation Note that the characteristic equation has three roots of S, these roots called poles 11

  12. 5- S-plane It is a plane the x-axis of which is real and the y-axis of which is imaginary. It is used to ascertain the nature or S root and hence the stability of the system. Imaginary axis ??( ??,??) ??(??,??) ??(?,??) ??( ??,?) ??(??,?) R??? ???? ??(?, ??) ??(??, ??) ??( ??, ??) 12

  13. Imaginary axis ??( ??,??) ??(??,??) ??(?,??) ??( ??,?) ??(??,?) R??? ???? (?, ??) ?? ( ??, ??) (??, ??) ?? ?? Root s ?1 Definition Example Response in time domain Notes ?1? ?1? S= - 4 S= 2 3? ? ?2?(?1cos?1? + ?2cos?2?) Cause the system to be stable Negative real Complex: negative real Cause the system to be stable ?2 ?2 S= 6? Pure imaginary (?1cos?3? + ?2cos?3?) Cause the system to be critical stable ?3 ?3 ??4?(?1cos?4? + ?2cos?4?) S=4 2? Complex: positive real Cause the system to be unstable ?4 ?4 ?1??5? S= 3 Positive real Cause the system to be unstable ?5 13

  14. Example 2 A closed system has the following characteristic equation. 2?8+ ?7+ 2?6 31?4 16 ?3 32 ?2 16 = 0 Find the roots of the equation and then locate them on the S-plane. Solution 2?8+ ?7+ 2?6 31?4 16 ?3 32 ?2 16 = 0 ?2 4 ?2+ 4 ?2+ ? + 1 2?2 ? + 1 = 0 ?1= 2 ?3= 2 ? 1 2+ 3 ?5= 2 j ?2= 2 ?4= 2 ? ?7=1 7 4+ 4 j 1 2 3 ?6= 2 j ?8=1 7 4 4 j 14

  15. Roots j ?1= 2 ?3 2 ?2= 2 ?7 ?3= 2 ? 1 ?5 ?4= 2 ? ?1 ?2 R??? ???? 1 2+ 3 ?5= 2 j ?6 -2 1 2 -1 1 2 3 ?6= 2 j -1 ?8 ?7=1 7 4+ 4 j -2 ?8=1 7 4 4 j ?4 15

  16. i???????? ???? j ?3 2 j ?7 1 j ?5 ?2 ?1 R??? ???? 1 2 -2 -1 ?6 -1j ?8 -2j ?4 16

  17. ?(s) ???(s) ? 4? 2? + 1 + ??(1+S) 0 Fig. 17

  18. Criteria of stability 1- The system is stable if and only if all roots are negative real or complex with negative real. For example the system has the following roots is stable: ?1= 4,?2,3= 2 3? 2- The system is unstable if one or more of the roots are positive real or complex with positive real. For example the system has the following roots is unstable: ?1= 3,?2= 2 ?3,4= 2 3? 3- The system is critically stable if one or more of the roots are pure imaginary on condition that no positive real or complex with positive real are exist, otherwise the system is unstable. For example the system has the following roots is critically stable: ?1,2= 3?,?3= 2, ?4,5= 2 3? 18

  19. Exercise Specify the stability of the system that has the following roots : a. ?1= 6,?2,3= 2 3?,?4= 2 b. ?1= 6,?2,3= 3?,?4= 2 c. ?1= 1,?2,3= 3?,?4= 2 d. ?1= 6,?2,3= 1 3?,?4= 2 e. ?1= 6,?2,3= 3?,?4= 2 f. ?1= 6,?2= 2,?3= 5,?4= 1 g. ?1= 5,?2,3= 2 3? 19

  20. Example 3 Consider the closed loop shown in the block diagram below. Test the stability of the system in the following cases: (a). Kc=1 (b). Kc=5/4 (c). Kc=2 Solution ???(s) ? 8? ? + 1 ?(s) + 0 ?? (a). Kc=1 1 + ???= 0 ? 8? ? + 1 1 + 1 = 0 1 4? 1 + 4? ? + 1 1 + 1 = 0 20

  21. 1 4? 1 + 1 = 0 (? + 1)(4? + 1) 4?2+ ? + 2 = 0 ?1,2= ? ?2 4?? 2? ?1,2= 1 1 4(4)(2) 2(4) ?1,2= 0.125 0.695? The system is stable Roots are complex numbers with real part is negative. 21

  22. (b). Kc= 5/4 1 + ???= 0 ? 8? ? + 1 5 4 1 + = 0 16?2+ 9 = 0 9 16 ?2= ?1,2= 3 4? The system is critically stable Roots are complex numbers with pure imaginary. 22

  23. (c). Kc= 2 1 + ???= 0 ? 8? ? + 1 1 + 2 = 0 4?2 3? + 3 = 0 ?1,2=3 9 4(4)(3) 2(4) ?1,2= 0.375 0.78? The system is unstable Roots are complex numbers with real part is positive. 23

  24. Homework 1 Consider the closed loop shown in the block diagram below. Test the stability of the system in the following cases: ?(s) (a). Kc = 1 (b). Kc = 4 ???(s) ? 4? 2? + 1 + ??(1+S) 0 Ans: (a). The roots are complex with negative real part. ?1,2= 3 (b). One of the roots is positive real S1=1.118, S2= - 1.118. The system is unstable 7 4 4 ?. The system is stable 24

  25. Homework 2 Test the stability of the closed loop shown in the block diagram below. ? + ? ? + 1 100.5 ? + 1 ? ? ? + 2? + 1 - ??? ??,?= ? ?? ? ? ? .The system is stable 25

  26. Example 4 Consider the closed loop shown in the block diagram below. Test the stability of the system in the following cases: (a). Kc=1 (b). Kc=8 (c). Kc=27 ysp 1 + 0.2 ?? 0.5? + 13 Solution ? y - 1 + ???= 0 5 0.2 5 ?? 0.5 ? + 13= 0 1 + 0.5 ? + 13+ ??= 0 ?3+ 6?2+ 12? + 8 1 + ?? = 0 ( )

  27. ? . ??= 1 ?3+ 6?2+ 12? + 8 1 + 1 = 0 ?3+ 6?2+ 12? + 16 = 0 .( ) In order to solve Eq.( ), we shall use trail and error to find one root that satisfies Eq.( ) , then dividing Eq.( ) by this root to reduce Eq.( ) to second order. ?2+ 2? + 4 Note that the root S1= - 4 satisfies Eq.(*) ?3+ 6?2+ 12? + 16 ?3 4?2 Dividing Eq.( ) by (S+4) , we obtain ?2+ 2? + 4 ? + 4 So, Eq.( ) can be written as 2?2+ 12?+16 2?2 8? ? + 4 ?2+ 2? + 4 = 0 4? + 16 22 4(4)(1 2 ) ?2.3= 2 12 2 = 1 ? 4? 16 12 2? The roots are: S1= - 4, S2,3 = 1 0 0 The system is stable 27

  28. ? . ??= 8 ?3+ 6?2+ 12? + 8 1 + ?? = 0 ( ) ?3+ 6?2+ 12? + 72 = 0 ( ) ???? ? ?? ?1= 6 satisfies Eq.(**) ?2+ 12 ?3+ 6?2+ 12? + 72 ?3 6?2 ?3+ 6?2+ 12? + 72 = ? + 6 ?2+ 12 = 0 ? + 6 0 + 12? + 72 ?1= 6 12? 72 ?2+ 12 = 0 0 0 ?2.3= 12 = 12 j One root is negative real and the other two roots are pure imaginary , so the system is critically stable The value of Kc that make the system critically stable is called maximum Kc,max or critical Kc, crit . ??,???= 8 28

  29. Kc,max make the system to be critically stable and then the response will oscillate sinusoidal as shown in Fig. below. y In this case , we can find the frequency of oscillation ? and the Time of one cycle?? as well. ?? ?? ? = ? ? ? ?? ? = ? Return to our example t ?2.3= 12 j ? = 12 ? = 2?? =2? ?? 2 ? 12= 1.813 ??=2? ?= 29

  30. ? . ??= 27 ?3+ 6?2+ 12? + 8 1 + ?? = 0 ( ) ?3+ 6?2+ 12? + 224 = 0 ..( ) ???? ? ?? ?1= 8 satisfies Eq.(**) Then Eq.(**) can be written as S + 8 ?2 2? + 28 = 0 ?2,3=2 4 112 ?1= 8 2 ?2.3= 1 27? The system is unstable Two of the roots are complex with the real part is positive 30

  31. Thank you for your attention Any ? 31

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