Process Control Course II Lecture 7: Controllers Overview
Explore the fundamentals of Proportional-Integral-Derivative (PID) controllers in the context of process control engineering. Learn about PID controller block diagrams, transfer functions, parameter identification, and practical examples to understand their application in regulating systems effectively.
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Process Control Course II Lecture 7 The Controllers Part IV By Prof. Alaa Kareem Mohammed 1
6. Proportional -IntegralDerivative controller (PID) A PID controller continuously calculates an error value as the difference between a desired set point and a measured process variable and applies a correction based on proportional, integral and derivative terms (denoted P, I, and D respectively). Figure below shows the block diagram of proportional-integral- derivative controller PID E(s) Gc (s) P(s) ? ??? ? + ?? ??? ? +?? ? ? ? ?? ?? R 0 I 2
? ? = ??? ? + ?? ??? ? +?? ? ? ? ?? ?? 0 Proportional part Derivative part Integral part Taking Laplace transform for both sides ? ? ? = ??? ? + ?? ? ?? ? +?? ? ? ? ?? ?? 0 ? ? = ??? ? + ?? ?? ? ? +?? ? ? ? ? 1 ? ? ? ? = ? ? ?? 1 + ?? + ? ? ? ? 1 = ?? 1 + ?? + ? ? 3
1 ??? = ?? 1 + ?? + ? ? ? ??? I is the controller integral time . R is the controller derivative time. ?? ?? ? ? ?????????? ???????????? ????. Exercise What type and parameters of the following controllers ? . ??(?) = 2(1 + 6? +1 4?) ???: ??= 2, ? = 6, ? = 4 2?2+ ? + 1 ? ???: ??= 1, ? = 2, ? = 1 ? . ??? = 24?2+ 2? + 6 ? ? =1 ? . ??? = ???: ??= 2, ? = 12, 3 4
Example 4 PID Controller has the transfer function ?? ? = 2 [1 + 0 5? +1 error signal (E) is introduced to this PID controller at time = 0. Find the response P(t) and sketch both the E(t) and P(t). ?]. A negative ramp change of slope (- 0.5) in PID E(s) Gc (s) P(s) Solution ) ? ? = 0 5 ? (Negative ramp ?? ? = 2 [1 + 0 5? +1 ?] ???: ??= 2 , ? = 0 5 , ? = 1 ? ? = ?? ? ? + ????? ? +?? ? ? ? ?? ?? ?( 0 5 ?) ?? +2 ? ? = 2 0 5 ? + 2 0 5 1 ( 0 5 ?)?? ? ? = ? 0 5 0 5?2 5
?(?) P ? = ? 0 5 0 5?2 0 ???? 0.5 E(t)= - 0.5 t ?(?) 0 0 ???? 0.5 6
Example 5 For the closed loop shown in Fig. below, if a unit step change occurs in ?? Find (a) the value of (T), ? ??? ? ? offset, (b) The response and sketch it. (c) Repeat (a) for values of I=0.2 and I=20 ?L(s) + ?(s) Solution ?sp 1 1 2?) . . 2(1 + 3? + ? + 1 - The change occurs in load variable, then the problem is regulator 1 ? + 1 ? ? = 1 + 2(1 + 3? +1 1 2?)( ? + 1 7
a. 1 ? + 1 ? ? = 1 + 2(2? + 6?2+ 1 1 )( 2? ? + 1 1 ? ? + 1 ? ? = = 1 +6?2+ 2? + 1 ?(? + 1 7?2+ 3? + 1 ) T= 2.645 , ?=0.568 1 ? ? ?????? = ? = lim ? 0 ? = 0 7?2+ 3? + 1 8
b. Response y t = 11 ? ? 7?2+ 3? + 1 1 y t = 1 7?2+ 3? + 1 1 1 1 ?? ? ??sin?? ?2?2+ 2??? + 1= ? 1 ?2= 1 0 5682= 0 82 A = 1 ?2 ? 0 82 2 64= 0 31 B = = 0 C = 0 568 2 64= 0 215 ? ?= 1 0 82 2 64 ? 0 215?sin0 31? ? ? = ? ? = ? ?? ? ? ???????? ??? 9
I = 0.2 c. 1 ? + 1 ? ? = 1 1 1 + 2(1 + 3? + 0.2?)( ? + 1 0.1 ? ? ? = 0.7?2+ 0.3 ? + 1 T= 0.836 , ?=0.179 1 ? 0.1 ? ?????? = ? = lim ? 0 ? = 0 0.7?2+ 0.3 ? + 1 10
I = 20 1 ? + 1 ? ? = 1 1 1 + 2(1 + 3? + 20?)( ? + 1 10 ? ? ? = 70 ?2+ 30? + 1 T= 8.366 , ?=1.792 1 ? 10 ? ?????? = ? = lim ? 0 ? = 0 70 ?2+ 30? + 1 11
i Controller symbol Parameters Kc offset T or R I 1 Proportional Integral-Derivative PID 0.2 2 3 0 0.836 0.179 2 Proportional Integral-Derivative PID 2 2.645 0.568 2 3 0 3 Proportional Integral-Derivative PID 20 8.366 1.792 2 3 0 12
Example 6 Consider the closed system shown in Fig. below ?o(s) ?sp(s) 3 + 2 + 2? +2 . 3? + 1 ? - 1. What is the type of the controller? 2. What are the parameters of the controller? 3. Determine T, and the offset if a unit step change occurs in the set point. 4. If the controller is replaced by proportional- Derivative (PD) controller with ??= 2. ? = 1, find the offset for a unit step change in ???. Find the response of the system and sketch it. 5. Repeat (4) if the controller is proportional with Kc=2 13
Solution 1. The controller is proportional Integral-derivative controller (PID). 2. The parameters of the controller are: ??? = 2 + 2? +2 ?= 2 (1 + ? +1 ?) Controller gain ??= 2 Derivative time ? = 1 Integral time ? = 1 3. The change occurs in the set point, then the problem is servo 14
2 1 + ? +1 3 ? ) ??(? ???(? 3? + 1 ? ? = = 1 + 2(1 + ? +1 3 ?)( 3? + 1 ?2+ ? + 1 1 5 ?2+ 1 16? + 1] ??? =1 ?[ ? = 1.224 ? = 0.473 Offset = ???? ?? ?2+ ? + 1 1 5 ?2+ 1 16? + 1 1 ? Offset = 1 lim ? 0 ? Offset 1 1 = 0 15
4. The controller is proportional derivative (PD) ) ??? = ??1 + ? ? = 2(1 + ? ?o(s) + ?sp(s) 3 . 2(1 + ?) 3? + 1 - 3 2 1 + ? ) ??(? ???(? 3? + 1 ? ? = = 3 1 + 2(1 + ?)( 3? + 1 ? ? =6(s + 1) = 0.857(? + 1) 1.285 ? + 1 9? + 7 Offset = ???? ?? 1 ? 0.857(? + 1) 1.285 ? + 1 Offset = 1 lim ? 0 ? Offset 1 0.857 = 0.143 16
Response ??(? = 11 ? 0.857(? + 1) 1.285 ? + 1 ) 1 1 ??(? = 0.857[ 1 1.285 ? + 1+ 1 ) ?(1.285 ? + 1)] ?? 1 ? ? 1.285? 1.285+ (1 ? ??(? = 0.857[ ) 1.285 )] 0.857 ? ??(? = 0.857 0.19 ? ) 1.285 0 ? 17
5. The controller is proportional (P) ??(?) = ??= 2 ?o(s) + ?sp(s) 3 . 3 2 2 3? + 1 ) ??(? ???(? 3? + 1 ? ? = = - 3 1 + 2 ( 3? + 1 0.857 ? ? = 0.428 ? + 1 Offset = ???? ?? 1 ? 0.857 Offset = 1 lim ? 0 ? 0.428 ? + 1 Offset 1 0.857 = 0.143 18
Response ??(? = 11 0.857 ) ? 0.428 ? + 1 ? ?? ??(? = 0.857(1 ? ) 0.428) 0.857 0 ? 19
Thank you for your attention Any ? 20
