Modelling and Control of Dynamic Systems PID Controllers Sven Laur - PowerPoint PPT Presentation

Modelling and Control of Dynamic Systems PID Controllers Sven Laur University of Tartu

Basic structure of a PID controller P System r [ k ] u [ k ] y [ k ] e [ k ] I + + g [ z ] ˆ D -1 PID controllers are unity feedback controllers with three components: ⊲ a proportional term P with an output u p [ k ] = K p · e [ k ] ; k ⊲ an integral term I with an output u i [ k ] = K i · � e k ; i =1 ⊲ a derivative term D with an output u d [ k ] = K d · ( e [ k ] − e [ k − 1]) . 1

PID controller is a linear controller Note that all parts of the PID controllers are linear systems ⊲ The transfer function of the proportional part is g 1 [ z ] = K p . K i ⊲ The transfer function of the integral part is g 2 [ z ] = z − 1 ⊲ The transfer function of the differential part is g 3 [ z ] = K d (1 − z ) z As a result, we can represent PID controller in an RST form r [ k ] u [ k ] y [ k ] System 1 T ( z ) + R ( z ) g [ z ] ˆ S ( z ) -1 2

Gentle Introduction

The basic assumption In the following, we assume that the dependency between the control signal and the output signal is monotone throughout the entire operating regime . ⊲ Let X op ⊆ R n be the set of plausible states for the system. ⊲ Let U op ⊆ R be the set of plausible inputs for the system. ⊲ Let y i [ k + 1] denote the output corresponding to x [ k ] and u i [ k ] A system has a monotone response if for any x [ k ] ∈ X op , u 1 , u 2 ∈ U op u 1 [ k ] ≤ u 2 [ k ] = y 1 [ k + 1] ≤ y 2 [ k + 1] . ⇒ If the output is differentiable then the condition can be rewritten as ∂y ( k + 1 , x , u ) ≥ 0 for x ∈ X op , u ∈ U op . ∂u 3

Proportional controllers If the system has monotone response then the error signal e [ k ] = r [ k ] − y [ k ] indicates in which direction the input signal should be changed. ⊲ If e [ k ] > 0 then we should increase the control signal. ⊲ If e [ k ] < 0 then we should decrease the control signal. Example The original zero-input response 1 g [ z ] = ˆ K p is too small to stabilize z − 1 . 1 K p creates a constant bias K p is too big 4

Drawbacks of proportional controllers Proportional controller has too few degrees of freedom in the design. ⊲ Too large gain K p creates oscillations ⊲ Too small gain K p slows down the response and introduces bias. %$! K p = 1 #$" K p = 1 . 8 #$! !$" !$! ! "! #!! #"! We need an additional term that would eliminate constant bias. 5

Integral correction term For obvious reasons, the output of an integral term I = e [1] + · · · + e [ k ] is initially small and then continuously increases if the bias is not eliminated. Secondly, note that the integral term I cannot go to zero unless the error signal changes a sign—overshoot becomes unavoidable. %$! K p = 1 K i = 0 . 3 #$" K p = 1 . 8 K i = 0 . 3 #$! !$" !$! ! "! #!! #"! As a result, the time needed to stabilise the output increases. 6

Anatomy of a PI response !"#$#"%&#'()*('+*&'%,-"()*.#/$#','%0 !&( !&! ! '&! ! "&! ! "! #! $! %! 1#'%"#)*0&-'() K p = 1 . 0 K i = 0 . 3 !&( !&! ! !&( ! '&! ! '&( ! "! #! $! %! 7

Differential correction term An integral correction term creates the necessary change in the control signal to removes the bias but it cannot detect changes in the error term. P = 0 . 6 P = 0 . 6 #%! #%! I = 14 . 9 I = 14 . 9 !%( !%( !%' !%' ! ! !%& !%& !%$ !%$ !%! !%! ! " #! #" $! ! " #! #" $! As a result, an introduction to differential term can significantly improve the behaviour of the controller and reduce the stabilisation time. 8

The effect of differential correction Sudden changes activate the differential correction term. If a PI controller already overshoots then setting K d > 0 only increases the overshoot. K p = 1 . 0 , K i = 0 . 3 , K d = 0 . 4 K p = 1 . 8 , K i = 0 . 3 , K d = 0 . 05 %$! %$! #$" #$" #$! #$! !$" !$" !$! !$! ! "! #!! #"! ! "! #!! #"! K p = 0 . 3 , K i = 0 . 3 , K d = 0 . 4 K p = 0 . 9 , K i = 0 . 3 , K d = 0 . 2 %$! %$! #$" #$" #$! #$! !$" !$" !$! !$! ! "! #!! #"! ! "! #!! #"! A positive differential correction works if a slow rising speed winds up the integral part and thus it must be discharged through oscillations. 9

Anatomy of an optimal PID response &'( The di ff erential term &'! must dominate !'( Uncharging of the integral term !'! ! "! #! $! %! &!! &"! The relation between coefficients K p and K i determines the controllers behaviour when the output is near the reference signal. ⊲ They must be chosen so that the integral part is quickly discharged. The size of coefficients K p and K d determine the initial rise speed. ⊲ They must be chosen so that the integral term does not grow too big. 10

Methods for Parameter Tuning

Ziegler-Nichols closed-loop method 1. Set gains K i = 0 and K d = 0 . 2. Gradually increase K p until the system starts to continuously oscillate. p be the corresponding gain and T ∗ the corresponding period. 3. Let K ∗ 4. Start fine-tuning with the initial parameters from the following table Controller Type K p K i K d K ∗ p P controller – – 2 K ∗ T ∗ p PI controller – 2 . 2 1 . 2 K ∗ T ∗ T ∗ p PID controller 1 . 7 2 8 11

Other tuning methods There are many other tuning methods for PID parameters. ⊲ Cohen-Coon method for processes with high inertia ⊲ Chien-Hrones-Reswick open-loop method. ⊲ Ziegler-Nichols open-loop method. A PID controller can be used as a starting point in further research. 1. Stabilise the system with an initial PID controller. 2. Find a linear model that provides the fit to the experimental data. 3. Compute the corresponding transfer function. 4. Design the corresponding linear controller. 5. If one is not satisfied then she can build a non-linear model and controller. 12