Modelling and Control of Dynamic Systems PID Controllers

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

[k]
Basic structure of a PID controller
+
e[k]
P
I
D
+
u[k]
-1
System
ˆg[z]
y[k]
PID controllers are unity feedback controllers with three components:
⊲ a proportional term P with an output up[k] = Kp · e[k];
⊲ an integral term I with an output ui[k] = Ki · k
ek;
⊲ a derivative term D with an output ud[k] = Kd · (e[k] − e[k − 1]).
i=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 g1[z] = Kp.
⊲ The transfer function of the integral part is g2[z] = K i
z−1
⊲ The transfer function of the differential part is g3[z] = K d(1−z)
z
As a result, we can represent PID controller in an RST form
r[k]
T(z)
+
1
R(z)
u[k] System
ˆg[z]
-1 S(z)
y[k]
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 Xop ⊆ R n be the set of plausible states for the system.
⊲ Let Uop ⊆ R be the set of plausible inputs for the system.
⊲ Let yi[k + 1] denote the output corresponding to x[k] and ui[k]
A system has a monotone response if for any x[k] ∈ Xop, u1, u2 ∈ Uop
u1[k] ≤ u2[k] =⇒ y1[k + 1] ≤ y2[k + 1] .
If the output is differentiable then the condition can be rewritten as
∂y(k + 1, x, u)
∂u
≥ 0 for x ∈ Xop, u ∈ Uop .
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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
ˆg[z] = 1
z−1.1
The original zero-input response
Kp is too small to stabilize
Kp creates a constant bias
Kp is too big
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Drawbacks of proportional controllers
Proportional controller has too few degrees of freedom in the design.
⊲ Too large gain Kp creates oscillations
⊲ Too small gain Kp slows down the response and introduces bias.
0.0 0.5 1.0 1.5 2.0
Kp = 1
Kp = 1.8
0 50 100 150
We need an additional term that would eliminate constant bias.
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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.
0.0 0.5 1.0 1.5 2.0
0 50 100 150
Kp = 1 Ki = 0.3
Kp = 1.8 Ki = 0.3
As a result, the time needed to stabilise the output increases.
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2.0 1.0 0.0 0.5
1.5 1.0 0.5 0.0 0.5
Anatomy of a PI response
Proportional and integral components
0 20 40 60 80
Control signal
Kp = 1.0
Ki = 0.3
0 20 40 60 80
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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.
0.0 0.2 0.4 0.6 0.8 1.0
P = 0.6
I = 14.9
0 5 10 15 20
0.0 0.2 0.4 0.6 0.8 1.0
P = 0.6
I = 14.9
0 5 10 15 20
As a result, an introduction to differential term can significantly improve
the behaviour of the controller and reduce the stabilisation time.
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The effect of differential correction
Sudden changes activate the differential correction term. If a PI controller
already overshoots then setting Kd > 0 only increases the overshoot.
0.0 0.5 1.0 1.5 2.0
0.0 0.5 1.0 1.5 2.0
Kp = 1.0, Ki = 0.3, Kd = 0.4
0 50 100 150
Kp = 0.3, Ki = 0.3, Kd = 0.4
0 50 100 150
Kp = 1.8, Ki = 0.3, Kd = 0.05
0.0 0.5 1.0 1.5 2.0
0.0 0.5 1.0 1.5 2.0
0 50 100 150
Kp = 0.9, Ki = 0.3, Kd = 0.2
0 50 100 150
A positive differential correction works if a slow rising speed winds up the
integral part and thus it must be discharged through oscillations.
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Anatomy of an optimal PID response
0.0 0.5 1.0 1.5
The differential term
must dominate
Uncharging of
the integral term
0 20 40 60 80 100 120
The relation between coefficients Kp and Ki 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 Kp and Kd determine the initial rise speed.
⊲ They must be chosen so that the integral term does not grow too big.
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Methods for Parameter Tuning

Ziegler-Nichols closed-loop method
1. Set gains Ki = 0 and Kd = 0.
2. Gradually increase Kp until the system starts to continuously oscillate.
3. Let K ∗ p be the corresponding gain and T ∗ the corresponding period.
4. Start fine-tuning with the initial parameters from the following table
Controller Type Kp Ki Kd
P controller
PI controller
PID controller
K ∗ p
2 – –
K ∗ p
2.2
K ∗ p
1.7
T ∗
1.2
T ∗
2
–
T ∗
8
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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.
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