80 Lab Experiment 11: Introduction to PID controller

CISE 302
Linear Control Systems
Lab Experiment 11: Introduction to PID controller

Objective: Study the three term (PID) controller and its effects on the feedback loop response.
Investigate the characteristics of the each of proportional (P), the integral (I), and the derivative
(D) controls, and how to use them to obtain a desired response.

List of Equipment/Software
Following equipment/software is required:

 MATLAB
 LabVIEW

Category Soft - Experiment

Deliverables
A complete lab report including the following:

 Summarized learning outcomes.

 LabVIEW programming files (Block diagram and Front Panel)
 Controller design and parameters for each of the given exercises.

Introduction: Consider the following unity feedback system:

R + e u Y
Controller Plant
-

Plant: A system to be controlled.

Controller: Provides excitation for the plant; Designed to control the overall system behavior.

The three-term controller: The transfer function of the PID controller looks like the following:

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81 Lab Experiment 11: Introduction to PID controller

KP = Proportional gain

KI = Integral gain

KD = Derivative gain

First, let's take a look at how the PID controller works in a closed-loop system using the
schematic shown above. The variable (e) represents the tracking error, the difference between the
desired input value (R) and the actual output (Y). This error signal (e) will be sent to the PID
controller, and the controller computes both the derivative and the integral of this error signal.
The signal (u) just past the controller is now equal to the proportional gain (KP) times the
magnitude of the error plus the integral gain (KI) times the integral of the error plus the
derivative gain (KD) times the derivative of the error.

( )
( ) ∫ ( )

This signal (u) will be sent to the plant, and the new output (Y) will be obtained. This new output
(Y) will be sent back to the sensor again to find the new error signal (e). The controller takes this
new error signal and computes its derivatives and its internal again. The process goes on and on.

Example Problem:

Suppose we have a simple mass, spring, and damper problem.

The modeling equation of this system is

̈ ̇

Taking the Laplace transform of the modeling equation (1), we get

( ) ( ) ( ) ( )

The transfer function between the displacement X(s) and the input F(s) then becomes

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82 Lab Experiment 11: Introduction to PID controller

( )
( )

Let

 M = 1kg
 b = 10 N.s/m
 k = 20 N/m
 F(s) = 1

Plug these values into the above transfer function

( )
( )
The goal of this problem is to show you how each of Kp, Ki and Kd contributes to obtain

 Fast rise time

 Minimum overshoot
 No steady-state error

Open-loop step response: Let's first view the open-loop step response.
num=1;
den=[1 10 20];
plant=tf(num,den);
step(plant)

MATLAB command window should give you the plot shown below.

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83 Lab Experiment 11: Introduction to PID controller

The DC gain of the plant transfer function is 1/20, so 0.05 is the final value of the output to a unit
step input. This corresponds to the steady-state error of 0.95, quite large indeed. Furthermore, the
rise time is about one second, and the settling time is about 1.5 seconds. Let's design a controller
that will reduce the rise time, reduce the settling time, and eliminates the steady-state error.

Proportional control:

The closed-loop transfer function of the above system with a proportional controller is:

P Controller Plant
R + e u Y
KP
- 𝑠 𝑠

( )
( ) ( )

Let the proportional gain (KP) equal 300:

Kp=300;
contr=Kp;
sys_cl=feedback(contr*plant,1);
t=0:0.01:2;
step(sys_cl,t)

MATLAB command window should give you the following plot.

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84 Lab Experiment 11: Introduction to PID controller

Note: The MATLAB function called feedback was used to obtain a closed-loop transfer function
directly from the open-loop transfer function (instead of computing closed-loop transfer function
by hand). The above plot shows that the proportional controller reduced both the rise time and
the steady-state error, increased the overshoot, and decreased the settling time by small amount.

Proportional-Derivative control:

The closed-loop transfer function of the given system with a PD controller is:

PD Controller Plant
R + e u Y
KP+KDs
- 𝑠 𝑠

( )
( ) ( ) ( )

Let KP equal 300 as before and let KD equal 10.

Kp=300;
Kd=10;
contr=tf([Kd Kp],1);
sys_cl=feedback(contr*plant,1);
t=0:0.01:2;
step(sys_cl,t)

MATLAB command window should give you the following plot.

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85 Lab Experiment 11: Introduction to PID controller

This plot shows that the derivative controller reduced both the overshoot and the settling time,
and had a small effect on the rise time and the steady-state error.

Proportional-Integral control:

Before going into a PID control, let's take a look at a PI control. For the given system, the
closed-loop transfer function with a PI control is:

PI Controller Plant
R + e u Y
KP+KI/s
- 𝑠 𝑠

( )
( ) ( )

Let's reduce the KP to 30, and let KI equal 70.

Kp=30;
Ki=70;
contr=tf([Kp Ki],[1 0]);
sys_cl=feedback(contr*plant,1);
t=0:0.01:2;
step(sys_cl,t)

MATLAB command window gives the following plot.

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86 Lab Experiment 11: Introduction to PID controller

We have reduced the proportional gain (Kp) because the integral controller also reduces the rise
time and increases the overshoot as the proportional controller does (double effect). The above
response shows that the integral controller eliminated the steady-state error.

Proportional-Integral-Derivative control:

Now, let's take a look at a PID controller. The closed-loop transfer function of the given system
with a PID controller is:

PID Controller Plant

R + e u Y
KP+KI/s+ KDs
- 𝑠 𝑠

( )
( ) ( ) ( )

After several trial and error runs, the gains Kp=350, Ki=300, and Kd=50 provided the desired
response. To confirm, enter the following commands to an m-file and run it in the command
window. You should get the following step response.

Kp=350;
Ki=300;
Kd=50;
contr=tf([Kd Kp Ki],[1 0]);
sys_cl=feedback(contr*plant,1);
t=0:0.01:2;
step(sys_cl,t)

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87 Lab Experiment 11: Introduction to PID controller

Now, we have obtained a closed-loop system with no overshoot, fast rise time, and no steady-
state error.

The characteristics of P, I, and D controllers:

The proportional controller (KP) will have the effect of reducing the rise time and will reduce,
but never eliminate, the steady state error. An integral controller (KI) will have the effect of
eliminating the steady state error, but it may make the transient response worse. A derivative
control (KD) will have the effect of increasing the stability of the system, reducing the overshoot
and improving the transient response.

Effect of each controller KP, KI and KD on the closed-loop system are summarized below

CL Response Rise Time Overshoot Settling Time S-S Error

KP Decrease Increase Small Change Decrease
KI Decrease Increase Increases Eliminate
KD Small Change Decreases Decreases Small Change

Note that these corrections may not be accurate, because KP, KI, and KD are dependent of each
other. In fact, changing one of these variables can change the effect of the other two. For this
reason the table should only be used as a reference when you are determining the values for KP,
KI, and KD.

Exersice:

Consider a process given below to be controlled by a PID controller,

400
G p ( s) 
s ( s  48.5)

a) Obtain the unit step response of Gp(s).

b) Try PI controllers with (Kp=2, 10, 100), and Ki=Kp/10. Investigate the unit step
response in each case, compare the results and comment.
c) Let Kp=100, Ki=10, and add a derivative term with (Kd=0.1, 0.9, 2). Investigate the unit
step response in each case, compare the results and comment.

Based on your results in parts b) and c) above what do you conclude as a suitable PID controller
for this process and give your justification.

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