© 2

009

The

Mat

hWor

ks, I

nc.

® ®

Introduction to PID Controller Designwith examples in MATLAB and Simulink

Dr. Bora Eryılmaz

Engineering Manager

Control and Estimation Tools Group

The MathWorks, Inc.

2

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What is a PID Controller?

A special type of controller C(s) with Proportional Integral Derivative

terms acting on the error signal E(s).

Step

P(s)

Process ModelPlant Output

C(s)

PID Controller

U(s)R(s) E(s) Y(s)

3

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What is a PID Controller? (cont.)

Ideal (standard) form:

Series (cascade) form:

Main point is: any second-order controller of the form

is a PID controller.

)(1

)( sEs

sK

s

KKsU

D

DIP

)(1

11)( sEsD

sD

s

IKsU

SS

SSS

12

012

2)(dsds

nsnsnsC

4

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PID Controllers Are Everywhere…

More than 90% of all controllers used in process industries are PID controllers.

A typical chemical plant has 100s or more PID controllers.

PID controllers are widely used in: Chemical plants Oil refineries Pharmaceutical industries Food industries Paper mills Electronic equipments

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Some History: Fluid Level Control

and at steady state

: in flow

: cross-sectional area

: liquid level

inq

A

h

( )

( ) ( )

G s

kA

kA

Y s R ss

Process Dynamics

1/A

sGain

k

Desired liquid level Actual level

r(t) error q_in(t)

h(t)=y (t)h(t)=y (t)

(0) 1G

1in

in

A dh q dt

h qA

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More History: Flyball Governor in Steam Engines

Proportional control Speed control for engines used

proportional control. See the flyball governor by James Watt in 1788.

dJw bw T T

w(t)

speedSet point - Desired Speed Gain

k

Engine Dynamics

1

J.s+b

Disturbance torque

r(t) error T(t)

Td(t)

: inertia

: friction coefficient

: disturbance torqued

J

b

T

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Many Types of “PID” Controllers…

Proportional (P):

Integral (I):

Proportional + Integral (PI):

Proportional + Derivative (PD):

You might see other combinations with different parameters than Kp, Ki, and Kd.

)()( sEKsU P

)()( sEs

KsU I

)()( sEs

KKsU IP

)(1

)( sEs

sKKsU

D

DP

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Low-Order Process Models

Many industrial processes can be modeled using simple stable transfer functions.

First-order process with delay:

Second-order process with delay:

There are many variations of these models, with or without time delays, with transfer functions zeros, ...

We can design PI/PID controllers based on these models.

1)(

0

00

s

eKsP

s

2000

20

2)(

0

ss

eKsP

s

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Desirable first-order responses with a tuning parameter K Remember the open-loop transfer function is given by

Design your PID controller so that L(s) looks like

Then the closed-loop transfer function will look like

)()()( sCsPsL

s

KsL )(

1)/(

1

)(1

)()(

KsKs

K

sL

sLsT

Step

P(s)

Process ModelPlant Output

C(s)

PID Controller

U(s)R(s) E(s) Y(s)

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Designing a PI controller for a first-order process model

PI controller for a first-order process model

Remember, given K, we want:

Our PI parameters:

Let’s put this in MATLAB and Simulink…

s

KKsC

s

KsP I

P

)( and 1

)(0

0

00

0 and K

KK

K

KK IP

s

KsCsPsL )()()(

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Desirable second-order responses with tuning parameters K and α Design your PID controller so that L(s) looks like

Then the closed-loop transfer function will look like

K and α are our design parameters.

1)/1()/(

1

)(1

)()(

22

sKsKKss

K

sL

sLsT

Step

P(s)

Process ModelPlant Output

C(s)

PID Controller

U(s)R(s) E(s) Y(s)

)1()()()(

ss

KsCsPsL

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Designing a PID controller for a second-order process model

PID controller for a second-order process model

Remember, given K and α, we want:

Let’s put this in MATLAB and Simulink…

1)( and

2)(

2000

20

s

sK

s

KKsC

ss

KsP

D

DIP

)1()()()(

ss

KsCsPsL

DDIP K

KK

K

KK

K

KK , 2-1 , , 2 2

02

0000

202

0000

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MATLAB and Simulink Helper

MATLAB commands useful for control design:• P = tf(num, den)• C = zpk(z, p, K)• L = minreal(P*C)• K = dcgain(P)• T = feedback(L,1)• bode(P,L), step(T)• sisotool(P)

Simulink blocks useful for control design:• Transfer Fcn• Zero-Pole• Integrator• Gain, Sum• Transport Delay (time

delay)• PID Controller

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Exam Question (10 points)

How to design PI/PID controllers for higher-order plants? Key idea is to shape first- or second-order “dominant” plant

dynamics. That is, you can ignore fast poles in the model. Find a first- or second-order model, P(s), that has a similar

response as the original model, Po(s). For example, you can use step(P, Po) or bode(P, Po) to compare responses. Similar poles and zeros can be ignored to simplify the model.

Question: Design a PI controller, using the technique of slides 9 & 10, for the plant

Can you get a closed-loop settling time less than 50 sec?

20 )1.01)(1)(201(

)151(2)(

sss

ssP