Chapter 7 Chapter 7 Adjusting Controller Adjusting Controller ParametersParameters

Professor Shi-Shang JangChemical Engineering Department

National Tsing-Hua UniversityHsin Chu, Taiwan

7-1 Basic Requirement of 7-1 Basic Requirement of a controllera controller

The closed loop system must be stableThe effects of disturbance must be

minimized, good disturbance rejection (regulation performance)

Rapid, smooth responses to set-point changes are required, good servo performance

Steady state error (offset) is eliminatedExcessive control action is avoided (control

action should avoid oscillation, input stable)

The control system robust, that is, insensitive to changes in process

7-1 Quarter Decay Ratio By 7-1 Quarter Decay Ratio By Ultimate PropertiesUltimate Properties

Ziegler-Nichols

Kc I D

P 0.5Ku - -

PI Ku/2.2 Pu/1.2 -

PID Ku/1.7 Pu/2 Pu/8

7-1 Example (Example 7-1 Example (Example 6-1.16-1.1

Transfer Fcn 6

1

0.75s+1Transfer Fcn 5

0.502 s+1

0.524

Transfer Fcn 4

-3.34

8.34s+1

Transfer Fcn 3

1

0.75s+1

Transfer Fcn 2

1

0.502 s+1

Transfer Fcn 1

8.34s+1

1.183

Transfer Fcn

0.2s+1

1.652

Subtract

Step

Scope

Gain

10 .37

Constant

0

Add

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Ku=10.4Pu=4.6min

Transfer Fcn 6

1

0.75s+1Transfer Fcn 5

0.524 s+1

0.502 s+1

Transfer Fcn 4

-3.34

8.34s+1

Transfer Fcn 3

1

0.75s+1

Transfer Fcn 2

1

0.502 s+1

Transfer Fcn 1

8.34s+1

1.183

Transfer Fcn

0.2s+1

1.652

To Workspace 1

simout 1

To Workspace

simout

Subtract

Step

Scope 1

ScopePID Controller(with Approximate

Derivative )

PID

Constant

0

Add

7-1 Example 7-1 Example (Example 6-1.1- (Example 6-1.1- Cont.Cont.

0 2 4 6 8 10 12 14 16 18 20-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

time

C %

TO

0 2 4 6 8 10 12 14 16 18 200

5

10

15

time

M %

CO

7-2 Open Loop 7-2 Open Loop CharacterizationCharacterization

7-2 Open Loop 7-2 Open Loop CharacterizationCharacterization

7-2 Open Loop Characterization -7-2 Open Loop Characterization -ExampleExample

7.2

0.80

10 1 30 1 3 1

0.8

54.3 1

s

G ss s s

e

s

0 50 100 150 200 2500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

t,s

T(t

)

7.2 61.5

7-2 Tuning for Quarter 7-2 Tuning for Quarter Decay RatioDecay Ratio

7-2 Tuning for Quarter 7-2 Tuning for Quarter Decay Ratio - ExampleDecay Ratio - Example

1 1

0

I 0

0

1.2 1.2 7.211.3125

0.8 54.3

2.0 2 7.2 14.4

7.23.6

2 2

c

D

tK

K

t

t

Transfer Fcn 2

1

3s+1

Transfer Fcn 1

1

30 s+1

Transfer Fcn

10 s+1

0.8

To Workspace

simout

Subtract

Step

ScopePID Controller

PID

0 50 100 150 200 2500

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

t,s

T(t

)

7-2.2 Tuning for 7-2.2 Tuning for Integral CriteriaIntegral Criteria

dttyydtteIAE set

00

dttyydtteISE set

0

2

0

2

• Integral of absolute value of the error (IAE)

• Integral of the square error (ISE)

• Integral of the time-weighted abolute error (ITAE)

dttyytdttetITAE set

00

7-2.2 Tuning for Integral 7-2.2 Tuning for Integral Criteria - IAECriteria - IAE

7-2.2 Tuning for Integral 7-2.2 Tuning for Integral Criteria - IAECriteria - IAE

7-2.2 Tuning for Integral Criteria – 7-2.2 Tuning for Integral Criteria – IAE : IAE : ExampleExample

0.921 0.921

0

0.749 0.749

0I

1.137 1.137

0

1.435 1.435 7.231.5589

0.8 54.3

54.3 7.213.6170

0.878 0.878 54.3

7.20.482 0.482 54.3 2.6313

54.3

c

D

tK

K

t

t

Transfer Fcn 2

1

3s+1

Transfer Fcn 1

1

30 s+1

Transfer Fcn

10 s+1

0.8

To Workspace

simout

Subtract

Step

ScopePID Controller

PID

Constant

0

Add

0 50 100 150 200 250-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time,s

T(t

)

Example – Example – Cont.Cont.

p p i s p

sM s

dTV c f t c T t UA T t T t f t c T t

dtdT

C w t UA T t T tdt

1

Out1

68*0.8

thou*cp

0.8s+1

0.1*0.8s+1

tds+1/v*tds+1

966

lambda

15

f(t)

42.2

W_base 2.1*241.5

U*A

1s

Ts,230F

1.652

0.2s+1

Transfer Fcn1

1

0.75s+1

Transfer Fcn

simout1

To Workspace1

simout

To Workspace

100

Ti(t)

151

T_set

1s

T, 150F

Subtract1

Subtract

Scope2

Scope1

Scope

Product1

Product

3.8

Kc

1s

Integrator

[Tm]

Goto3

[W]

Goto2

[Ts]

Goto1

[T]

Goto

[T]

From4

[Tm]

From3

[W]

From2

[Ts]

From1

[T]

From

150

Constant1

150

Constant

Add8

Add7

Add6

Add5

Add4

Add3

Add2

Add1

Add

1/3.2

1/taui

1/265.7

1/CM

1/(120*68*0.8)

1/(V*thou*cp)

2

In2

1

In1

Example – Example – Cont.Cont.

1100200

0100;

1

100

ln;

1

Dynamics Valve Control

TT

TT

vv

vv

Ks

KsG

rTransmitteSensor

wK

s

KsG

Example – Example – Cont.Cont.

Example – Example – Cont.Cont.

1

2

1 2

1 1

1.652 1.183 1.0

0.2 1 8.34 1 0.502 1 0.75 1

3.34 0.524 1 1.0

8.34 1 0.502 1 0.75 1

1 1

v s

F

c

c c

G s G s G s H ss s s s

sG s G s H s

s s s

G s G s G sC s R s F s

G s G s G s G s

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Step Response Test; FOPDT Fit 2

0 2 4 6 8 10 12 14 16 18 20-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Response;

C

Time (min)

Regression Test

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

time

Response;

C

Step Response Test; SOPDT, Smith’s Method

0 2 4 6 8 10 12 14 16 18 20-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Response;

C

Time (min)

1 2

1min

τ τ1 1 8.32 0.482.57

K τ 1.9 1 0.8

c

cp c

K

1 2τ τ τ 8.8I

1 2

1 2

τ ττ 0.453

τ τD

0 2 4 6 8 10 12 14 16 18 20150

150.2

150.4

150.6

150.8

151

151.2

151.4

151.6

151.8

152

FOPTD

SOPTD

PID Control Comparison

Tem

perature; C

Time, min

7-3 Summary7-3 SummaryA control loop should be stable, fast

responding and robustZ-N QD tuning and IAE tuning are

widely used in the industriesOn-Line tuning is also widely used

HomeworksHomeworksText p 2717-3, , 7-12, 7-15, 7-19, 7-22

Supplemental MaterialsSupplemental Materials

Synthesis of Feedback Synthesis of Feedback ControllersControllersController synthesis

◦Given the transfer functions of the components of a feedback loop, synthesize the controller required to produce a specific closed-loop response

Formula derivation

Ch

apte

r 7

1 ( ) / ( )( )

1 ( ) / ( )( )c

C s R sG s

C s R sG s

For perfect control

◦This says that in order to force the output to equal the set point at all times, the controller gain must be infinite.

◦In other words, perfect control cannot be achieved with feedback control.

◦This is because any feedback corrective action must be based on an error.

Ch

apte

r 7

( ) ( )C s R s

1 ( ) / ( ) 1 1( )

1 ( ) / ( ) 0( ) ( )c

C s R sG s

C s R sG s G s

( ) / ( ) 1C s R s

Specification of the Closed-Specification of the Closed-Loop ResponseLoop Response

The simplest achievable closed-loop response is a first-order lag response

τc is the time constant of the closed-loop response◦ The single tuning parameter for the synthesized

controller◦ Design parameter τc provides a convenient

controller tuning parameter that can be used to make the controller more aggressive (small τc) or less aggressive (large τc).

Ch

apte

r 7

This controller has integral mode◦No offset◦i.e. unity gain

Ch

apte

r 7

111 ( ) / ( ) 1 1 1

( )11 ( ) / ( ) 1 1( ) ( ) ( )1

1

cc

c

c

sC s R sG s

C s R s sG s G s G ss

~

NotesNotesAlthough second- and higher-

order closed-loop responses could be specified, it is seldom necessary to do so.

When the process contains dead time, the closed-loop response must also contain a dead-time term, with the dead time equal to the process dead time.

Ch

apte

r 7

θ

τ 1

s

c

C e

R s

• If the process transfer function contains a known time delay θ, a reasonable choice for the desired closed-loop transfer function is:

• The time-delay term is essential because it is physically impossible for the controlled variable to respond to a set-point change at t = 0, before t = θ.

• If the time delay is unknown, θ must be replaced by an estimate.

θ

θ

1

τ 1

s

c sc

eG

G s e

Ch

apte

r 7

• Although this controller is not in a standard PID form, it is physically realizable.

• Using a truncated Taylor series expansion:

θ 1 θse s

Note that this controller also contains integral control action.

Ch

apte

r 7

θ θ

θ

1 1

τ θτ 1

s s

c scc

e eG

sG Gs e

FOPDT ModelConsider the standard FOPDT model,

θ

τ 1

sKeG s

s

1 τ, τ τ

θ τc Ic

KK

Ch

apte

r 7

θ θ

θ

1 τ 1 1 τ 1(1 )

τ θ τ θ θ τ τ

s s

c sc c c

e s eG

s s K sG Ke

1 1/ τc IK s

Ch

apte

r 7

SOPDT ModelConsider a SOPTD model,

θ

1 2τ 1 τ 1

sKeG s

s s

11 τ

τc c DI

G K ss

where:

1 2 1 21 2

1 2

τ τ τ τ1, τ τ τ , τ

τ τ τc I Dc

KK

ExampleUse the DS design method to calculate PID controller settings for the process:

2

10 1 5 1

seG

s s

Ch

apte

r 7 Consider three values of the desired closed-loop time constant: .

Evaluate the controllers for unit step changes in both the set point and the disturbance, assuming that Gd = G. Repeat the evaluation for two cases:

1, 3, and 10c

a. The process model is perfect ( = G).

b. The model gain is = 0.9, instead of the actual value, K = 2. Thus,

0.9

10 1 5 1

seG

s s

GK

The controller settings for this example are:

3.75 1.88 0.682

8.33 4.17 1.51

15 15 15

3.33 3.33 3.33

τ 1c τ 3c 10c

2cK K

0.9cK K

τI

τD

Ch

apte

r 7

1 210, 5, 1, 2,K

If 1,c

1 2τ τ1 1 153.75

τ 2 2cc

KK

1 2τ τ τ 15I

1 2

1 2

τ ττ 3.33

τ τD

Ch

apte

r 7

The values of Kc decrease as increases, but the values of and do not change.

τcτI τD

Perfect process modelPerfect process modelC

hap

ter

7

With model mismatchWith model mismatchC

hap

ter

7

Comparison to quarter decay Comparison to quarter decay ratio responseratio response

Ch

apte

r 7

Ch

apte

r 7

Set the parameters of the PID controlleraccording to Table 7-1.1

Ch

apte

r 7