Presented at Reno AIChE Meeting – Nov. 6, 2001

Getting More Information

From Relay Feedback Tests

William L. LuybenLehigh University

Scope

• Relay-Feedback Test – method, advantages, normal results = Ku and Pu

• Curve Shapes – change with deadtime

• Proposed Curvature Factor

• Proposed Identification Method D, Kp and

• Effectiveness

• Conclusion

Relay-Feedback Test

• Insert relay in feedback loop, specify height “h” (high gain P controller, Hi and Lo limits on OP

OPhi – OPlo = 2 h)

• Produces limit cycle in PV signalwith period = ultimate gain = Pu

and amplitude “a” PVmax – PVmin = 2 a

• Calculate ultimate gain = Ku = 4h/a

0 5 10 15-0.2

-0.1

0

0.1

0.2

0 5 10 15

-1

-0.5

0

0.5

1

OP

Pu=3.5a=0.17

PV

Time (minutes)

Time (minutes)

5.717.0

)1(44

ahKu

h=1

Advantages• Fast

• Only need to specify “h”

• Gives accurate dynamic information at frequency that is important for feedback controller design.

• Closedloop test. Keeps process in linear region.

• Detect load changes from asymmetric pulses.

Limitation - only get two parameters.

Reference - Yu, C. C. Autotuning of PID Controllers, 1999, Springer, London

Proposed Ways to Get More Information

• Run two tests: conventional and then with dynamic element

inserted to get another point on the Nyquist plot.

(Li et al; Ind. Eng. Chem. Research 1991, 30 1530)

• Two-channel test: use two relays in parallel, one with

conventional and one with integrator.

(Friman and Waller; Ind. Eng. Chem. Research 1997, 36, 2662)

Curve Shapes

Deadtime in process affects shape of curve.

Mentioned by (1) Astrom and (2) Friman and Waller.

Nothing quantitative proposed by these authors.

Basic Idea – use curve shape to find a third

process parameter.

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

Time (minutes)

PV

Deadtime = 1

Gain = Kp = 0.5 With OP = 1

Time constant = 2

63%

First-Order with Deadtime1)(

seK

OPPVG

o

Dsp

s

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.1

-0.05

0

0.05

0.1

0.15

Time

0 1 2 3 4 5 6 7 8 9 10-1

-0.5

0

0.5

1

Time

0 10 20 30 40 50 60 70 80 90-1

-0.5

0

0.5

1

Time

D=0.1

D=1

D=10

First-order/deadtime process

Y

Y

Y

uPbF 4

Curvature Factor =

Y

0

a

a/2

b

Pu

time

t1 t2

Y0

aa/2

b

Pu

time1)(

seK

GDs

ps

)/()(

D

eKG Dsps

)0/(

)/1()(

DseG

Ds

s

time

Y0

a

-a

aPu

221

F

Pb u

Y 0

Pu

time

a

5.081

F

Pb u

t1 t2

F=2

F=0.5

• Range of Possible F Factors

is from 0.5 to 2.

• Can relate F Curvature Factor

to D Deadtime.

D/ 0.1 1 10

Relay-feedbackTest: Pu 0.382 2.98 21.4 a 0.0952 0.632 1.0 b 0.0488 0.620 9.31 Ku 13.4 2.01 1.27 F=4b/Pu 0.511 0.832 1.74

IMC: 0.2 1.7 17 Kc 5.25 0.882 0.353 I 1.05 1.5 5.5

0.5 1 1.5 2-1.5

-1

-0.5

0

0.5

1

1.5

2

Curvature Factor = F = 4b/Pu

Log 1

0(D

/)

PureIntegrator (D=0)

PureDeadtime(D=)

D = 1

)(6788.2)(8974.9

)(7147.122783.5)/(log32

10

FF

FD

)arctan(arg )( uui DGu

Proposed Procedure

1. Run relay-feedback test a, b and Pu

2. Calculate Ku=4h/a , u =2/Pu and F=4b/Pu

3. Calculate D/ from correlation with F.

4. Define D/ = c (c is known constant)

)arctan(arg )( uui cGu

One equation in unknown .

uu

pi K

KG

u

1

)(1 2)(

5. Calculate D = c

6. Calculate Kp

7. Now three process parameters are known: D, and Kp.

8. Use IMC tuning rules.

 

2/2

2)2.0,7.1max(

D

DKK

D

I

pc

 

Effectiveness

First-order with various deadtimes

Third-order

Inverse response

Openloop unstable

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

First-Order; D=0.05

Time

Y00

5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-5

0

5

10

15

20

Time

M00

5ZN

TL

ZN

TL

IMC

IMC

Fig. 5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5First-Order; D=0.1

Time

Y01

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-2

0

2

4

6

8

10

Time

M01

ZN

TL

ZN

TL

IMC

IMC

Fig. 6

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

First-Order; D=1

Time

Y1

0 1 2 3 4 5 6 7 8 9 100.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Time

M1

ZN

TL

ZN

TL

IMC

IMC

Fig. 7

0 10 20 30 40 50 600

0.2

0.4

0.6

0.8

1First-Order; D=10

Time

Y10

0 10 20 30 40 50 600.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time

M10

ZN

TL

ZN

TL

IMC

IMC

Fig. 8

Now test on third-order process.

Actual process:

Approximate process:

3)( )1(81

s

eG

Ds

s

1)(

seK

GDs

ps

0 2 4 6 8 10 12 14 16 18 20-0.03

-0.02

-0.01

0

0.01

0.02

0.03 Third-Order; D=0.1

Time

0 5 10 15 20 25 30-0.1

-0.05

0

0.05

0.1 Third-Order; D=1

Time

0 10 20 30 40 50 60 70 80 90-0.2

-0.1

0

0.1

0.2 Third-Order; D=10

Time

Y

Y

Y

Fig. 9

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

Third-Order; D=0.1

Time

Y01

0 5 10 15 20 25 30-5

0

5

10

15

20

25

Time

M01

Fig. 10

ZN

IMC

ZN

IMC

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4Third-Order; D=1

Time

Y1

0 5 10 15 20 25 304

5

6

7

8

9

10

11

Time

M1

Fig. 11

ZN

IMC

ZN

IMC

0 10 20 30 40 50 60 70 80 900

0.2

0.4

0.6

0.8

1

1.2

1.4

Third-Order; D=10

Time

Y10

0 10 20 30 40 50 60 70 80 903

4

5

6

7

8

9

10

Time

M10

Fig. 12

ZN

IMC

ZN

IMC

0 2 4 6 8 10 12 14 16 18 20-1.5

-1

-0.5

0

0.5

1

1.5

Time

D=0.1; Inverse response

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

-1

0

1

2

Time

D=1; Inverse response

0 10 20 30 40 50 60 70 80 90-2

-1

0

1

2

Time

D=10; Inverse response

Tauz=0.8

Tauz=0.8

Tauz=0.8

Tauz=1.6

Tauz=1.6

Tauz=1.6

Y

Y

Y

Fig. 13

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.2

-0.1

0

0.1

0.2 First-Order Unstable; D=0.1

Time

Y1

0 0.5 1 1.5 2 2.5 3 3.5 4-0.4

-0.2

0

0.2

0.4First-Order Unstable; D=0.2

Time

Y2

0 1 2 3 4 5 6-0.4

-0.2

0

0.2

0.4 First-Order Unstable; D=0.3

Time

Y3

Fig. 14

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2First-Order Unstable; D=0.1

Time

Y1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-6

-4

-2

0

2

4

6

8

Time

M1

Fig. 15

ZN

IMC TL

ZN

IMC

TL

0 1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2

3

4

First-Order Unstable; D=0.2

Time

Y2

0 1 2 3 4 5 6 7 8 9 10-4

-3

-2

-1

0

1

2

3

4

Time

M2

Fig. 16

ZNIMC

TL

ZN

IMC

TL

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

First-Order Unstable; D=0.3

Time

Y3

0 2 4 6 8 10 12 14 16 18 20-5

-4

-3

-2

-1

0

1

2

3

Time

M3

Fig. 17

ZN

TL

ZN

TL

Conclusion

Only one simple test required.

Use shape of curve to deduce deadtime.

Works for first and higher-order systems.

Does not work for inverse response or unstable.