presented at reno aiche meeting – nov. 6, 2001 getting more information from relay feedback tests
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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.
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