Fractional Order Relay Feedback Experiments for MIMO Process Identification
and Decoupling
Zhuo LiPhD Student, EECS, UC Merced
Member of the MESA [email protected]
6/12/2013
2
Outlines
• Background• Identification
– The relay feedback technique– relay meets fractional calculus– relay meets fractional order systems
• Decoupling – The experiment platform– When decoupling meet fractional order systems
• Some random thinking
3
Background
4
MEMS Micro-electro-mechanical systems
Inside an accelerometerhttp://memsblog.wordpress.com/2011/01/05/chipworks-2/
5
Nano fabrication, wafer processing
Demand:High precision High yieldRepeatabilityEfficiency Massive production
Challenges:Difficult to senseHigh nonlinearityMulti variableSynchronization
Fabrication of SiC nano-pillars by inductively coupled SF6/O2 plasma etchingJ H Choi1,2, L Latu-Romain2, E Bano1, F Dhalluin2, T Chevolleau2 and T Baron2
2012 J. Phys. D: Appl. Phys. 45 235204
6
Mission for control engineers
• Temperature • Pressure • Gas flow• RF power • etc ……
• Advanced modeling techniques • Advanced control technologies
7
The relay feedback technique
8
1942
Z-N Critical Oscillation
P feedback
1984
Astrom & Hugglund
Relay feedback
tuning
Luyben
Using relay for identification
1997
Waller
Two channel Relay
1987 1996
K.K Tan
Modified Relay CC Yu
Biased relay
1992
Astrom, 1984, Automatic Tuning of Simple Regulators with Specifications on Phase and Amplitude MarginsLuyben, 1987, Derivation of Transfer Functions for Highly Nonlinear Distillation ColumnsLi, 1991, An improved auto tune identification method……
Ramirez, R. WUse FFT for relay
1985 1991
W Li
Relay with time delay
The time lineA Leva
1993 2011
J Lee et. al
Relay with FO integrator behind
9
Varieties of relay feedbacksRelay type Process phase Phase pre-know Describing function Phase shift range
Ideal Yes One point
With hysteresis, , No 3rd and 4th quadrant
With time delay Yes 3rd and 4th quadrant
delay , behind Same as above - - -
With an integrator, Yes One point
integrator, , behind - - - -
TC relaytan
Yes 3rd quadrant
Biased ideal relay Yes One point
Biased with hysteresis
, No - 3rd and 4th quadrant
10
Ideal Relay
2 channel relay
Relay plus an integrator
Im
Re
Relay with hysteresis
Relay plus time delay
-180
-90
The frequency response
11
When relay feedback meets with fractional order integrator
12
1
𝑠𝛼
1
𝑠𝛼
13
Block diagrams
1
𝑠𝛼𝑒 𝑣 𝑢
-1 0 1 2 3 4 5 6 7 8 9 10-1.5
-1
-0.5
0
0.5
1
1.5
𝐴𝑠𝑖𝑛(𝜔𝑡)−𝐴𝜔𝛼
𝑠𝑖𝑛(𝜔𝑡− 𝜋2𝛼)
-1 0 1 2 3 4 5 6 7 8 9 10-1.5
-1
-0.5
0
0.5
1
1.5
−2𝐻𝜋
𝑒𝜋2
(1−𝛼) 𝑗
1𝜔𝛼 𝑒
−𝜋2𝛼 𝑗
𝐴𝜔𝛼
𝑠𝑖𝑛(𝜔𝑡− 𝜋2𝛼)
𝐴2 𝑗
4 𝐻𝜔𝛼
𝜋 𝐴4𝐻𝜋 𝐴
𝑒−𝜋
2𝛼 𝑗
Relay with integer order integrator
Relay with fractional order integrator
𝐻-H+ -
𝑒 𝑢1𝑠𝑣
𝐴𝜔
−𝐴𝜔
𝐴𝑠𝑖𝑛(𝜔𝑡) −𝐴𝜔𝑐𝑜𝑠 (𝜔𝑡 )
−𝐴2𝑗 −
𝐴2𝜔
1𝑗 𝜔
−2𝐻𝜋
4 𝐻𝜔𝜋 𝐴
−4𝐻𝜋 𝐴
𝑗
-1 0 1 2 3 4 5 6 7 8 9 10-1.5
-1
-0.5
0
0.5
1
1.5
14
Varieties of relay feedbacksRelay type Process phase Phase pre-know Describing function Phase shift range
Ideal Yes One point
With hysteresis, , No 3rd and 4th quadrant
With time delay Yes 3rd and 4th quadrant
delay , behind Same as above - - -
With an integrator, Yes One point
integrator, , behind - - - -
TC relaytan
Yes 3rd quadrant
With FO integrator, Yes 3rd and 4th quadrant
Biased ideal relay Yes One point
Biased with hysteresis
, No - 3rd and 4th quadrant
15
Ideal Relay
2 channel relay
Relay plus an integrator
Relay plus an FO integrator
Im
Re
Relay with hysteresis
Relay plus time delay
-180
-90
The frequency response
16
Ideal Relay
2 channel relay
Relay plus an integrator
Relay plus an FO integrator
Im
Re
Relay with hysteresis
Relay plus time delay
-180
-90
The frequency response
17
Simulation Eg.1
𝐺 (𝑠)= 22𝑠+1
𝑒− 0.1𝑠0 5 10-1
0
1
Ideal relay, H=1
0 5 10-1
0
1
With Hysteresis, eps=0.3
0 5 10-1
0
1
With time delay, L=1
0 5 10-1
0
1
With integrator
0 5 10-2
0
2
TC relay, Hi=H
p=1
Time [sec]0 5 10
-1
0
1
With FO integrator, =0.1
Time [sec]
A To T Error (%) L
Ideal 0.097 0.39 1.623 18.86 0.099
Hysteresis 0.383 1.552 1.624 18.79 0.201
Delay 0.846 3.612 1.632 18.4 0.135
Integrator 0.7 2.853 1.589 20.53 0.126
TC 0.195 0.743 2.183 9.16 0.099
FO int 0.133 0.536 1.627 18.65 0.102
R. K. WOOD and M. W. BERRY Model“Terminal composition control of a binary distillation column”
8
0 200 400 600 800-20
0
201/s
, = 1.3
0 200 400 600 800-20
0
201/s
, = 1.4
0 200 400 600 800-20
0
201/s
, = 1.5
0 200 400 600 800-20
0
201/s
, = 1.6
0 200 400 600 800-20
0
201/s
, = 1.7
0 200 400 600 800-20
0
201/s
, = 1.8
0 20 40 60 80 100 120 140 160 180 200-1
0
11/s
, = 0.1
0 20 40 60 80 100 120 140 160 180 200-1
0
11/s
, = 0.2
0 20 40 60 80 100 120 140 160 180 200-2
0
21/s
, = 0.3
0 20 40 60 80 100 120 140 160 180 200-2
0
21/s
, = 0.4
0 20 40 60 80 100 120 140 160 180 200-2
0
21/s
, = 0.5
0 20 40 60 80 100 120 140 160 180 200-2
0
21/s
, = 0.6
0 20 40 60 80 100 120 140 160 180 200-5
0
51/s
, = 0.7
0 20 40 60 80 100 120 140 160 180 200-5
0
51/s
, = 0.8
0 20 40 60 80 100 120 140 160 180 200-5
0
51/s
, = 0.9
0 20 40 60 80 100 120 140 160 180 200-10
0
101/s
, = 1
0 20 40 60 80 100 120 140 160 180 200-10
0
101/s
, = 1.1
0 20 40 60 80 100 120 140 160 180 200-10
0
101/s
, = 1.2
𝟏𝐬𝛂
Simulation Eg.2A To T Error (%) L
0.1 0.853 4.42 13.4 19.73 1.03
0.3 1.09 5.7 13.53 19.01 1.06
0.5 1.5 7.86 13.54 18.91 1.1
0.7 2.29 12.1 13.55 18.87 1.18
0.9 3.81 20.5 13.57 18.72 1.29
1.1 7.17 42 13.64 18.31 2
1.3 10.8 81.8 14.8 11.4 3.25
1.4 11.9 111 16.47 1.35 1.05
1.5 12.5 148 19.63 17.56 2.1
1.6 12.7 196 24.98 49.56 1.43
1.7 12.8 243 30.51 82.68 7.62
1.8 12.8 235 29.44 76.31 13.2
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Advantages• Wider phase range
• Phase can be predetermined,
• Non-zero initial part (efficient)
Relay with time delay Relay with FO integrator
Save a quarter cycle time !Think about some slow processese.g. distillation column
21
When relay feedback meets with fractional order system
22
Equations for relay identification For integer order systems For fractional order system (Proposed method)
Equations for IO are special cases of those for FO
23
Simulation
𝐺 (𝑠)= 12.8
16.7 𝑠0.5+1𝑒− 𝑠
0 5 10 15 20 25 30 35 40 45 50-2
0
2Relay with integrator
0 5 10 15 20 25 30 35 40 45 50-2
0
2Relay with time delay
0 5 10 15 20 25 30 35 40 45 50-1
0
1Ideal relay
0 5 10 15 20 25 30 35 40 45 50-1
0
1Relay with hysteresis
Time [sec]
Ideal relay With
delayWith
integrator
With hyst
0.70491.2369 1.0962
0.9002
2.45208.1960 6.2240
4.1150
13.994914.2198 14.0771
14.0679
Error16.2%
14.85% 15.71%15.76%
L0.9314
1.0945 0.8251 1.1830
24
Experimental implementation
0 20 40 60 80 100 120 140 16021
22
23
24
25
26
27
28
29Order: = 0.8
Time [sec]
Tem
pera
ture
C
Raw data
Model response
Identified by curve fittingUsing Dr.Podlubny’s mlf
0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10.06
0.08
0.1
0.12
0.14
0.16
0.18
Fractioanl order
Fitt
ing
erro
r: le
ast
mea
n sq
uare
s
Order scanning
0 10 20 30 40 50 60 70 80 90 100
-50
0
50
100
150
200
250
300
Rel
ay s
igna
l: P
WM
dut
y cy
cle
/255
relay with hyst
0 10 20 30 40 50 60 70 80 90 100
18
19
20
21
22
23
24
25
Tem
pera
ture
[C
]
Identified by relay feedback0 50 100 150 200 250 300 350
0
10
20
30
40
50
0 50 100 150 200 250 300 35020
22
24
26
28
30
0 50 100 150
-50
0
50
100
150
200
250
300
0 50 100 15019
20
21
22
23
24
25
26
Raw Data fromPlatform on slide 27
25
Future work• Other model structures • Using relay transient
𝐺 (𝑠)=𝐾¿¿
26
The experiment platform
27
The development highlights
• Thermoelectric modules• H-bridge, heating/cooling • IR thermo meters• Two inputs four outputs• Real-time control• Product of multiple failures
Peltier
Power
Peltier I 2C Bus
Metal plateArduino
SerialPC(Matlab)
IR Thermometers
MOSFET
Side product
28
The hardware configuration
Load
Heat sink
Peltier
Electric powerHeat
29
A video demo
30
The four modesPower on cooling – heat pumping Power off cooling – annealing/natural
dissipation
Power on heating – electrical heating Power off heating – thermo cycle
Performance testing• PID control with anti-windup• Testing with actuator only having cooling capability
Set point
Control signal
Temperature
The non-minimum phase temperature data
Fitting using second order model
0 20 40 60 80 100 120 140 160 180 200-2
-1
0
1
2
3
4
Time [sec]
Tem
pera
ture
C
Temperature data
Model response
0 50 100 150 200 250 300
0
50
100
Inpu
t/25
5 [D
uty
cycl
e]
0 50 100 150 200 250 30015
20
25
30
Time [sec]
Tem
pera
ture
C
𝐺 (𝑠)=𝐾 (𝑇3 𝑠+1)𝑇1𝑠
2+𝑇2𝑠+1𝑒−𝐿𝑠
[K T1 T2 T3] = [1.7048 198.8152 53.7816 -39.3604]
Fitting using fractional order model Commemorate order
0 20 40 60 80 100 120 140 160 180 200-2
-1
0
1
2
3
4
Time [sec]
Tem
pera
ture
C
Temperature data
Model response
[K T1 T2 T3] = [2716 -877 349.3 -6.1]
33
Decoupling
34
35
The conventional techniques
36
Conventional Decoupling
• Ideal decoupling• Simple decoupling• Inverted decoupling
Example – simplified decoupling• System
• Decoupler• D
-1.5
-1
-0.5
0
0.5
1From: In(1)
To:
Out
(1)
0 2 4 6 8 10 12 14-3
-2
-1
0
1
2
To:
Out
(2)
From: In(2)
0 2 4 6 8 10 12 14
Step Response
Time (seconds)
Am
plitu
de
Original responseAfter decoupling
Example – modified simplified• System
• Decoupler• D
-1.5
-1
-0.5
0
0.5
1From: In(1)
To:
Out
(1)
0 2 4 6 8 10 12 14-3
-2
-1
0
1
2
To:
Out
(2)
From: In(2)
0 2 4 6 8 10 12 14
Step Response
Time (seconds)
Am
plitu
de
Original responseAfter decoupling
39
What if the process is fractional order
40
Fractional order decoupler
Random thinkings
41
Another example
42
Credit: Dr.Richard Migan
Zhuo Li
Some diffusion data
43
Temperature in a sealed room – bounded diffusion
• Half order plus delay• Using NILT/Mittag leffler
• [K T L] = 6.0031 5.2222 14.7917• Fitting error (least mean squares): 0.2214
0 20 40 60 80 100 120 140 160 180-1
0
1
2
3
4
5
Time [sec]
Tem
pera
ture
C
Raw data
Model response
• Half order plus delay• Using NILT/Mittag leffler
• [K T L] = 2.1232 22.8021 9.7312• Fitting error (least mean squares): 0.0700
0 20 40 60 80 100 120 140 160 180-1
0
1
2
3
4
5
Time [sec]
Tem
pera
ture
C
Raw data
Model response
0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
5
10
15
20
25
order:
fitt
ing e
rror
44
45
Thank you