Fractional Order Relay Feedback Experiments for MIMO Process Identification and

Decoupling

Zhuo LiPhD Student, EECS, UC Merced

Member of the MESA Labzli32@ucmerced.edu

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 & HugglundRelay feedbacktuning

LuybenUsing relay for identification

1997

Waller Two channel Relay

1987 1996

K.K Tan Modified Relay

CC YuBiased 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 LiRelay with time delay

The time lineA Leva

1993 2011

J Lee et. alRelay 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

1𝑠𝛼

1𝑠𝛼

13

Block diagrams

1𝑠𝛼

𝑒 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− 𝜋

2𝛼)

-1 0 1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5

− 2𝐻𝜋 𝑒

𝜋2 (1−𝛼) 𝑗

1𝜔𝛼 𝑒

−𝜋2 𝛼 𝑗

𝐴𝜔𝛼 𝑠𝑖𝑛(𝜔𝑡− 𝜋

2𝛼)

𝐴2 𝑗

4 𝐻𝜔𝛼

𝜋 𝐴4𝐻𝜋 𝐴 𝑒

−𝜋2 𝛼 𝑗

Relay with integer order integrator

Relay with fractional order integrator

𝐻-H+ -

𝑒 𝑢1𝑠𝑣

𝐴𝜔− 𝐴𝜔

𝐴𝑠𝑖𝑛(𝜔𝑡) − 𝐴𝜔 𝑐𝑜𝑠 (𝜔𝑡 )

− 𝐴2 𝑗 − 𝐴

2𝜔

1𝑗 𝜔

− 2𝐻𝜋

4 𝐻𝜔𝜋 𝐴 − 4𝐻

𝜋 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

1Ideal relay, H=1

0 5 10-1

0

1With Hysteresis, eps=0.3

0 5 10-1

0

1With time delay, L=1

0 5 10-1

0

1With integrator

0 5 10-2

0

2

TC relay, Hi=Hp=1

Time [sec]0 5 10

-1

0

1With 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

20

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 )= 12.816.7 𝑠0.5+1

𝑒− 𝑠0 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 dataModel 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

Fitti

ng e

rror:

leas

t 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 dataFitting 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 dataModel response

0 50 100 150 200 250 300

0

50

100

Inpu

t/255

[Dut

y cy

cle]

0 50 100 150 200 250 30015

20

25

30

Time [sec]

Tem

pera

ture

C

𝐺 (𝑠 )=𝐾 (𝑇3 𝑠+1)𝑇 1𝑠

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 dataModel 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: O

ut(1

)

0 2 4 6 8 10 12 14-3

-2

-1

0

1

2

To: O

ut(2

)

From: In(2)

0 2 4 6 8 10 12 14

Step Response

Time (seconds)

Ampl

itude

Original responseAfter decoupling

Example – modified simplified• System

• Decoupler• D

-1.5

-1

-0.5

0

0.5

1From: In(1)

To: O

ut(1

)

0 2 4 6 8 10 12 14-3

-2

-1

0

1

2

To: O

ut(2

)

From: In(2)

0 2 4 6 8 10 12 14

Step Response

Time (seconds)

Ampl

itude

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 dataModel 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 dataModel response

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

5

10

15

20

25

order:

fittin

g er

ror

44

45

Thank you