adaptive control for level process

MODEL REFERENCE ADAPTIVE CONTROL OF TANK LEVEL USING NEURAL NETWORKS :

MODEL REFERENCE ADAPTIVE CONTROL OF TANK LEVEL USING NEURAL NETWORKS Presentation By Dr.N.S.Bhuvaneswari , Professor, Easwari Engineering College,Chennai

Contents:

Contents Process Description Collection of data Process modeling using statistical model identification technique Design of Conventional Control system Design of Neuro Control system Need and Description of MRAS Design of Neuro based MRAS Results and Conclusions

HARDWARE SETUP OF CYLINDICAL TANK LEVEL PROCESS :

HARDWARE SETUP OF CYLINDICAL TANK LEVEL PROCESS

Process I/O dynamic Data:

Process I/O dynamic Data

Dynamic Data:

Dynamic Data

Dynamic Data:

Dynamic Data

Process Modelling by Statistical model Identification:

Process Modelling by Statistical model Identification Transfer Function = 4.722 ------------- 277.5s + 1

Transfer function for process parameter variations:

Transfer function for process parameter variations (1)When the outlet valve is fully opened condition, 0.214 Transfer function = ________ 371.17s+1 (2). When the outlet valve is in almost closed position, 4.915 Transfer function= _______ 70.17s+1

Controller Tuning by GA:

Controller Tuning by GA

Response using Conventional Control:

Response using Conventional Control

Closed Loop data:

Closed Loop data

Closed Loop data:

Closed Loop data

Closed Loop data:

Closed Loop data

Closed Loop data:

Closed Loop data

Closed Loop data:

Closed Loop data

Closed Loop data:

Closed Loop data

Neural network architecture used for training :

Neural network architecture used for training

Response using Neuro control system:

Response using Neuro control system

Response for Neuro control system when outlet valve is slightly opened from the normal mid position:

Response for Neuro control system when outlet valve is slightly opened from the normal mid position

Response for Neuro control system when outlet valve is fully opened:

Response for Neuro control system when outlet valve is fully opened

Block diagram of MRAS:

Block diagram of MRAS

Design of the model:

Design of the model

Mathematical model of the designed model:

Mathematical model of the designed model 1.011 Transfer function of the model= ------------ 20.8s+1

Response without adaptive mechanism:

Response without adaptive mechanism

Response with adaptive mechanism:

Response with adaptive mechanism

Process and model output when the outlet valve is at normal position:

Process and model output when the outlet valve is at normal position

Controller output :

Controller output

Response when the outlet valve is slightly closed from the normal:

Response when the outlet valve is slightly closed from the normal

Controller output:

Controller output

Response when the outlet valve is opened fully :

Response when the outlet valve is opened fully

Controller output:

Controller output

Comparision:

Comparision Methodology Mean Square error Average Settling time in sec Conventional Control 900 units 200 Neuro Control 850 units 180 Neural network based MRACS 400 units 100

Conclusion:

Conclusion The designed adaptive control is able to track the positive and negative changes in the setpoint smoothly despite the changes in the process parameters. The average settling time is found to be less in adaptive control methodology.

References:

References [1].Anandanatarajan R., Chidambaram M. and Jayasingh T. (2006), ‘Limitations of a PI controller for a first order non-linear process with dead time’ , ISA Transactions , Vol. 45, pp. 185-200. [2]Bhuvaneswari.N.S.Uma.G,Rangaswamy.T.R.(2008),’Neuro based model reference adaptive control of a conical tank level process”, International Journal of Control and Intelligent System, Actapress Publications, Canada, vol.36,no.1,2008 [3].Lee I.B. and Sung S.W. (1996), ‘Limitations and countermeasures of PID controllers’, Ind Eng. Chem ., Vol. 35, pp. 2596-2610. [4].Madhubala T.K., Boopathy M., Sarat Chandra J. and Radhakrishnan T.K. (2004), ‘Development and Tuning of Fuzzy Controller for a Conical Level System’, Proc of IEEE International Conference, ICISIP, pp. 450-455. [5].Noureddine Goléa, Amar Golea and Khier Benmahammed (2002), ‘Fuzzy Model Reference Adaptive Control’, IEEE Transactions on Fuzzy Systems , Vol. 10, No. 4, pp. 436-444. [6]. Pauli Viljamaa and Heikki N. Koivo (1995), ‘Fuzzy logic in PID gain scheduling’, Proceeding of Third European Congress on Fuzzy and Intelligent Technologies ; EUFIT’95, Aachen, Germany. [7].Young-Moon Park, Un-chul Moon and Kwang Y. Lee (1995), ‘A Self- Organizing Fuzzy Logic Controller for Dynamic Systems using a Fuzzy Auto-Regressive Moving Average (FARMA) Model’, IEEE Transactions on Fuzzy Systems, Vol. 3, No. 1, pp. 75-82.

Thank you:

Thank you Any Queries ?

Statistical model identification:

Statistical model identification Example homework: Practice with Matlab by inputting this exact example to become familiar with Matlab . (You need not turn anything in regarding this example). Assume that the system input signal named up is: up=[u(0) u(1) u(2) u(3) u(4) u(5) u(6) u(7) u(8) u(10) u(11)]' =[4 5 5 6 7 8 9 10 11 12 13]' Assume that the measured system output signal named yp is: yp =[y(0) y(1) y(2) y(3) y(4) y(5) y(6) y(7) y(8) y(9) y(10) y(11)]' =[0.62 0.72 0.76 0.52 0.91 0.92 0.97 1.52 1.42 1.99 1.95]'

Statistical model identification:

Statistical model identification Model Structure:y(k)=a 1 y(k-1)+a 2 y(k-2)+b 1 u(k-1)+b 2 u(k-2) Model parameters: a 1 , a 2 , b 1 , b 2 Solution: Vector Least squares regression

Statistical model identification:

Statistical model identification

Statistical model identification:

Statistical model identification % input Y vector into the Matlab command window >> Y=[0.62 0.72 0.76 0.52 0.91 0.92 0.97 1.52 1.42 1.99 1.95]' % Matlab prints out of Y vector Y = 0.6200 0.7200 0.7600 0.5200 0.9100 0.9200 0.9700 1.5200 1.4200 1.9900 1.9500 >> P=FI'*FI

Statistical model identification:

Statistical model identification % Matlab prints out the P vector P = 12.6531 10.0502 90.1600 78.7100 10.0502 8.6930 73.9200 66.2800 90.1600 73.9200 661.0000 577.0000 78.7100 66.2800 577.0000 517.0000 % we compute as the intermediate value PINV % remember P= , so PINV = >> PINV=inv(P) % Matlab prints out of PINV matrix PINV = 2.8603 0.4888 -0.3872 -0.0660 0.4888 5.1942 -0.0506 -0.6839 -0.3872 -0.0506 0.1112 -0.0586 -0.0660 -0.6839 -0.0586 0.1651 % we compute as the intermediate value FIValue % remember FI = , PINV = >> FIValue =PINV*FI'

Statistical model identification:

Statistical model identification FIValue = Columns 1 through 8 0 0.2244 0.1625 0.2598 -0.7943 -0.2493 -0.4833 -0.7885 0 0.1008 0.5840 0.4391 0.4790 -1.3115 -0.0153 -0.6734 0 0.2046 0.0112 -0.0680 0.1341 0.0477 0.0766 0.1093 0 -0.2754 -0.1043 -0.0102 -0.0804 0.1645 0.0035 0.0999 Columns 9 through 11 0.3559 -0.1145 1.0138 -0.8793 1.1942 0.2189 -0.0537 0.0098 -0.1533 0.1358 -0.1272 0.0100 % we compute as Theta % remember the FIValue =

Statistical model identification:

Statistical model identification >> Theta= FIValue *Y Theta = 0.0550 0.4905 0.1710 -0.0904 Since Theta= , we have Then we have the model structure as: y(k)=0.0550y(k-1)+0.4905y(k-2)+0.1710u(k-1)-0.0904u(k-2)