Introduction to Database Modeling

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Data Base Dynamic Modeling Summer Semester 2011
Assignment

Process System Engineering (PSE)
Solution

Identify Continuous and ARX models

Name:

Hossein

Sure Name: ShirGir Matr.Nr.
1

146074
Hossein ShirGir, 146074 , Dortmund ,Germany

a) Identify continuous models that match the observed output. Please proceed with the following steps: 1- Consider that the continuous models should be built as products of the following four transfer functions:

For starting I ran the program by initial point and G1 which defined by assignment and I got the figure G1. Its starting point for getting the map, the first transfer function is second order but there is deviation in some points so we should improve the transfer function.

2

Hossein ShirGir, 146074 , Dortmund ,Germany

FIG.1 G1

Comment: Transfer function is second order transfer function there is deviation should be improve and eliminate with effect of other transfer function.

3

Hossein ShirGir, 146074 , Dortmund ,Germany

FIG.2 G1.G2

Comment: This transfer function is resulted by multiplication of G1 and G2 some deviation of first map eliminated by effect of denominator of second transfer function the higher order of denominator seems to improve function convergence.

4

Hossein ShirGir, 146074 , Dortmund ,Germany

FIG.3 G1.G4

Comment : This map resulted by G1 and G3 multiplication it seems better than G1 alone because the denominator order improved in resulted transfer function but because of lack of nominator order it doesn’t has stability as G1.G2 so the deviation increase in compare of previous map and transfer function.

5

Hossein ShirGir, 146074 , Dortmund ,Germany

FIG.3 G1.G4

Comment: as you see there is a great deviation between true value and PT. So it is easy to recognize that selected transfer function is not suitable as enough as to met the process control demand. It shows even we have denominator but in this case it makes first transfer function (G1) unstable. And also you can find out that G4 has bad effect on system so for last attempt I am going to ignore it and also increase my denominator order and nominator order by the G1, G2 and G3 multiplication.

6

Hossein ShirGir, 146074 , Dortmund ,Germany

For G1.G2.G3 system: We can reach our Matlab code:
G_sim_12=K_12*(tf([TN_12 1],[(1/wo_12).^2 2*d_12/wo_12 1])).*(tf ([TN2_12 1],[TD2_12 1])).*(tf ([1] , [TD3_12 1])); %structure of function G_sim=K*(tf([TN 1],[(1/wo).^2 2*d/wo 1])).*(tf ([TN2 1] , [TD2 1])).*(tf ([1] , [TD3 1]));

FIG.5 G1.G2.G3

7

Hossein ShirGir, 146074 , Dortmund ,Germany

Comment: As we expected the system seems to be a robust and stable system, it shows the control system is good behavior system. And also it can be find out when you compare J factor of transfer function the system of G1.G2.G3 has lowest J factor.



Whic h m od el w ould yo u c hoose to rep resent the p la nt? Why?

The Last model which generated by G1.G2.G3 is best one because it has higher order so it controls process properly. It has good stability as well.

8

Hossein ShirGir, 146074 , Dortmund ,Germany

Now I should define the G1,G2,G3,G4 and then I put it in the solution file and get result to put it.

Case K d

G1
0.3655 -0.5504 0.3808 3.8750 0 0 0 0

G1.G2
0.2289 -0.5747 0.4145 3.8531 5.8376 9.6480 0 0 0.2005

G1.G3
0.3827 -0.4604 0.2914 4.7612 0 0 0.1685 0 0.1413

G1.G4
1 -1 0.3000 5 0 0 0 5 8.2120

G1.G2.G3
0.3013 -0.4810 0.2593 5.3122 0.5830 1.4956 0.5830 0 0.0328

J

0.2431

For Zero and Poles I complete the chart below:

Zero and Poles Pole 1 Zero 1

G1
-1.4755 + 3.5830i -1.4755 - 3.5830i

G1.G2
-1.5971 + 3.5065i -1.5971 - 3.5065i -0.1713 1.7399 -0.1036

G1.G3
-5.9341 -1.3875 + 4.5545i -1.3875 - 4.5545i

G1.G4
-1.5000 + 4.7697i -1.5000 - 4.7697i -0.2000

G1.G2.G3
-1.3773 + 5.1306i -1.3773 - 5.1306i -1.7152 + 0.0000i -1.7152 - 0.0000i 2.0790 -0.6686

1.8168

2.1720

1

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