president universityerwin sitompulsmi 9/1 dr.-ing. erwin sitompul president university lecture 9...

President University Erwin Sitompul SMI 9/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 9

System Modeling and Identification

http://zitompul.wordpress.com


Chapter 5 Discrete-Time Process Models

Homework 8(a) Find the discrete-time transfer functions of the following

continuous-time transfer function, for Ts = 0.25 s and Ts = 1 s. Use the Forward Difference Approximation

2

10( )

2 10G s

s s

(b) Calculate the step response of both transfer functions for 0 ≤ t ≤ 5 s.

(c) Compare the step response of both transfer functions with the step response of the continuous-time transfer function G(s) in one plot.


Solution of Homework 8

(a)2

10( ) ,

2 10G s

s s

s

1zs

T

2

s s

10( )

1 12 10

G zz zT T

2

s2 2

s s s

10

2 2 10 2 1

T

z T z T T

2 2

s1 2 2

s s s

10

1 2 2 10 2 1

T z

T z T T z


2

s2 2

s s

10

2 1 2 2 10

T

z z z T T


s 0.25 sT

2 2

1 2 2

10(0.25)( )

1 2(0.25) 2 10(0.25) 2(0.25) 1

zG z

z z



2

1 2

0.625( )

1 1.5 1.125

zG z

z z

s 1 sT

2 2

1 2 2

10(1)( )

1 2(1) 2 10(1) 2(1) 1

zG z

z z

2

2

10( )

1 9

zG z

z


(b) The step response of both transfer functions for 0 ≤ t ≤ 5 s.



Using the following command in Matlab workspace:

Y1 = dlsim([0.625],[1 –1.5 1.125],ones(1,21))Y1 = [0 0 0.6250 1.5625 2.2656 2.2656 1.4746 0.2881 –0.6018

–0.6018 0.3993 1.9010 3.0273 3.0273 1.7602 –0.1403 –1.5658

–1.5658 0.0378 2.4433 4.2473]

Using the following command in Matlab workspace:Y2 = dlsim([10],[1 0 9],ones(1,6))Y2 = [ 0 0 10 10 –80 –80 ]



Solution of Homework 8(c) Comparing the step responses

Step Response

Time (sec)0 1 2 3 4 5

-2

-1

0

1

2

3

4

5Step Response

Time (sec)0 1 2 3 4 5

-80

-70

-60

-50

-40

-30

-20

-10

0

10

• FDA delivers bad results• Possible solutions can be the

use of smaller sampling time Ts or the use of ZOH or TA

Ts = 0.25 s Ts = 1 s


Solution of Homework 8Chapter 5 Discrete-Time Process Models

s 0.05 sT FDA, s 0.01 sT FDA,

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

• FDA with smaller sampling time Ts


Step Response

Time (sec)A

mpl

itude

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4



s 0.25 sT TA, s 0.25 sT ZOH,

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

• Using TA or ZOH, with reasonably large sampling time Ts


Industry processes can be modeled in various ways, such as in state-space description or in transfer functions.

The models mostly used for control purposes are in form of linear differential or difference equations, with parameters assumed as known and constant.

In real conditions, it is often necessary to measure or estimate these parameters from input and output signals of the process.

This case is referred to as parameter estimation or process identification.

Chapter 6 Process Identification

Process Identification



Process Identification The objective of process identification is to find a model that

can describe the process. The information provided to do that is the inputs and the

outputs of the process.

Process( )ty( )tu

Model M( )ty( )tu

independent, arbitrary,

measurable, known

dependent, measurable,

known

( ) ( ( ))t ty f u

M M( ) ( ( ))t ty f u

The ideal result of a process identification will be:

M( ( )) ( ( ))t tf u f u


Identification ProcedureChapter 6 Process Identification

A general procedure in process identification includes: Determination of model structure

→ Based on mathematical origin or artificial intelligence Estimation of model parameter

→ Based on the chosen model structure Model verification

→ A model must be able to produce accurate output if “unseen” input data is given to it


Classification of Identification Methods Based on input signals

Natural, generated during the process and measured Artificial, generated especially for the identification purpose

Based on mathematics point of view Deterministic, assuming exact knowledge about process

outputs, inputs, disturbance, etc, and do not consider random sources and influences

Stochastic, assuming some properties and some knowledge of random disturbances, statistical approach

Based on data processing Batch method, one calculation using the whole data at once,

off-line Recursive method, gradual use of data, estimated parameters

are improved from each experiment, can be on-line or off-line



Identification from Step ResponseChapter 6 Process Identification

The methods in this category aim to provide first estimate of the process and provide approximate information about the process gain, dominant time constant, and time delay.

The input signal used to excite the process is a step change of the process input.

It is necessary that the process is in a steady-state before the step change occurs.

The measured step response needs to be normalized for unit step change and zero initial conditions.


“First Order + Time Delay” Approximation The approximation model for the identified process is given

in s-Domain as:

d( )1

T sKG s e

s

Chapter 6 Identification from Step Response

where K is the process gain, τ denotes time constant, and Td is the time delay.

The step response of the transfer function G(s) given above in time domain is:

d

d

( )

d

0,

( )1 ,

t T

t T

y tK e t T


Chapter 6

( )K y

Unit step response

Approximation of unit step responseFirst order + time delay

If the step response is a normalized one, the process gain K is equal to the new steady-state output, K = y(∞).

The actual unit step response and its approximation will always have two crossing points.

Time constant τ and time delay Td can be calculated if the two crossing points are already chosen.

The two crossing points should be chosen thoughtfully, to avoid large difference between the two step responses.

dT

0.632 ( )y

“First Order + Time Delay” ApproximationIdentification from Step Response


Chapter 6

( )K y

Unit step response

Approximation of unit step responseFirst order + time delay

From two freely-chosen points (t1,y1) and (t2,y2), after some manipulations, we can also obtain τ and Td through calculations as follows:

dT

0.632 ( )y

“First Order + Time Delay” Approximation

1t

1y

2y

2t

2 1

1

2

ln

t t

K yK y

2 1d 1

t tT

2 1d 1

t tT

2

1

ln,

ln

K y

KK y

K

1

2

ln

ln

K y

KK y

K

Identification from Step Response


Chapter 6

“First Order + Time Delay” Approximation Advantage:

Easy calculation, straightforward after two points are chosen

Disadvantage: Low accuracy, the higher the process order, the lower

the accuracy of the model Time delay will always present in the model



Time-Percent Value MethodChapter 6

The approximation model for the identified process is given in s-Domain as:

( )

1n

KG s

s

From the unit step response, empirical values h∞, t10, t30, t50, t70, and t90 are obtained.

h

10t

0.1h

30t 50t 70t 90t

0.3h

0.5h

0.7h

0.9h

Step response



Chapter 6

The values of parameters K, τ, and n are determined as follows: K is obtained from the steady-state value of the step

response of the process divided by the magnitude of the input step.

Using the “t/t Table”, up to 6 points of ti/tj can be located → the model order n can be determined.

Using the “t/τ Table”, up to 5 points of ti/τ for the previously determined model order n can be located → the time constant τ can be determined.

Time-Percent Value Method

( )

hK

u t



Chapter 6

Time-Percent Value Method

t/t Table t/τ Table



Chapter 6

Example: Time-Percent Value MethodA step function u(t) = 3(t) is fed in a process. As the step response, the following graph is obtained.

Determine the approximate transfer function of the process by using the Time-Percent Value Method.



Example: Time-Percent Value MethodChapter 6

150h

10 12 st

0.1h

30 18 st

50 23.5 st 70 29 st

90 40 st

0.3h

0.5h

0.7h

0.9h( )

hK

u t

150

3 50

10

90

120.3

40

t

t

10

70

120.41

29

t

t

10

50

120.51

23.5

t

t

30

70

180.62

29

t

t

10

30

120.67

18

t

t

30

50

180.77

23.5

t

t



Example: Time-Percent Value MethodChapter 6

t/t Table

10

90

0.3t

t

10

70

0.41t

t

10

50

0.51t

t

30

70

0.62t

t

10

30

0.67t

t

30

50

0.77t

t

From 6 ti/tj points, the most representative order for the model is 5



t/τ Table

5 values of ti/τ can be located for n = 5

10 2.5t

30 3.6t

50 4.8t

70 6t

90 8.1t

404.94

8.1

294.83

6

23.54.90

4.8

185

3.6

124.80

2.5

avg

(4.80 5 4.90 4.83 4.94)4.89

5

5

50( )

(4.89 1)G s

s

Result:

Example: Time-Percent Value MethodChapter 6 Identification from Step Response


Homework 9Chapter 6 Identification from Step Response

Time Percent Value MethodDetermine the approximation of the model in the last example, if after examining the t/t table, the model order is chosen to be 4 instead of 5.


Homework 9Chapter 6 Identification from Step Response

“First Order + Time Delay” ApproximationDetermine the approximation of the model in the last example, using the data from t1= 15 s and t2 = 40 s.

Print the graph and draw the response of your model on it.

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president universityerwin sitompulsmi 9/1 dr.-ing. erwin sitompul president university lecture 9...

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