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Pedro BeiroInstituto Superior de Engenharia de Coimbra

Departamento de Engenharia MecnicaRua Pedro Nunes, 3030 - 199 Coimbra, Portugal

pbeirao@isec.pt

Instrumentao e ControloMdulo de Controlo de Sistemas

2

Launching Simulink

In the MATLAB command window type>> simulink

23

Creating a new model

Click the New model iconin the upper left corner tostart a new Simulink file

Select the Simulink icon toshow elements on the file

4

Library of elements

New model is created here

Workspace

35

Saving the new model

Create a new folder (eg.: simulink_files)

Use .mdl suffix when saving

6

Example 1

Build a Simulink model to solve the differential equation

Initial condition x(0) = -1

Input function 3sin(2t)

Output solution of differential equation x(t)

=x(t) 3sin(2t)

47

Example 1

Build this model in Simulink

1s

Integrator

x(0) = -1

input3sin(2t)

outputx(t)

x(t) x(t)

8

Example 1

Drag a Sine Wave block from Sources library to themodel window

Select an input block

59

Example 1

Drag an Integrator block from Continuous library to themodel window

Select an operator block

10

Example 1

Drag a Scope block from Sinks library to the modelwindow

Select an output block

611

Example 1

Place the cursor on Sine Wave block output and dragfrom there to the Integrator input

Place the cursor on Integrator block output and drag fromthere to the Scope input

Arrows indicate the direction of signal flowConnect blocks with signals

12

Example 1

Double-click on Sine Wave block to set amplitude = 3and frequency = 2

This produces the desired input function 3sin(2t)Select simulation parameters

713

Example 1

Double-click on Integrator block to set initial condition = -1

Select simulation parameters

14

Example 1

Run the simulation in the model window select Startfrom Simulation menu

View the output x(t) in Scope window

Run the simulation

815

Example 1

Double-click on Scope block to view simulation results

Simulation results

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Example 2

Build a Simulink model to solve the differential equation

Second order mass-spring-damper system

Initial conditions zero

Input f(t) step with magnitude 3

Parameters m = 0.25, c = 0.5, k = 1

+ + = mx(t) cx(t) kx(t) f(t)

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Example 2

Solve for the term with highest-order derivative

Make the left-hand side of this equation the output of asumming block

= mx(t) f(t) cx(t) kx(t)

Sum

mx(t)+-

-

18

Example 2

Drag a Sum block from Math library to the model window

10

19

Example 2

Double-click to change block parameters to rectangularand + - -

20

Example 2

Add a gain (multiplier) block to eliminate the coefficient mand produce the highest-derivative alone

1m

mx(t) x(t)

Sum Gain

+-

-

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21

Example 2

Drag a Gain block from Math library to the model window

22

Example 2

Double-click to set gain block parameter = =1 1 4m 0.25

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23

Example 2

Add integrators to obtain the desired output variable

1m

mx(t) x(t)

Sum Gain

1s

Integrator

x(t) 1s

x(t)

Integrator

+-

-

24

Example 2

Drag Integrator blocks from Continuous library to themodel window

Initial conditions are = 0

13

25

Example 2

Connect gain blocks to integrated signals

Obtain the terms of the right-hand side of the motionequation

c

k

1m

1s

1s

x(t) x(t) x(t)mx(t)

Sum Gain Integrator Integrator

Gain

Gain

+-

-

26

Example 2

Drag new Gain blocks from Math library to the modelwindow

14

27

Example 2

Drag a Scope block from Sinks library to the modelwindow

Connect from the gain block input backwards up to thebranch point

28

Example 2

Double-click on gain blocks to set parameter c = 0.5 andparameter k = 1

15

29

Example 2

Connect all signals and inputs to the summing block

c

k

1m

1s

1s

x(t) x(t) x(t)mx(t)

Sum Gain Integrator Integrator

Gain

Gainkx(t)

cx(t)

+-

-

f(t)

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Example 2

Drag a Step block from Sources library to the modelwindow

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31

Example 2

Double-click on Step block to set parameter

For a step input of magnitude = 3 set Final value = 3

32

Example 2

Final Simulink model

17

33

Example 2

Run the simulation in the model window select Startfrom Simulation menu

34

Example 2

Underdamped response with a overshoot of 0.5 and afinal value of 3

Is this expected?

18

35

Example 2

Analysis based on motion equations

Standard form

Natural frequency

Damping ratio

Static gain

+ + =

x(t) c 1

x(t) x(t) f(t)k k km

= =n

k 2

m

= =n

c2 0.5 k

= =

1K 1k

36

Example 2

Damping ratio equal to 0.5 less than 1

Expect underdamped system

Expect overshoot

Static gain is 1

Expect output magnitude match input magnitude

Input and output magnitude 3

Simulation results confirm expectations

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37

Single tank level system

Qin [m3/s] inlet volumetric flow of the tank

Qout [m3/s] outlet volumetric flow of the tank

h [m] water level in the tank

Example 3

38

Build a Simulink model to predict changes in the tanklevel due to changes in flow conditions

Develop a mass balance

Volumetric flows a density term is needed to convertfrom volume units to mass units

It is also needed to know the cross-sectional area of thetank in order to determine the volume holdup and hencethe mass holdup in the tank

The mass balance of the simple tank system is given by:

Example 3

( )= in out

d AhQ Q

dt

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The tank has a cylindrical shape constant crosssectional area A = 20 m2

There is no heating effect density assumed constant

The flow Qout through the discharge valve (with crosssection a) is related with the water level in the tank:

The mass balance equation can be rewritten:

Example 3

= = in out in outdh dh

A Q Q A Q Qdt dt

( )= = = =out d out d outQ C a 2gh Q C a 2g h Q k h 2 h

= = = inin out inQ k hdh dh dhA Q Q A Q k h

dt dt dt A

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Final Simulink model

Example 3

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41

Null initial conditions (set up through the integrator block)

Unit step change in the inlet flow rate of the tank system

Simulation gives the following response

Example 3

( )( )in

H sQ s

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Two tank level system without interaction

Qin1 [m3/s] inlet volumetric flow of the first tank

Qout1 [m3/s] outlet volumetric flow of the first tank

Qin2 [m3/s] inlet volumetric flow of the second tank

Qout2 [m3/s] outlet volumetric flow of the second tank

h1 [m] water level in the first tank

h2 [m] water level in the second tank

Example 4

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Example 4

44

Build a Simulink model to predict changes in both tanklevels due to changes in flow conditions

Develop a mass balance

Volumetric flows a density term is needed to convertfrom volume units to mass units

It is also needed to know the cross-sectional area of bothtanks in order to determine the volume holdup and hencethe mass holdup in both tanks

The mass balance of the two tank level system is givenby:

Example 4

( ) ( )= =

1 1 2 2in1 out1 in2 out2

d A h d A hQ Q and Q Q

dt dt

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45

Both tanks have a cylindrical shape constant crosssectional area A1 = A2 = A = 20 m2

There is no heating effect density assumed constant

The flows Qout1 and Qout2 through the discharge valves(with cross section a) are related with the water level inboth tanks:

The volumetric flow exiting the first tank feeds the secondtank Qout1 = Qin2The mass balance equations of both tanks are given by:

Example 4

= =in1 1 1 1 1 2 21 2Q k h k h k hdh dh

and dt A dt A

= = = =out1 1 1 1 out2 2 2 2Q k h 2 h and Q k h 2 h

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Final Simulink model

Example 4

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Final Simulink model

Example 4

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Final Simulink model subsystem

Example 4

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Null initial conditions (set up through the integrator block)

Unit step change in the inlet flow rate of the tank system

Simulation gives the following responses

Example 4

( )( )

( )( )

1 2

in1 in2

H s H s and Q s Q s

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Two tank level system with interaction

Qin1 [m3/s] inlet volumetric flow of the first tank

Qout1 [m3/s] outlet volumetric flow of the first tank

Qin2 [m3/s] inlet volumetric flow of the second tank

Qout2 [m3/s] outlet volumetric flow of the second tank

h1 [m] water level in the first tank

h2 [m] water level in the second tank

Example 5

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51

Example 5

52

Build a Simulink model to predict changes in both tanklevels due to changes in flow conditions

Develop a mass balance

Volumetric flows a density term is needed to convertfrom volume units to mass units

It is also needed to know the cross-sectional area of bothtanks in order to determine the volume holdup and hencethe mass holdup in both tanks

The mass balance of the two tank level system is givenby:

Example 5

( ) ( )= =

1 1 2 2in1 out1 in2 out2

d A h d A hQ Q and Q Q

dt dt

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53

Both tanks have a cylindrical shape constant crosssectional area A1 = A2 = A = 20 m2

There is no heating effect density assumed constant

Due to the interaction between the two tanks Qout1 isobtained by the level difference of the two tanks:

The volumetric flow exiting the first tank feeds the secondtank Qout1 = Qin2

The mass balance equations of both tanks are given by:

Example 5

= = out1 1 1 2 1 2Q k h h 2 h h

= =in1 1 1 2 1 1 2 2 21 2Q k h h k h h k hdh dh

and dt A dt A

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Example 5

Final Simulink model

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55

Tank 1 is empty and Tank 2 has a level of 1 m (set upthrough the corresponding integrator block)

Unit step change in the inlet flow rate of the tank system

Simulation gives the following responses

Example 5

( )( )

( )( )

1 2

in1 in2

H s H s and

Q s Q s

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Mass-spring-damper system

Mass m = 4 kg

Damping coefficient c = 2 Ns/m

Spring constant k = 36 N/m

Position initial condition x(0) = 0.5 mVelocity initial condition = 0 m

Example 6

x(0)

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57

Build a Simulink model of the mass-spring-dampersystem

Develop a force balance using 2nd Newtons law:

Find the system responses to the following inputs:Constant input of 0 N or no inputConstant input of 100 NStep input of 100 N with an initial value of 0, final valueof 100 and step time of 20 sSinusoidal input with amplitude of 100 N and frequencyof 3 rad/s

Write an m-file to plot all the responses in a single figure

Example 6

= mx(t) f(t) cx(t) kx(t)

58

Example 6

Final Simulink model

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59

Example 6

m-file to plot all the system responses in a single figure

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Example 6

0 5 10 15 20 25 30 35 40-0.4

-0.2

0

0.2

0.4

0.6Free Response

time (sec)

posi

tion (m

)

0 5 10 15 20 25 30 35 400

1

2

3

4

5Constant Forced Response

time (sec)

posi

tion (m

)

0 5 10 15 20 25 30 35 40-1

0

1

2

3

4

5Step Forced Response

time (sec)

posi

tion (m

)

0 5 10 15 20 25 30 35 40-0.4

-0.2

0

0.2

0.4

0.6Sinusoidal Forced Response

time (sec)

posi

tion (m

)

System responses to the given inputs

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Example 7

Industrial mixing process

Qhot [m3/s] hot volumetric flowQcold [m3/s] cold volumetric flowThot [C] hot temperatureTcold [C] cold temperature

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Example 7

Build a Simulink model to simulate the temperaturecontrol of a mixing process

The temperature of the mixed flow is given by:

Control flow Q by manipulating Qcold

Control temperature T by manipulating Qhot

Consider volume V = 0.5

( )+ + + = =hot hot cold cold hot cold hot hot cold cold

Q T Q T Q Q T Q T Q T QTdT dTdt V dt V

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63

Example 7

Final Simulink model

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Example 7

Final Simulink model

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65

Example 7

Final Simulink model subsystem

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Example 7

Final Simulink model subsystem

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67

Example 7

Run the simulation and plot the responses of T and Q

Modify the values of the PI controllers to improve theresponse: