HES3310 Control Engineering Lab 1 - Simulation Using MATLAB and SIMULINK

8/2/2019 HES3310 Control Engineering Lab 1 - Simulation Using MATLAB and SIMULINK

1/22

SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS)

FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE

HES3310 Control EngineeringSemester 1, 2012

SIMULATION USING MATLAB AND SIMULINK

By

Stephen, P. Y. Bong (4209168)

Chang Kin Yung (4224744)

Lecturer: Dr. Wallance Wong

Due Date: 13th April 2012 (Friday)


2/22

Simulation Using MATLAB and SIMULINK 4209168; 4224744

HES3310 Control Engineering, Semester 1, 2012 Page 2 of22

OBJECTIVES

To write differential equations and transfer functions describing the dynamics of electriccircuits.

To introduce MATLAB and SIMULINK as tools for control solution. To find the output response using MATLAB and SIMULINK.PART A PRELIMINARY WORKS

1. For the circuits shown in figures below, verify that the transfer function for Circuit A andCircuit B are:

Circuit A

Circuit B

Circuit A:( )

( ) 1

12

++

=

RCsLCssE

sE

i

o

Circuit B:( )

( ) ( ) 1

1

212211

2

2211 ++++

=

sCRCRCRsCRCRsE

sE

i

o


3/22



The fundamental laws utilized to govern the current flow in an electric circuit are the

Kirchhoffs Current Law (KCL) and Kirchhoffs Voltage Law (KVL). According to Ogata, K.

(2002); KCL can be defined as the sum of currents entering a node is equal to the sum of

currents leaving the same node; whereas KVL can be interpreted as at any given instant, the

algebraic sum of the voltages around any loop in an electrical circuit is zero .

Circuit A

Applying KVL to the system, two equations can be obtained from the 2 loops:

Loop (1):iedti

CiR

dt

diL =++

1(1)

Loop (2):oedti

C=

1(2)

Taking the Laplace transforms to (1) and (2) gives:

{ } { } ( ) ( ) ( ) ( )sEsIsC

sIRssILedtiC

iRdt

diL ii =++=

++

11

1

LLLL

(1): ( ) ( ) ( ) ( )sEsIsC

sIRssIL i=++11

(3)

{ } ( ) ( )sEsIsC

edtiC

oo ==

11

1

LL

(2): ( ) ( )sEsIsC

o=11

(4)

Substituting (4) into (3) by eliminatingI(s) yields:

( )

( ) 1

12

++

=

RCsLCssE

sE

i

o


4/22



Circuit B

Applying KVL to the circuit, 3 loops equations can be obtained:

Loop (1): ( ) ieRidtiiC

=+ 11211

1

(5)

Loop (2): ( ) 01

1

2

2

2221

1

=++ dtiCRidtii

C(6)

Loop (3):oedti

C=

12

2

(7)

Taking Laplace transform through (5) to (7) gives:

( ) { } { } ( ) ( )[ ] ( ) ( )sEsIRsIsIsC

eRidtiiC

ii =+=+

1121

1

1121

1

11 LLL

(5): ( ) ( )[ ] ( ) ( )sEsIRsIsIsC

i=+ 1121

1

1(8)

( ) { } ( ) ( )[ ] ( ) ( ) 011

01

1

2

2

2221

1

2

2

2221

1

=++=

++

sIsCsIRsIsIsCdtiCRidtiiC LLL

(6): ( ) ( )[ ] ( ) ( ) 011 22

2221

1

=++ sIsC

sIRsIsIsC

(9)

{ } ( ) ( )sEsIsC

edtiC

oo ==

22

2

2

1

1LL

(7): ( ) ( )sEsIsC

o=2

2

1(10)

Solving equations (8), (9) & (10) gives: ( )( ) ( ) 1

1

212211

2

2211 ++++

=

sCRCRCRsCRCRsE

sE

i

o


5/22



2. Using inverse Laplace transform, find the equation describing the time response of circuit A toa unit impulse input whenL = 1 H, C= 0.04 F forR = 4, 6 and 12 . Plot the response for time

t= 0 to t= 3s.

ForR = 4 and (L = 1 H & C= 0.04 F), the transfer function for Circuit A becomes:

( )( ) ( )( ) ( )( ) 116.004.0

1104.0404.01

122

++

=

++

=

sssssEsE

i

o

For Unit Impulse Input,Ei(s) = 1, thus,

( )( ) ( ) ( )

( ) ( ) ( )222

222

04.01

22

212

21

21

25

212

25

2522425404.0

1

116.004.0

1

++

=

++

=

+++

=

++

=

++

=

ss

sssssssEo

We know that( )

teas

at

sin

22

1 =

++

L , thus, the inverse Laplace transform ofEo(s) is

as follows:

( )( ) ( )

( )tes

tet

o 21sin21

25

212

21

21

25 222

1 =

++

=L


( )

( ) ( )( ) ( )( ) 124.004.0

1

104.0604.01

122

++

=

++

=

sssssE

sE

i

o

For Unit Impulse Input,Ei(s) = 1, hence,

( )( ) ( ) ( )

( ) ( ) 222

222

04.01

22

434

425

16325

2533425604.0

1

124.004.0

1

++

=

++

=

+++

=

++

=

++

=

ss

sssssssE

o

Since( )

teas

at

sin

22

1 =

++

L , thus, the inverse Laplace transform of Eo(s) is as

follows:

( )( )

( )tes

te to 4sin4

25

43

4

4

25 322

1 =

++

=L -


6/22




( )

( ) ( )( ) ( )( ) 148.004.0

1

104.01204.01

122

++

=

++

=

sssssE

sE

i

o

For Unit Impulse Input,Ei(s) = 1, therefore,

( )( )

( ) ( ) ( ) ( )( ) ( )( )11611625

116

25

116

25

256612251204.0

1

148.004.0

1

222

222

04.01

22

+++

=

+

=

+

=

+++

=

++

=

++

=

ssss

sssssssEo

Let( )( ) ( )( ) ( ) ( )116116116116

25

++

+

+

=

+++ s

B

s

A

ss

Cross-multiplication of the two fractions gives:

( )( ) ( )( )11611625 ++++= sBsA

By comparing the coefficient ofs2,

A +B = 0 Eq. (A)

By comparing the constant,

11

25= BA Eq. (B)

Solving Eq. (A) and Eq. (B), gives112

25=A and

112

25=B

Therefore, the partial fraction decomposition ofEo(s) is:

( )

( )( ) ( )( ) ( ) ( )

++

+

=

+++

=

116

1

116

1

112

25

116116

25

ssss

sEo

Taking inverse Laplace transforms gives:

( )( ) ( )

( ) ( )[ ]tto eess

te 11611611

112

25

116

1

112

25

116

1

112

25+

=

++

+

= LL

( ) ( ) ( )[ ]tto eete 116116112

25+

=


7/22



The output, eo(t) for time t= 0 to t= 3s, with various resistance (R = 4 , 6 and 12 ) and

Unit Impulse voltage input are computed by using Microsoft Excel and tabulated in Table 1

below:

Time

(s)

Output, eo(t)

R = 4 R = 6 R = 12

0 0 0 0

0.1 0.43118 1.73093 1.01131

0.2 0.6332 2.36216 1.17165

0.3 0.64083 2.27363 1.05281

0.4 0.51663 1.80639 0.86684

0.5 0.32911 1.21735 0.68717

0.6 0.137 0.66992 0.53504

0.7 -0.0194 0.24613 0.41289

0.8 -0.1204 -0.0318 0.31721

0.9 -0.1637 -0.1784 0.243141 -0.1598 -0.2261 0.18615

1.1 -0.1249 -0.2106 0.14243

1.2 -0.0763 -0.1633 0.10894

1.3 -0.0283 -0.1073 0.08332

1.4 0.00956 -0.0568 0.06371

1.5 0.03301 -0.0186 0.04872

1.6 0.04207 0.00576 0.03726

1.7 0.03965 0.01807 0.02849

1.8 0.03003 0.02151 0.02178

1.9 0.01752 0.01943 0.01666

2 0.0056 0.01471 0.01274

2.1 -0.0035 0.00942 0.00974

2.2 -0.0089 0.00477 0.00745

2.3 -0.0107 0.00135 0.00569

2.4 -0.0098 -0.0008 0.00435

2.5 -0.0072 -0.0018 0.00333

2.6 -0.004 -0.002 0.00255

2.7 -0.001 -0.0018 0.00195

2.8 0.00115 -0.0013 0.00149

2.9 0.00238 -0.0008 0.00114

3 0.00273 -0.0004 0.00087

Table 1: Output for time t= 0 to t= 3s, with various resistance (R = 4 , 6 and 12 ) and

Unit Impulse Voltage Input


8/22



Based on the output tabulated in Table 1, the respond of Circuit A to Unit Impulse voltage input are

plotted by using Microsoft Excel as shown in Fig. 1 below:

Fig. 1: Responds of Circuit A to Unit Impulse Voltage Input

-0.3

0.2

0.7

1.2

1.7

2.2

0 0.5 1 1.5 2 2.5 3

Amplitu

de,eo

(t)(V)

Time (s)

Respond of Circuit A to Unit Impulse Voltage Input

R = 4 Ohms

R = 6 Ohms

R = 12 Ohms


9/22



3. Repeat 2. to find the response of Circuit A to Unit Step Voltage Input. For Unit Step voltageinput,Ei(s) = 1/s. Thus, the transfer function becomes for Circuit A becomes:

( )( )

( )11

1

1

1 22 ++=

++

=

RCsLCsssE

RCsLCs

s

sEo

o

ForR = 4 and (L = 1 H & C= 0.04 F),

( )( )116.004.0

12

++

=

ssssEo

Let ( )( )

( ) ( )( )116.004.0

116.004.0

116.004.0116.004.0

12

2

22++

++++=

++

++=

++

=

sss

CBssssA

ss

CBs

s

A

ssssEo

By comparing the numerator and the denominator, we get,

( ) ( )( ) ( ) AsCAsBA

CBssssA

++++=

++++=

16.004.01

116.004.01

2

2

By comparing the coefficient of s2, s and constant, a system of equations as follows can be

obtained:

=

=

=

=

=+

=+

16.0

04.0

1

1

0

0

001

1016.0

0104.0

1

0

0

001

1016.0

0104.0

1

016.0

004.01

C

B

A

C

B

A

A

CA

BA

Therefore, the partial-fraction decomposition ofEo(s) is:

( )( )

( ) ( )22222222

222

21)2(

2

21)2(

)2(1

21)2(

2)2(1

25224

)4(1

254

)4(1

116.004.0

16.004.01

116.004.0

1

++

+

++

++=

++

++=

+++

++=

++

++=

++

+=

++

=

ss

s

ss

s

sss

s

s

ss

s

sss

s

sssssEo

Taking Laplace transforms gives:

( )( ) ( )

( ) ( )tete

ss

s

ste

tt

o

21sin21

221cos1

21)2(

21

21

2

21)2(

21

22

22

1

22

11

=

++

++

+

= LLL

Therefore, ( ) ( ) ( )tetete tto 21sin21

221cos1 22 =


10/22



ForR = 6 and (L = 1 H & C= 0.04 F),

( )( )

( ) ( )( )124.004.0

124.004.0

124.004.0124.004.0

12

2

22++

++++=

++

++=

++

=

sss

CBssssA

ss

CBs

s

A

ssssE

o

By comparing the numerator and the denominator, we get:

( ) ( )( ) ( ) ACAsBA

CBssssA

++++=

++++=

24.004.01

124.004.01

2

2

By comparing the coefficients of s2, s and constant, a system of equations as follows can be

obtained:

=

=

=

=

=+

=+

24.0

04.0

1

1

0

0

001

1024.0

0104.0

1

0

0

001

1024.0

0104.0

1

024.0

004.01

C

B

A

C

B

A

A

CA

BA


( )( )

22222222

222

4)3(

3

4)3(

)3(1

16)3(

3)3(1

25336

)6(1

256

)6(1

124.004.0

24.004.01

124.004.0

1

++

+

++

++=

++

++=

+++

++=

++

++=

++

+=

++

=

ss

s

ss

s

sss

s

s

ss

s

sss

s

sssssEo


( )

( ) ( )tete

ss

s

ste

tt

o

4sin4

34cos1

4)3(

4

4

3

4)3(

31

33

22

1

22

11

=

++

++

+

= LLL

Therefore, ( ) ( ) ( )tetete tto 4sin4

34cos1 33 =

ForR = 12 and (L = 1 H & C= 0.04 F),

( )( )

( ) ( )( )148.004.0

148.004.0

148.004.0148.004.0

12

2

22++

++++=

++

++=

++

=

sss

CBssssA

ss

CBs

s

A

ssssE

o

By comparing the numerator and the denominator, we get:

( ) ( )( ) ( ) ACAsBA

CBssssA

++++=

++++=

48.004.01

148.004.01

2

2


11/22



By comparing the coefficients of s2, s and constant, a system of equations as follows can be

obtained:

=

=

=

=

=+

=+

48.0

04.0

1

1

0

0

001

1048.0

0104.0

1

0

0

001

1048.0

0104.0

1

048.0

004.01

C

B

A

C

B

A

A

CA

BA


( )( )

( ) ( )22222222

222

11)3(

6

11)3(

)6(1

11)6(

6)6(1

256612

)12(1

2512

)12(1

148.004.0

48.004.01

148.004.0

1

+

+

+

++=

+

++=

+++

++=

++

++=

++

+=

++

=

ss

s

ss

s

sss

s

s

ss

s

sss

s

sssssEo


( )( ) ( )

( ) ( )tete

ss

s

ste

tt

o

11sinh11

311cosh1

11)3(

11

11

6

11)6(

61

66

22

1

22

11

=

++

++

+

= LLL

Therefore, ( ) ( ) ( )tetete tto 11sinh11

611cosh1 66 =


12/22



The output, eo(t) for time t= 0 to t= 3s, with various resistance (R = 4 , 6 and 12 ) and

Unit Step voltage input are computed by using Microsoft Excel and tabulated in Table 2 below:

Time

(s)

Output, eo(t)

R = 4 R = 6 R = 12

0 0 0 0

0.1 0.10767 0.1013 0.08537

0.2 0.35992 0.32237 0.24155

0.3 0.65817 0.56847 0.39678

0.4 0.92708 0.783 0.52958

0.5 1.12206 0.94069 0.63668

0.6 1.22812 1.03815 0.72077

0.7 1.25318 1.08462 0.78593

0.8 1.21888 1.09453 0.83609

0.9 1.15172 1.08257 0.87458

1 1.07609 1.0608 0.90406

1.1 1.01003 1.03766 0.92663

1.2 0.96373 1.01802 0.94389

1.3 0.94001 1.00393 0.95709

1.4 0.93622 0.99547 0.96719

1.5 0.94655 0.99166 0.97491

1.6 0.96426 0.99111 0.98082

1.7 0.98334 0.99244 0.98533

1.8 0.9995 0.99456 0.98878

1.9 1.01043 0.99673 0.991422 1.01565 0.99852 0.99344

2.1 1.01599 0.99978 0.99499

2.2 1.01299 1.00051 0.99617

2.3 1.00836 1.00081 0.99707

2.4 1.00357 1.00083 0.99776

2.5 0.99964 1.00069 0.99829

2.6 0.99707 1.00048 0.99869

2.7 0.99595 1.00028 0.999

2.8 0.99601 1.00012 0.99923

2.9 0.99686 1.00001 0.99941

3 0.99806 0.99995 0.99955

Table 2: Output for time t= 0 to t= 3s, with various resistance (R = 4 , 6 and 12 ) and

Unit Step Voltage Input


13/22



Fig. 2: Responds of Circuit A to Unit Step Voltage Input

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

0 0.5 1 1.5 2 2.5 3

Amplitude,eo

(t)(V)

Time (s)

Respond of Circuit A to Unit Step Input

R = 4 Ohms

R = 6 Ohms

R = 12 Ohms


14/22



PART B EXPERIMENTAL RESULTS

Using the transfer function of Circuit A shown in Figure in PART A, the m-file lab301.m is

modified to get the unit impulse and step responses. Fig. 1 shows the plot of Unit Impulse of Circuit

A (forR = 4, 6 and 12 Ohms) and Fig. 2 shows the plot of Unit Step Responses of Circuit A (for R

= 4, 6 and 12 Ohms).

Respond of Circuit A to Unit Impulse Input (forR = 4, 6 and 12 Ohms)

Fig. 3: Plot of Reponses of Circuit A to Unit Impulse input via MATLAB

Fig. 4: Plot of Responses of Circuit A to Unit Step voltage input via MATLAB

Plot of Unit Impulse Input of Circuit A (for R = 4, 6 and 12 Ohm)

Time (s ec)

Amplitude

0 0.5 1 1.5 2 2.5 3-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

R = 4 Ohm

R = 6 Ohm

R = 12 Ohm

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

Plot of Unit Step Responses of Circuit A (for R = 4, 6 and 12 Ohm)

Time (sec )

Amplitude

R = 4 Ohm

R = 6 Ohm

R = 12 Ohm


15/22



For Unit Step voltage input,

(R = 4 )


16/22



(R = 6 )


17/22



(R = 12)


18/22



PART C COMPARISONS OF RESULTS

Comparisons of Output Functions

Unit Impulse Voltage Input

Resistance(Ohms,)

Theoretical Experimetal

4 ( ) ( )tete to 21sin21

25 2=

6 ( ) ( )tete to 4sin425 3=

12 ( )( ) ( )[ ]tt

o eete116116

112

25+

=


19/22



Unit Step Voltage Input

Resistance

(Ohms, )Theoretical Experimetal

4 ( ) ( ) ( )tetete tto 21sin21

221cos1 22 =

6 ( ) ( ) ( )tetetett

o 4sin4

3

4cos1

33 =

12 ( ) ( ) ( )tetete tto 11sinh11

611cosh1

66 =


20/22



Comparisons of Graphs

Theoretical

Experimental

Fig. 5: Comparison of plots of responds of Circuit A to Unit Impulse voltage input

Based on our knowledge of oscillation, the period of oscillation is inversely proportional to the

damping ratio. Based on the plots of responds of Circuit A as shown in Fig. 5 above, the system

takes a longer time to stop the oscillation with resistance of 4 Ohms, however, for the case of

utilizing resistance of 12 Ohms, the system is critically damped.

-0.3

0.2

0.7

1.2

1.7

2.2

0 0.5 1 1.5 2 2.5 3

Amplitude,eo

(t)(V)

Time (s)

Respond of Circuit A to Unit Impulse Voltage Input

R = 4 Ohms

R = 6 Ohms

R = 12 Ohms

Plot of Unit Impulse Input of Circuit A (for R = 4, 6 and 12 Ohm)

Time (s ec)

Amplitude

0 0.5 1 1.5 2 2.5 3-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

R = 4 Ohm

R = 6 Ohm

R = 12 Ohm


21/22



Theoretical

Experimental

Fig. 6: Comparison of plots of responds of Circuit A to Unit Step voltage input

Based on the plots shown in Fig. 6 above, the system with resistance of 4 Ohms has the longest

period of oscillation. On contrary, for the case of utilizing a resistance of 12 Ohms, there has no

oscillation occurred or it can be referred as critically damped situation.

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

0 0.5 1 1.5 2 2.5 3

Amplitude,eo

(t)(V)

Time (s)

Respond of Circuit A to Unit Step Input

R = 4 Ohms

R = 6 Ohms

R = 12 Ohms

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

Plot of Unit Step Responses of Circuit A (for R = 4, 6 and 12 Ohm)

Time (sec)

Amplitude

R = 4 Ohm

R = 6 Ohm

R = 12 Ohm


22/22


CONCLUSION

According to the results which obtained theoretically and experimentally as shown in previous

section, it can be proved the results of simulation using MATLAB are very closed to analytical

solutions. Therefore, as a verdict, it can be concluded that MATLAB is a powerful mathematical

software which can be employed in the analysis of dynamics of electric circuits. Apart from that,

based on the computations and plots obtained from simulation of MATLAB, it can be verified thatMATLAB is a sensitive tool of writing differential equations and transfer functions. Besides, the

output response of a system can be obtained from SIMULINK as well.

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HES3310 Control Engineering Lab 1 - Simulation Using MATLAB and SIMULINK

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