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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)
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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
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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
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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
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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
ForR = 6 and (L = 1 H & C= 0.04 F), the transfer function for Circuit A becomes:
( )
( ) ( )( ) ( )( ) 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 -
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ForR = 12 and (L = 1 H & C= 0.04 F), the transfer function for Circuit A becomes:
( )
( ) ( )( ) ( )( ) 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+
=
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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
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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
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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 =
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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
Therefore, the partial-fraction decomposition ofEo(s) is:
( )( )
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
Taking Laplace transforms gives:
( )
( ) ( )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
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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
Therefore, the partial-fraction decomposition ofEo(s) is:
( )( )
( ) ( )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
Taking Laplace transforms gives:
( )( ) ( )
( ) ( )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 =
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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
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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
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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
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For Unit Step voltage input,
(R = 4 )
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(R = 6 )
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(R = 12)
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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+
=
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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 =
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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
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HES3310 Control Engineering, Semester 1, 2012 Page 21 of22
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
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Simulation Using MATLAB and SIMULINK 4209168; 4224744
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.